force problem solving with solution

Forces in Physics, tutorials and Problems with Solutions

Free tutorials on forces with questions and problems with detailed solutions and examples. The concepts of forces, friction forces, action and reaction forces, free body diagrams, tension of string, inclined planes, etc. are discussed and through examples, questions with solutions and clear and self explanatory diagrams. Questions to practice for the SAT Physics test on forces are also included with their detailed solutions. The discussions of applications of forces engineering system are also included.

Forces: Tutorials with Examples and Detailed Solutions

Forces in Physics .
Components of a Force in a System of Coordinates .
Free Body Diagrams, Tutorials with Examples and Explanations .
Forces of Friction .
Addition of Forces .
What is the Tension of a String or rope? .
Newton's Laws in Physics .
Hooke's Law, Examples with solutions .

Problems on Forces with Detailed Solutions

Inclined Planes Problems in Physics with Solutions .
Tension, String, Forces Problems with Solutions .

SAT Questions on Forces with Solutions

Physics Practice Questions with Solutions on Forces and Newton's Laws .

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Physics Formulas Reference
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High School Physics : Calculating Force

Study concepts, example questions & explanations for high school physics, all high school physics resources, example questions, example question #1 : net force.

$force problem solving with solution$

Plug these into the equation to solve for acceleration.

$force problem solving with solution$

Example Question #2 : Calculating Force

$force problem solving with solution$

Plug in the values given to us and solve for the force.

$force problem solving with solution$

Example Question #1 : Calculating Force

$force problem solving with solution$

Plug in the given values to solve for the mass.

$force problem solving with solution$

(Assume the only two forces acting on the object are friction and Derek).

$force problem solving with solution$

Plug in the information we've been given so far to find the force of friction.

$force problem solving with solution$

Friction will be negative because it acts in the direction opposite to the force of Derek.

$force problem solving with solution$

Newton's third law states that when one object exerts a force on a second object, the second object exerts a force equal in size, but opposite in direction to the first. That means that the force of the hammer on the nail and the nail on the hammer will be equal in size, but opposite in direction.

$force problem solving with solution$

Example Question #6 : Calculating Force

$force problem solving with solution$

We can find the net force by adding the individual force together.

$force problem solving with solution$

If the object has a constant velocity, that means that the net acceleration must be zero.

$force problem solving with solution$

In conjunction with Newton's second law, we can see that the net force is also zero. If there is no net acceleration, then there is no net force.

$force problem solving with solution$

Since Franklin is lifting the weight vertically, that means there will be two force acting upon the weight: his lifting force and gravity. The net force will be equal to the sum of the forces acting on the weight.

$force problem solving with solution$

We know the mass of the weight and we know the acceleration, so we can solve for the lifting force.

$force problem solving with solution$

We are given the mass, but we will need to calculate the acceleration to use in the formula.

$force problem solving with solution$

Plug in our given values and solve for acceleration.

$force problem solving with solution$

Now we know both the acceleration and the mass, allowing us to solve for the force.

$force problem solving with solution$

Example Question #9 : Calculating Force

$force problem solving with solution$

We can calculate the gravitational force using the mass.

$force problem solving with solution$

Example Question #10 : Calculating Force

$force problem solving with solution$

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Work done by force – problems and solutions

1. A person pulls a block 2 m along a horizontal surface by a constant force F = 20 N. Determine the work done by force F acting on the block.

Work done by a force – problems and solutions 1

Force (F) = 20 N

Displacement (s) = 2 m

Angle (θ ) = 0

Wanted : Work (W)

W = F d cos θ = (20)(2)(cos 0) = (20)(2)(1) = 40 Joule

2. A force F = 10 N acting on a box 1 m along a horizontal surface. The force acts at a 30 o angle as shown in figure below. Determine the work done by force F!

Work done by a force – problems and solutions 2

Force (F) = 10 N

The horizontal force (F x ) = F cos 30 o = (10)(0.5√3) = 5√3 N

Displacement (d) = 1 meter

Wanted : Work (W) ?

W = F x d = (5√3)(1) = 5√3 Joule

3. A body falls freely from rest, from a height of 2 m. If acceleration due to gravity is 10 m/s 2 , determine the work done by the force of gravity !

Object’s mass (m) = 1 kg

Height (h) = 2 m

Acceleration due to gravity (g) = 10 m/s 2

Wanted : Work done by the force of gravity (W)

W = F d = w h = m g h

W = (1)(10)(2) = 20 Joule

W = work, F = force, d = distance, w = weight , h = height, m = mass, g = acceleration due to gravity.

4. An 1-kg object attached to a spring so it is elongated 2 cm. If acceleration due to gravity is 10 m/s 2 , determine (a) the spring constant (b) work done by spring force on object

Mass (m) = 1 kg

Elongation (x) = 2 cm = 0.02 m

Weight (w) = m g = (1 kg)(10 m/s 2 ) = 10 kg m/s 2 = 10 N

Wanted : Spring constant and work done by spring force

(a) Spring constant

Formula of Hooke’s law :

k = F / x = w / x = m g / x

k = (1)(10) / 0.02 = 10 / 0.02

k = 500 N/m

(b) work done by spring force

W = – ½ k x 2

W = – ½ (500)(0.02) 2

W = – (250)(0.0004)

W = -0.1 Joule

The minus sign indicates that the direction of spring force is opposite with the direction of object displacement.

5. A force F = 10 N accelerates a box over a displacement 2 m. The floor is rough and exerts a friction force F k = 2 N. Determine the net work done on the box.

Work done by a force – problems and solutions 3

Force of kinetic friction (F k ) = 2 N

Displacement (d) = 2 m

Wanted : Net work (W net )

Work done by force F :

W 1 = F d cos 0 = (10)(2)(1) = 20 Joule

Work done by force of kinetic friction (F k ) :

W 2 = F k d = (2)(2)(cos 180) = (2)(2)(-1) = -4 Joule

W net = W 1 – W 2

W net = 20 – 4

W net = 16 Joule

6 . What is the work done by force F on the block.

Work done by force – problems and solutions 1

Force (F) = 12 Newton

Displacement (d) = 4 meters

Wanted: Work (W)

W = F d = (12 Newton)(4 meters) = 48 N m = 48 Joule

7 . A block is pushed by a force of 200 N. The block’s displacement is 2 meters. What is the work done on the block?

Force (F) = 200 Newton

Displacement (d) = 2 meters

W = (200 Newton)(2 meters)

W = 400 N m

W = 400 Joule

8 . The driver of the sedan wants to park his car exactly 0.5 m in front of the truck which is at 10 m from the sedan’s position. What is the work required by the sedan?

Work done by force – problems and solutions 2

Displacement (d) = 10 meters – 0.5 meters = 9.5 meters

Force (F) = 50 Newton

W = (50 Newton)(9.5 meters)

W = 475 N m

W = 475 Joule

Work done by force – problems and solutions 3

Work done by Tom and Jerry so the car can move as far as 4 meters. Forces exerted by Tom and Jerry are 50 N and 70 N.

Displacement (s) = 4 meters

Net force (F) = 50 Newton + 70 Newton = 120 Newton

W = F s = (120 Newton)(4 meters) = 480 N m = 480 Joule

10 . A driver pulling a car so the car moves as far as 1000 cm. What is the work done on the car?

Force (F) = 250 Newton

Displacement (s) = 1000 cm = 1000/100 meters = 10 meters

W = F s = (250 Newton)(10 meters) = 2500 N m = 2500 Joule

11. Based on figure below, if work done by net force is 375 Joule, determine object’s displacement.

Work done by force – problems and solutions 11

Work (W) = 375 Joule

Net force ( ΣF) = 40 N + 10 N – 25 N = 25 Newton ( rightward )

Wanted : Displacement ( d )

The equation of work :

Object’s displacement :

d = W / F = 375 Joule / 25 Newton

d = 15 meter s

12. The activities below w hich do not do work is …

A. Push an object as far as 10 meters

B. Push a car until a move

C. Push a wall

D. Pulled a box

W = work , F = force , d = displacement

B ased on the above formula, work done by force and there is a displacement.

The correct answer is C.

1 3 . Andrew pushes an object with force of 20 N so the object moves in circular motion with a radius of 7 meters. Determine the work done by Andrew for two times circular motion .

B. 1400 Joule

C. 1540 Joule

D. 1760 Joule

If the person pushes wheelchair for two times circular motion then the person and wheelchair return to the original position, so the displacement of the person is zero.

Displacement = 0 so work = 0.

The correct answer is A.

14 . Someone push an object on the floor with force of 350 N. The floor exerts a friction force 70 N. Determine the work done by force to move the object as far as 6 meters.

The force of push (F) = 350 Newton

Friction force (F fric ) = 70 Newton

Displacement of object (s) = 6 meters

There are two forces that act on the object, the push force (F) and friction force (F fric ). The push force has the same direction as the displacement of the object because the push force does a positive work. In another hand, the friction force has the opposite direction with a displacement of the object so that the friction force does a negative work.

Work done by push force :

W = F d = (350 Newton)(6 meters) = 2100 Newton-meters = 2100 Joule

Work done by friction force :

W = – (F fric )(s) = – (70 Newton)(6 meters) = – 420 Newton-meters = – 420 Joule

The net work :

W net = 2100 Joule – 420 Joule

W net = 1680 Joule

15 . An object is pushed by a horizontal force of 14 Newton on a rough floor with the friction force of 10 Newton. Determine the net work of move the object as far as 8 meters.

A. 0.5 Joule

C. 32 Joule

D. 192 Joule

Push force (F) = 14 Newton

Friction force (F fric ) = 10 Newton

Displacement of object (d) = 8 meters

There are two forces that act on an object, push force (F) and friction force (F fric ).

The push force has the same direction as the displacement of the object so that the push force does a positive work. In another hand, the friction force has the opposite direction as the displacement of the object so that the friction force does a negative work.

W = F s = (14 Newton)(8 meters) = 112 Newton meters = 112 Joule

W = – (F fric )(s) = – (10 Newton)(8 meters) = – 80 Newton meters = – 80 Joule

W net = 112 Joule – 80 Joule

W net = 32 Joule

16 . Determine the net work based on figure below.

Work

B. 450 Joule

C. 600 Joule

D. 750 Joule

Work = Force (F) x displacement (d)

Work = Area of triangle 1 + area of rectangle + area of triangle 2

Work = 1/2(40-0)(3-0) + (40-0)(9-3) + 1/2(40-0)(12-9)

Work = 1/2(40)(3) + (40)(6) + 1/2(40)(3)

Work = (20)(3) + 240 + (20)(3)

Work = 60 + 240 + 60

Work = 360 Joule

17 . A piece of wood with a length of 60 cm plugged vertically into the ground. Wood hit with a 10-kg hammer from a height of 40 cm above the top of the wood. If the average resistance force of the ground is 2 x 10 3 N and the acceleration due to gravity is 10 m/s 2 , then the wood will enter entirely into the ground after…. hits.

Mass of hammer (m) = 10 kg

Acceleration due to gravity (g) = 10 m/ s 2

Weight of hammer (w) = m g = (10)(10) = 100 kg m/s 2

Displacement of hammer before hits the wood (d) = 40 cm = 0.4 meters

The resistance of wood (F) = 2 x 10 3 N = 2000 N

Length of wood (s) = 60 cm = 0.6 meters

Wanted : T he wood will enter entirely into the ground after…. hits.

Work done on the hammer when hammer moves as far as 0.4 meters is :

W = F d = w s = (100 N)(0.4 m) = 40 Nm = 40 Joule

Work done by the resistance force of the ground :

W = F d = (2000 N)(0.6 m) = 1200 Nm = 1200 Joule

T he wood will enter entirely into the ground after…. hits.

1200 Joule / 40 Joule = 30

The correct answer is D.

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Application of conservation of mechanical energy for motion on an inclined plane
Application of conservation of mechanical energy for projectile motion

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force problem solving with solution

6.1 Solving Problems with Newton’s Laws

Learning objectives.

By the end of this section, you will be able to:

Apply problem-solving techniques to solve for quantities in more complex systems of forces
Use concepts from kinematics to solve problems using Newton’s laws of motion
Solve more complex equilibrium problems
Solve more complex acceleration problems
Apply calculus to more advanced dynamics problems

Success in problem solving is necessary to understand and apply physical principles. We developed a pattern of analyzing and setting up the solutions to problems involving Newton’s laws in Newton’s Laws of Motion ; in this chapter, we continue to discuss these strategies and apply a step-by-step process.

Problem-Solving Strategies

We follow here the basics of problem solving presented earlier in this text, but we emphasize specific strategies that are useful in applying Newton’s laws of motion . Once you identify the physical principles involved in the problem and determine that they include Newton’s laws of motion, you can apply these steps to find a solution. These techniques also reinforce concepts that are useful in many other areas of physics. Many problem-solving strategies are stated outright in the worked examples, so the following techniques should reinforce skills you have already begun to develop.

Problem-Solving Strategy

Applying newton’s laws of motion.

Identify the physical principles involved by listing the givens and the quantities to be calculated.
Sketch the situation, using arrows to represent all forces.
Determine the system of interest. The result is a free-body diagram that is essential to solving the problem.
Apply Newton’s second law to solve the problem. If necessary, apply appropriate kinematic equations from the chapter on motion along a straight line.
Check the solution to see whether it is reasonable.

Let’s apply this problem-solving strategy to the challenge of lifting a grand piano into a second-story apartment. Once we have determined that Newton’s laws of motion are involved (if the problem involves forces), it is particularly important to draw a careful sketch of the situation. Such a sketch is shown in Figure 6.2 (a). Then, as in Figure 6.2 (b), we can represent all forces with arrows. Whenever sufficient information exists, it is best to label these arrows carefully and make the length and direction of each correspond to the represented force.

As with most problems, we next need to identify what needs to be determined and what is known or can be inferred from the problem as stated, that is, make a list of knowns and unknowns. It is particularly crucial to identify the system of interest, since Newton’s second law involves only external forces. We can then determine which forces are external and which are internal, a necessary step to employ Newton’s second law. (See Figure 6.2 (c).) Newton’s third law may be used to identify whether forces are exerted between components of a system (internal) or between the system and something outside (external). As illustrated in Newton’s Laws of Motion , the system of interest depends on the question we need to answer. Only forces are shown in free-body diagrams, not acceleration or velocity. We have drawn several free-body diagrams in previous worked examples. Figure 6.2 (c) shows a free-body diagram for the system of interest. Note that no internal forces are shown in a free-body diagram.

Once a free-body diagram is drawn, we apply Newton’s second law. This is done in Figure 6.2 (d) for a particular situation. In general, once external forces are clearly identified in free-body diagrams, it should be a straightforward task to put them into equation form and solve for the unknown, as done in all previous examples. If the problem is one-dimensional—that is, if all forces are parallel—then the forces can be handled algebraically. If the problem is two-dimensional, then it must be broken down into a pair of one-dimensional problems. We do this by projecting the force vectors onto a set of axes chosen for convenience. As seen in previous examples, the choice of axes can simplify the problem. For example, when an incline is involved, a set of axes with one axis parallel to the incline and one perpendicular to it is most convenient. It is almost always convenient to make one axis parallel to the direction of motion, if this is known. Generally, just write Newton’s second law in components along the different directions. Then, you have the following equations:

(If, for example, the system is accelerating horizontally, then you can then set a y = 0 . a y = 0 . ) We need this information to determine unknown forces acting on a system.

As always, we must check the solution. In some cases, it is easy to tell whether the solution is reasonable. For example, it is reasonable to find that friction causes an object to slide down an incline more slowly than when no friction exists. In practice, intuition develops gradually through problem solving; with experience, it becomes progressively easier to judge whether an answer is reasonable. Another way to check a solution is to check the units. If we are solving for force and end up with units of millimeters per second, then we have made a mistake.

There are many interesting applications of Newton’s laws of motion, a few more of which are presented in this section. These serve also to illustrate some further subtleties of physics and to help build problem-solving skills. We look first at problems involving particle equilibrium, which make use of Newton’s first law, and then consider particle acceleration, which involves Newton’s second law.

Particle Equilibrium

Recall that a particle in equilibrium is one for which the external forces are balanced. Static equilibrium involves objects at rest, and dynamic equilibrium involves objects in motion without acceleration, but it is important to remember that these conditions are relative. For example, an object may be at rest when viewed from our frame of reference, but the same object would appear to be in motion when viewed by someone moving at a constant velocity. We now make use of the knowledge attained in Newton’s Laws of Motion , regarding the different types of forces and the use of free-body diagrams, to solve additional problems in particle equilibrium .

Example 6.1

Different tensions at different angles.

Thus, as you might expect,

This gives us the following relationship:

Note that T 1 T 1 and T 2 T 2 are not equal in this case because the angles on either side are not equal. It is reasonable that T 2 T 2 ends up being greater than T 1 T 1 because it is exerted more vertically than T 1 . T 1 .

Now consider the force components along the vertical or y -axis:

This implies

Substituting the expressions for the vertical components gives

There are two unknowns in this equation, but substituting the expression for T 2 T 2 in terms of T 1 T 1 reduces this to one equation with one unknown:

which yields

Solving this last equation gives the magnitude of T 1 T 1 to be

Finally, we find the magnitude of T 2 T 2 by using the relationship between them, T 2 = 1.225 T 1 T 2 = 1.225 T 1 , found above. Thus we obtain

Significance

Particle acceleration.

We have given a variety of examples of particles in equilibrium. We now turn our attention to particle acceleration problems, which are the result of a nonzero net force. Refer again to the steps given at the beginning of this section, and notice how they are applied to the following examples.

Example 6.2

Drag force on a barge.

The drag of the water F → D F → D is in the direction opposite to the direction of motion of the boat; this force thus works against F → app , F → app , as shown in the free-body diagram in Figure 6.4 (b). The system of interest here is the barge, since the forces on it are given as well as its acceleration. Because the applied forces are perpendicular, the x - and y -axes are in the same direction as F → 1 F → 1 and F → 2 . F → 2 . The problem quickly becomes a one-dimensional problem along the direction of F → app F → app , since friction is in the direction opposite to F → app . F → app . Our strategy is to find the magnitude and direction of the net applied force F → app F → app and then apply Newton’s second law to solve for the drag force F → D . F → D .

The angle is given by

From Newton’s first law, we know this is the same direction as the acceleration. We also know that F → D F → D is in the opposite direction of F → app , F → app , since it acts to slow down the acceleration. Therefore, the net external force is in the same direction as F → app , F → app , but its magnitude is slightly less than F → app . F → app . The problem is now one-dimensional. From the free-body diagram, we can see that

However, Newton’s second law states that

This can be solved for the magnitude of the drag force of the water F D F D in terms of known quantities:

Substituting known values gives

The direction of F → D F → D has already been determined to be in the direction opposite to F → app , F → app , or at an angle of 53 ° 53 ° south of west.

In Newton’s Laws of Motion , we discussed the normal force , which is a contact force that acts normal to the surface so that an object does not have an acceleration perpendicular to the surface. The bathroom scale is an excellent example of a normal force acting on a body. It provides a quantitative reading of how much it must push upward to support the weight of an object. But can you predict what you would see on the dial of a bathroom scale if you stood on it during an elevator ride? Will you see a value greater than your weight when the elevator starts up? What about when the elevator moves upward at a constant speed? Take a guess before reading the next example.

Example 6.3

What does the bathroom scale read in an elevator.

From the free-body diagram, we see that F → net = F → s − w → , F → net = F → s − w → , so we have

Solving for F s F s gives us an equation with only one unknown:

or, because w = m g , w = m g , simply

No assumptions were made about the acceleration, so this solution should be valid for a variety of accelerations in addition to those in this situation. ( Note: We are considering the case when the elevator is accelerating upward. If the elevator is accelerating downward, Newton’s second law becomes F s − w = − m a . F s − w = − m a . )

We have a = 1.20 m/s 2 , a = 1.20 m/s 2 , so that F s = ( 75.0 kg ) ( 9.80 m/s 2 ) + ( 75.0 kg ) ( 1.20 m/s 2 ) F s = ( 75.0 kg ) ( 9.80 m/s 2 ) + ( 75.0 kg ) ( 1.20 m/s 2 ) yielding F s = 825 N . F s = 825 N .
Now, what happens when the elevator reaches a constant upward velocity? Will the scale still read more than his weight? For any constant velocity—up, down, or stationary—acceleration is zero because a = Δ v Δ t a = Δ v Δ t and Δ v = 0 . Δ v = 0 . Thus, F s = m a + m g = 0 + m g F s = m a + m g = 0 + m g or F s = ( 75.0 kg ) ( 9.80 m/s 2 ) , F s = ( 75.0 kg ) ( 9.80 m/s 2 ) , which gives F s = 735 N . F s = 735 N .

Thus, the scale reading in the elevator is greater than his 735-N (165-lb.) weight. This means that the scale is pushing up on the person with a force greater than his weight, as it must in order to accelerate him upward. Clearly, the greater the acceleration of the elevator, the greater the scale reading, consistent with what you feel in rapidly accelerating versus slowly accelerating elevators. In Figure 6.5 (b), the scale reading is 735 N, which equals the person’s weight. This is the case whenever the elevator has a constant velocity—moving up, moving down, or stationary.

Check Your Understanding 6.1

Now calculate the scale reading when the elevator accelerates downward at a rate of 1.20 m/s 2 . 1.20 m/s 2 .

The solution to the previous example also applies to an elevator accelerating downward, as mentioned. When an elevator accelerates downward, a is negative, and the scale reading is less than the weight of the person. If a constant downward velocity is reached, the scale reading again becomes equal to the person’s weight. If the elevator is in free fall and accelerating downward at g , then the scale reading is zero and the person appears to be weightless.

Example 6.4

Two attached blocks.

For block 1: T → + w → 1 + N → = m 1 a → 1 T → + w → 1 + N → = m 1 a → 1

For block 2: T → + w → 2 = m 2 a → 2 . T → + w → 2 = m 2 a → 2 .

Notice that T → T → is the same for both blocks. Since the string and the pulley have negligible mass, and since there is no friction in the pulley, the tension is the same throughout the string. We can now write component equations for each block. All forces are either horizontal or vertical, so we can use the same horizontal/vertical coordinate system for both objects

When block 1 moves to the right, block 2 travels an equal distance downward; thus, a 1 x = − a 2 y . a 1 x = − a 2 y . Writing the common acceleration of the blocks as a = a 1 x = − a 2 y , a = a 1 x = − a 2 y , we now have

From these two equations, we can express a and T in terms of the masses m 1 and m 2 , and g : m 1 and m 2 , and g :

Check Your Understanding 6.2

Calculate the acceleration of the system, and the tension in the string, when the masses are m 1 = 5.00 kg m 1 = 5.00 kg and m 2 = 3.00 kg . m 2 = 3.00 kg .

Example 6.5

Atwood machine.

We have For m 1 , ∑ F y = T − m 1 g = m 1 a . For m 2 , ∑ F y = T − m 2 g = − m 2 a . For m 1 , ∑ F y = T − m 1 g = m 1 a . For m 2 , ∑ F y = T − m 2 g = − m 2 a . (The negative sign in front of m 2 a m 2 a indicates that m 2 m 2 accelerates downward; both blocks accelerate at the same rate, but in opposite directions.) Solve the two equations simultaneously (subtract them) and the result is ( m 2 − m 1 ) g = ( m 1 + m 2 ) a . ( m 2 − m 1 ) g = ( m 1 + m 2 ) a . Solving for a : a = m 2 − m 1 m 1 + m 2 g = 4 kg − 2 kg 4 kg + 2 kg ( 9.8 m/s 2 ) = 3.27 m/s 2 . a = m 2 − m 1 m 1 + m 2 g = 4 kg − 2 kg 4 kg + 2 kg ( 9.8 m/s 2 ) = 3.27 m/s 2 .
Observing the first block, we see that T − m 1 g = m 1 a T = m 1 ( g + a ) = ( 2 kg ) ( 9.8 m/s 2 + 3.27 m/s 2 ) = 26.1 N . T − m 1 g = m 1 a T = m 1 ( g + a ) = ( 2 kg ) ( 9.8 m/s 2 + 3.27 m/s 2 ) = 26.1 N .

Check Your Understanding 6.3

Determine a general formula in terms of m 1 , m 2 m 1 , m 2 and g for calculating the tension in the string for the Atwood machine shown above.

Newton’s Laws of Motion and Kinematics

Physics is most interesting and most powerful when applied to general situations that involve more than a narrow set of physical principles. Newton’s laws of motion can also be integrated with other concepts that have been discussed previously in this text to solve problems of motion. For example, forces produce accelerations, a topic of kinematics , and hence the relevance of earlier chapters.

When approaching problems that involve various types of forces, acceleration, velocity, and/or position, listing the givens and the quantities to be calculated will allow you to identify the principles involved. Then, you can refer to the chapters that deal with a particular topic and solve the problem using strategies outlined in the text. The following worked example illustrates how the problem-solving strategy given earlier in this chapter, as well as strategies presented in other chapters, is applied to an integrated concept problem.

Example 6.6

What force must a soccer player exert to reach top speed.

We are given the initial and final velocities (zero and 8.00 m/s forward); thus, the change in velocity is Δ v = 8.00 m/s Δ v = 8.00 m/s . We are given the elapsed time, so Δ t = 2.50 s . Δ t = 2.50 s . The unknown is acceleration, which can be found from its definition: a = Δ v Δ t . a = Δ v Δ t . Substituting the known values yields a = 8.00 m/s 2.50 s = 3.20 m/s 2 . a = 8.00 m/s 2.50 s = 3.20 m/s 2 .
Here we are asked to find the average force the ground exerts on the runner to produce this acceleration. (Remember that we are dealing with the force or forces acting on the object of interest.) This is the reaction force to that exerted by the player backward against the ground, by Newton’s third law. Neglecting air resistance, this would be equal in magnitude to the net external force on the player, since this force causes her acceleration. Since we now know the player’s acceleration and are given her mass, we can use Newton’s second law to find the force exerted. That is, F net = m a . F net = m a . Substituting the known values of m and a gives F net = ( 70.0 kg ) ( 3.20 m/s 2 ) = 224 N . F net = ( 70.0 kg ) ( 3.20 m/s 2 ) = 224 N .

This is a reasonable result: The acceleration is attainable for an athlete in good condition. The force is about 50 pounds, a reasonable average force.

Check Your Understanding 6.4

The soccer player stops after completing the play described above, but now notices that the ball is in position to be stolen. If she now experiences a force of 126 N to attempt to steal the ball, which is 2.00 m away from her, how long will it take her to get to the ball?

Example 6.7

What force acts on a model helicopter.

The magnitude of the force is now easily found:

Check Your Understanding 6.5

Find the direction of the resultant for the 1.50-kg model helicopter.

Example 6.8

Baggage tractor.

∑ F x = m system a x ∑ F x = m system a x and ∑ F x = 820.0 t , ∑ F x = 820.0 t , so 820.0 t = ( 650.0 + 250.0 + 150.0 ) a a = 0.7809 t . 820.0 t = ( 650.0 + 250.0 + 150.0 ) a a = 0.7809 t . Since acceleration is a function of time, we can determine the velocity of the tractor by using a = d v d t a = d v d t with the initial condition that v 0 = 0 v 0 = 0 at t = 0 . t = 0 . We integrate from t = 0 t = 0 to t = 3 : t = 3 : d v = a d t , ∫ 0 3 d v = ∫ 0 3.00 a d t = ∫ 0 3.00 0.7809 t d t , v = 0.3905 t 2 ] 0 3.00 = 3.51 m/s . d v = a d t , ∫ 0 3 d v = ∫ 0 3.00 a d t = ∫ 0 3.00 0.7809 t d t , v = 0.3905 t 2 ] 0 3.00 = 3.51 m/s .
Refer to the free-body diagram in Figure 6.8 (b). ∑ F x = m tractor a x 820.0 t − T = m tractor ( 0.7805 ) t ( 820.0 ) ( 3.00 ) − T = ( 650.0 ) ( 0.7805 ) ( 3.00 ) T = 938 N . ∑ F x = m tractor a x 820.0 t − T = m tractor ( 0.7805 ) t ( 820.0 ) ( 3.00 ) − T = ( 650.0 ) ( 0.7805 ) ( 3.00 ) T = 938 N .

Recall that v = d s d t v = d s d t and a = d v d t a = d v d t . If acceleration is a function of time, we can use the calculus forms developed in Motion Along a Straight Line , as shown in this example. However, sometimes acceleration is a function of displacement. In this case, we can derive an important result from these calculus relations. Solving for dt in each, we have d t = d s v d t = d s v and d t = d v a . d t = d v a . Now, equating these expressions, we have d s v = d v a . d s v = d v a . We can rearrange this to obtain a d s = v d v . a d s = v d v .

Example 6.9

Motion of a projectile fired vertically.

The acceleration depends on v and is therefore variable. Since a = f ( v ) , a = f ( v ) , we can relate a to v using the rearrangement described above,

We replace ds with dy because we are dealing with the vertical direction,

We now separate the variables ( v ’s and dv ’s on one side; dy on the other):

Thus, h = 114 m . h = 114 m .

Check Your Understanding 6.6

If atmospheric resistance is neglected, find the maximum height for the mortar shell. Is calculus required for this solution?

Interactive

Explore the forces at work in this simulation when you try to push a filing cabinet. Create an applied force and see the resulting frictional force and total force acting on the cabinet. Charts show the forces, position, velocity, and acceleration vs. time. View a free-body diagram of all the forces (including gravitational and normal forces).

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StickMan Physics

StickMan Physics

Animated Physics Lessons

F=ma Practice Problems

F=ma problem set.

Practice solving for net force, using Newtons second law (F=ma), and relating F=ma to the acceleration equations.

In these practice problems we will either use F=ma or our 1D motion acceleration equations to solve force problems.

F=ma Equations

1. What is the acceleration of the 15 kg box that has 500 N of force applied to the right?

a = 33.33 m/s 2 Right

2. What is the acceleration of the 25 kg box that has 50 N of force applied to the right?

a=2.0 m/s 2 Right

3. What is the acceleration of the 3 kg box that has 25 N of force applied to the right and 55 N left?

a = 10.0 m/s 2 Left

4. What is the acceleration of the 5 kg box that has a 25 N force and 50 N force applied both right?

a = 15.0 m/s 2 Right

5. What is the acceleration of the 25 kg box that has a 100 N force north and 50 N force east applied?

a= 4.47 m/s 2

5b. What direction would this box accelerate?

63.43° North of East

6. Does a 795 kg Lorinser speedy, 6300 kg elephant, or 8.6 kg wagon have more inertia and why?

6300 kg Elephant

The more mass the more inertia

different masses

7. How much force is required to accelerate a 795 kg Lorinser Speedy by 15 m/s 2 ?

F = 11925 N

8. How much force is required to accelerate an 8.6 kg wagon by 15 m/s 2 ?

F = 129 N

9. How much does a 6300 kg elephant accelerate when you apply 500 N of force?

a = 0.0794 m/s 2

10. What is the mass of an object if it takes a net force of 40 N to accelerate at a rate of 0.88 m/s 2 ?

m = 45.45 kg

11. How much force is required to accelerate a 0.142 kg baseball to 44.7 m/s during a pitchers 1.5 meter delivery?

F = 94.58 N

12. A 0.050 kg golf ball leaves the tee at a speed of 75.0 m/s. The club is in contact with the ball for 0.020 s. What is the net force of the club on the ball?

F = 187.5 N

13. A 90.0 kg astronaut receives a 30.0 N force from her jetpack. How much faster is she be moving after 2.00 seconds?

0.667 m/s faster

14. A 795 kg car starts from rest and travels 41 m in 3.0 s. How much force did the car engine provide?

F = 7242 N

15. Joe and his sailboat have a combined weight of 450 kg. How far has Joe sailed when he started at 5 m/s and a gust of wind provided 600 Newtons of force for 4 seconds?

x = 30.64 m

16. Tom pulls a 45 kilogram wagon with a force of 200 Newtons at a 15° angle to the horizontal from rest. How much faster will the wagon be moving after 2 seconds?

force at an angle

v f = 8.58 m/s

Back to the Newtons Second Law Lesson
Continue to Mass and Weight
Back to the Main Forces Page
Back to the Stickman Physics Home Page
Equation Sheet

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Force Examples

force problem solving with solution

A force of one object has a result to push or have an interaction with another object. Here are few force examples to learn how to calculate force by applying mass and acceleration.

Force Mass Acceleration Problems with Solutions

Let us consider the problem: Find the force of an object with mass(m) as 600 kg and acceleration as 50 m/s^2

We can calculate force the using the given formula.

Substituting the values in the formula,

= 600 x 50 = 30000 N Hence, force of the object is 30000 Newtons.

Let us consider the problem: Find the mass of an object with force 200 Newtons and acceleration as 10 m/s^2

We can calculate the mass using the given formula.

= 200 / 10 = 20 Kgs Hence, mass of the object is 20 Kgs

Let us consider the problem: Find the acceleration of an object with mass(m) as 300 kg and force as 1000 N

We can calculate the acceleration using the given formula.

Acceleration:

= 1000 / 300 = 3.333 m/s^2 Hence, acceleration of the object is 3.333 m/s^2

Related Examples:

Potential Energy Examples
Young's Modulus Calculation Examples
Kinetic Energy Example
Work Physics Problems With Solutions
Newton Second Law Of Motion Example Problems With Answers
Synthetic Division Examples

Calculators and Converters

Classical Physics

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Centripetal Force

Practice problem 1.

at the bottom (and rising)
halfway to the top
45° past the top
Determine the minimum coefficient of static friction needed to complete the stunt as planned.

The only forces present in this problem are weight (which always points down), normal (which always points toward the center of the loop), and friction (which is always tangential to the loop). The weight of the motorcycle never changes. The normal varies from a maximum at the bottom of the loop to a minimum at the top. As the motorcycle drives up the loop, the friction force acts along the direction of motion to keep gravity from slowing it down. As the motorcycle drives down the loop, the friction force acts opposite the direction of motion to keep gravity from speeding it up. The bike isn't speeding up or slowing down, but it is changing direction. This means the net force always points toward the center of motion.

Weight points down and normal points up, so the net force is their difference. The normal force points toward the center, so it should be given the positive value. The net force is the centripetal force.

Friction isn't necessary since the motorcycle isn't accelerating horizontally.

f i = 0 N

The normal force points horizontally, toward the center of the loop. There isn't anything to balance this force at this position. This is our net force — our centripetal force.

Weight points down and normal points inward (toward the center of the loop). They aren't opposite anymore, so they can't cancel. The motorcycle isn't accelerating vertically at this moment, so there must be something to balance the weight. That's where friction steps in. Friction and weight are equal.

Both weight and normal point down, so the net force is their sum.

Once again, friction isn't necessary since the motorcycle isn't accelerating horizontally.

f iii = 0 N

The normal and friction forces are at right angles to each other. This makes them "good" vectors. Normal points toward the center and contributes to the centripetal force. Friction is tangential to the circle and contributes nothing to the centripetal force. Weight is the "bad" vector. It needs to be broken up into components. The component pointing toward the center contributes to the centripetal force.

The component of the weight tangent to the track is balanced by the friction (so that the speed stays constant).

As I was solving this problem, I realized there is a conceptually easier way to solve it. What stays constant in this problem? Weight and centripetal force. What changes? Direction. Compute weight and centripetal force once, then combine them as appropriate to the location.

Weight points down and normal points up. The net force (their difference) is the centripetal force.

Normal points toward the center, which makes it the centripetal force.

Friction counteracts weight to keep the motorcycle moving with constant speed around the loop.

Both weight and normal point down toward the center of the loop. The net force (their sum) is the centripetal force.

Again, friction isn't necessary since the motorcycle isn't accelerating horizontally.

The position for this part of the problem was rigged so the two components would have the same magnitude and there'd be fewer things to calculate.

The normal force and a component of the weight point toward the center of the loop (the component perpendicular to the loop), so they get to be the centripetal force.

The other component of the weight (the component parallel to the loop) is balanced by friction so the net force tangential to the loop is zero and the speed does not change.

Static friction needs to be greatest when the motorcycle is vertical. The tires need to grip the loop to balance out the weight and keep the bike from accelerating tangential to the loop. At these two points (one where the bike is going up and the other where the bike is going down) friction equals weight and normal provides the centripetal force. I'll solve this both ways — once using the fancy, formal method…

and again using the conceptually easier method…

practice problem 2

meters per second
Earth days per rotation
rotations per Earth year

Use the centripetal acceleration equation and solve for speed. Substitute values for the acceleration due to gravity on Earth and the radius of the Earth's orbit (also known as an astronomical unit).

Sounds fast, but things in space tend to move fast anyway. How does this compare to the yearly motion of the Earth around the Sun? That's the goal of the next parts of the question.

We'll solve this practice problem two ways. First we'll use the definition of speed and substitute the value calculated above and the distance traveled in one rotation (the circumference).

The fancier way is to start from the centripetal acceleration equation, replace speed with circumference over period, and simplify. This gives us an equation that some people like so much they memorize it.

Solve for period and substitute.

That's a quick "year" — if we use the astronomical definition of the year as the period of one trip around the Sun. How does it compare to a regular calendar year?

This problem is best solved by dimensional analysis.

We will return to this problem in the section on power .

practice problem 3

The curve radii of modern high-speed systems result in dependence on the speed and the maximum possible superelevation of the guideway to compensate for the centrifugal forces occurring. The Transrapid's guideway can have a maximum superelevation of 12 degree (up to 16 degree in special cases) which allows smaller radii at higher speeds than in the case of conventional wheel-on-rail systems. Minimal radius: 350 m 200 km/h: 0 705 m 400 km/h: 2825 m 500 km/h: 4415 m ThyssenKrupp Transrapid GmbH, 2002

the maximum centripetal acceleration (in m/s 2 and g) implied by these specifications
the speed limit (in m/s and km/h) on a curved section of track with the minimal radius

Once you get past reading the awkward translation from German to English, this is a conceptually easy question. Set up a table like the one below and complete the missing parts. Use a graphing calculator as it eliminates the drudgery of repeated calculation.

Start by converting the maximum speeds from km/h to m/s.

Then apply the equation for centripetal acceleration.

The resulting values will be in m/s 2 . To convert to g, divide by the standard value for the acceleration due to gravity.

a c [g] = a c [m/s 2 ] ÷ 9.80665 m/s 2

This procedure gives a set of numbers that are reasonably close to one another: three values in m/s 2 and three in g. Find the mean of both of these triplets. A partially completed table is provided below.

For the second part of this question, follow the logic of the first part in reverse order. Assume that the maximum permissible centripetal acceleration is the same for all curves, regardless of size. Use the mean value we just calculated to determine the speed limit on a curve with a 350 m radius.

Then convert from m/s to km/h.

Add these results to the table.

A more advanced technique is to solve the problem graphically.

Starting from the equation for centripetal acceleration.

Make v 2 the subject and compare to the equation for a straight line.

The equation is telling us we should put speed squared on the y axis and radius on the x axis.

This makes the centripetal acceleration equal the slope of the line of best fit.

The intercept value of 7.28 m 2 /s 2 is effectively equal to zero. Since the speed squared values are all on the order of several thousand, an intercept that's less than ten is "small" in comparison.

Use the coefficients from the line of best fit to find the speed limit on the minimum radius curve. Finish by converting the speed back to km/h from m/s.

practice problem 4

Complete the first two columns using astronomical data from a reliable source. Be sure to specify the units used for each entry.
Complete the last two columns using a calculator. Be sure to state your answers in SI units.

The data for the moon and planets came from a page in this book. The data for the Sun came from a page in The Physics Factbook. The distance from the Earth to the Sun is called an astronomical unit (au). I plan on discussing the au in the section of this book on miscellaneous units (1 au = 1.496 × 10 11 m). I know you know the period of Earth's orbit. You may just have to pause and think for a second. The first two columns of the table are done.

Speed is the rate of change of distance with time, which in circular motion means circumference divided by period.

We are going to use this equation over and over again. Please watch those units. Convert to meters and seconds as appropriate.

Now that we have all the speeds, we can compute the centripetal accelerations.

Again we'll do it over and over again…

The table is now ready for completion.

› Force and Motion
› This article

Solving problems which involve forces, friction, and Newton's Laws: A step-by-step guide

This step-by-step guide is meant to show you how to approach problems where you have to deal with moving objects subject to friction and other forces, and you need to apply Newton's Laws. We will go through many problems, so you can have a clear idea of the process involved in solving them.

The problems we will examine include objects that

are pushed/pulled horizontally with an angle
move up or down an incline
hang from ropes attached to the ceiling
hang from ropes that run over pulleys
move connected by a string
are pushed in contact with each other (Coming soon!)
Box pulled at an angle over a horizontal surface
Block pushed over the floor with a downward and forward force
Object moving at constant velocity over a horizontal surface
Block pushed up a frictionless ramp
Mass pulled up an incline with friction
A mass hanging from two ropes
Two hanging objects connected by a rope
Two masses on a pulley
Two blocks connected by a string are pulled horizontally

force problem solving with solution

Newton's Laws
Einstein's Theory of Special Relativity
About Concept Checkers
School Pricing
Newton's Laws of Motion
Newton's First Law
Newton's Third Law
Sample Problems and Solutions
Kinematic Equations Introduction
Solving Problems with Kinematic Equations
Kinematic Equations and Free Fall
Kinematic Equations and Kinematic Graphs

Check Your Understanding

Answer: d = 1720 m

Answer: a = 8.10 m/s/s

Answers: d = 33.1 m and v f = 25.5 m/s

Answers: a = 11.2 m/s/s and d = 79.8 m

Answer: t = 1.29 s

Answers: a = 243 m/s/s

Answer: a = 0.712 m/s/s

Answer: d = 704 m

Answer: d = 28.6 m

Answer: v i = 7.17 m/s

Answer: v i = 5.03 m/s and hang time = 1.03 s (except for in sports commericals)

Answer: a = 1.62*10 5 m/s/s

Answer: d = 48.0 m

Answer: t = 8.69 s

Answer: a = -1.08*10^6 m/s/s

Answer: d = -57.0 m (57.0 meters deep)

Answer: v i = 47.6 m/s

Answer: a = 2.86 m/s/s and t = 30. 8 s

Answer: a = 15.8 m/s/s

Answer: v i = 94.4 mi/hr

Solutions to Above Problems

d = (0 m/s)*(32.8 s)+ 0.5*(3.20 m/s 2 )*(32.8 s) 2

Return to Problem 1

110 m = (0 m/s)*(5.21 s)+ 0.5*(a)*(5.21 s) 2

110 m = (13.57 s 2 )*a

a = (110 m)/(13.57 s 2 )

a = 8.10 m/ s 2

Return to Problem 2

d = (0 m/s)*(2.60 s)+ 0.5*(-9.8 m/s 2 )*(2.60 s) 2

d = -33.1 m (- indicates direction)

v f = v i + a*t

v f = 0 + (-9.8 m/s 2 )*(2.60 s)

v f = -25.5 m/s (- indicates direction)

Return to Problem 3

a = (46.1 m/s - 18.5 m/s)/(2.47 s)

a = 11.2 m/s 2

d = v i *t + 0.5*a*t 2

d = (18.5 m/s)*(2.47 s)+ 0.5*(11.2 m/s 2 )*(2.47 s) 2

d = 45.7 m + 34.1 m

(Note: the d can also be calculated using the equation v f 2 = v i 2 + 2*a*d)

Return to Problem 4

-1.40 m = (0 m/s)*(t)+ 0.5*(-1.67 m/s 2 )*(t) 2

-1.40 m = 0+ (-0.835 m/s 2 )*(t) 2

(-1.40 m)/(-0.835 m/s 2 ) = t 2

1.68 s 2 = t 2

Return to Problem 5

a = (444 m/s - 0 m/s)/(1.83 s)

a = 243 m/s 2

d = (0 m/s)*(1.83 s)+ 0.5*(243 m/s 2 )*(1.83 s) 2

d = 0 m + 406 m

Return to Problem 6

(7.10 m/s) 2 = (0 m/s) 2 + 2*(a)*(35.4 m)

50.4 m 2 /s 2 = (0 m/s) 2 + (70.8 m)*a

(50.4 m 2 /s 2 )/(70.8 m) = a

a = 0.712 m/s 2

Return to Problem 7

(65 m/s) 2 = (0 m/s) 2 + 2*(3 m/s 2 )*d

4225 m 2 /s 2 = (0 m/s) 2 + (6 m/s 2 )*d

(4225 m 2 /s 2 )/(6 m/s 2 ) = d

Return to Problem 8

d = (22.4 m/s + 0 m/s)/2 *2.55 s

d = (11.2 m/s)*2.55 s

Return to Problem 9

(0 m/s) 2 = v i 2 + 2*(-9.8 m/s 2 )*(2.62 m)

0 m 2 /s 2 = v i 2 - 51.35 m 2 /s 2

51.35 m 2 /s 2 = v i 2

v i = 7.17 m/s

Return to Problem 10

(0 m/s) 2 = v i 2 + 2*(-9.8 m/s 2 )*(1.29 m)

0 m 2 /s 2 = v i 2 - 25.28 m 2 /s 2

25.28 m 2 /s 2 = v i 2

v i = 5.03 m/s

To find hang time, find the time to the peak and then double it.

0 m/s = 5.03 m/s + (-9.8 m/s 2 )*t up

-5.03 m/s = (-9.8 m/s 2 )*t up

(-5.03 m/s)/(-9.8 m/s 2 ) = t up

t up = 0.513 s

hang time = 1.03 s

Return to Problem 11

(521 m/s) 2 = (0 m/s) 2 + 2*(a)*(0.840 m)

271441 m 2 /s 2 = (0 m/s) 2 + (1.68 m)*a

(271441 m 2 /s 2 )/(1.68 m) = a

a = 1.62*10 5 m /s 2

Return to Problem 12

(NOTE: the time required to move to the peak of the trajectory is one-half the total hang time - 3.125 s.)

First use: v f = v i + a*t

0 m/s = v i + (-9.8 m/s 2 )*(3.13 s)

0 m/s = v i - 30.7 m/s

v i = 30.7 m/s (30.674 m/s)

Now use: v f 2 = v i 2 + 2*a*d

(0 m/s) 2 = (30.7 m/s) 2 + 2*(-9.8 m/s 2 )*(d)

0 m 2 /s 2 = (940 m 2 /s 2 ) + (-19.6 m/s 2 )*d

-940 m 2 /s 2 = (-19.6 m/s 2 )*d

(-940 m 2 /s 2 )/(-19.6 m/s 2 ) = d

Return to Problem 13

-370 m = (0 m/s)*(t)+ 0.5*(-9.8 m/s 2 )*(t) 2

-370 m = 0+ (-4.9 m/s 2 )*(t) 2

(-370 m)/(-4.9 m/s 2 ) = t 2

75.5 s 2 = t 2

Return to Problem 14

(0 m/s) 2 = (367 m/s) 2 + 2*(a)*(0.0621 m)

0 m 2 /s 2 = (134689 m 2 /s 2 ) + (0.1242 m)*a

-134689 m 2 /s 2 = (0.1242 m)*a

(-134689 m 2 /s 2 )/(0.1242 m) = a

a = -1.08*10 6 m /s 2

(The - sign indicates that the bullet slowed down.)

Return to Problem 15

d = (0 m/s)*(3.41 s)+ 0.5*(-9.8 m/s 2 )*(3.41 s) 2

d = 0 m+ 0.5*(-9.8 m/s 2 )*(11.63 s 2 )

d = -57.0 m

(NOTE: the - sign indicates direction)

Return to Problem 16

(0 m/s) 2 = v i 2 + 2*(- 3.90 m/s 2 )*(290 m)

0 m 2 /s 2 = v i 2 - 2262 m 2 /s 2

2262 m 2 /s 2 = v i 2

v i = 47.6 m /s

Return to Problem 17

( 88.3 m/s) 2 = (0 m/s) 2 + 2*(a)*(1365 m)

7797 m 2 /s 2 = (0 m 2 /s 2 ) + (2730 m)*a

7797 m 2 /s 2 = (2730 m)*a

(7797 m 2 /s 2 )/(2730 m) = a

a = 2.86 m/s 2

88.3 m/s = 0 m/s + (2.86 m/s 2 )*t

(88.3 m/s)/(2.86 m/s 2 ) = t

t = 30. 8 s

Return to Problem 18

( 112 m/s) 2 = (0 m/s) 2 + 2*(a)*(398 m)

12544 m 2 /s 2 = 0 m 2 /s 2 + (796 m)*a

12544 m 2 /s 2 = (796 m)*a

(12544 m 2 /s 2 )/(796 m) = a

a = 15.8 m/s 2

Return to Problem 19

v f 2 = v i 2 + 2*a*d

(0 m/s) 2 = v i 2 + 2*(-9.8 m/s 2 )*(91.5 m)

0 m 2 /s 2 = v i 2 - 1793 m 2 /s 2

1793 m 2 /s 2 = v i 2

v i = 42.3 m/s

Now convert from m/s to mi/hr:

v i = 42.3 m/s * (2.23 mi/hr)/(1 m/s)

v i = 94.4 mi/hr

Return to Problem 20

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21.9: Sample problems and solutions

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Exercise \(\PageIndex{1}\)

A cathode ray tube in a television accelerates an electron using a potential difference of \(\Delta V=500\text{V}\) . The electron must be deflected upwards by a distance \(h=3\text{cm}\) using a uniform magnetic field, \(\vec B\) , before striking the phosphorescent screen, which is a distance \(d= 5\text{cm}\) away. What direction and magnitude must the magnetic field have in order to steer the electron towards its destination?

First, we determine the velocity of the electron that were accelerated over a potential difference of \(\Delta V=500\text{V}\) . Their kinetic energy is given by their charge times the potential difference::

\[\begin{aligned} K &= e\Delta V \\[4pt] \frac{1}{2} mv^2 &= e\Delta V\\[4pt] \therefore v &= \sqrt{\frac{2e\Delta V}{m}}= \sqrt{\frac{2(1.602\times 10^{-19}\text{C})(500\text{V})}{(9.109\times 10^{-31}\text{kg})}}\\[4pt] &= 1.326\times 10^{7}\text{ms}^{-1}\end{aligned}\]

Now that we have the velocity, we must determine the direction of the magnetic field. We know that the electron is moving directly towards the phosphorescent screen (which we will define as \(\vec x\) ) and the electron must be deflected directly upwards (which we will define as \(\vec z\) ). Knowing this, we can use the right hand rule to quickly determine that the magnetic force will be acting in the \(-\vec y\) direction.

In the region with a magnetic field, the electron will undergo uniform circular motion with a radius give by the cyclotron radius, \(R\) :

\[\begin{aligned} R=\frac{mv}{qB}\end{aligned}\]

We thus need to determine the radius of that circle for the electron to arrive that desired location on the screen. A section of the circle about which the electron moves is illustrated in Figure \(\PageIndex{1}\).

Figure \(\PageIndex{1}\) : Deflection of an electron noving in a uniform magnetic field.

From geometry and Pythagoras’ Theorem, we have:

\[\begin{aligned} R^2 &= (R-h)^2+d^2\\[4pt] R^2 &= R^2-2Rh+h^2+d^2\\[4pt] \therefore R &= \frac{h^2+d^2}{2h}=\frac{(3\text{cm})^2+(5\text{cm})^2}{2(3\text{cm})}=5.67\text{cm}\end{aligned}\]

The strength of the magnetic field is then given by:

\[\begin{aligned} B&=\frac{mv}{qR}=\frac{(9.11\times 10^{-31}\text{kg})(1.326\times 10^{7}\text{ms}^{-1})}{(1.6\times 10^{-19}\text{C})(0.0567\text{m})}=0.00135\text{T}\end{aligned}\]

Exercise \(\PageIndex{2}\)

A galvanometer has a square coil with a side length of \(a=2.5\text{cm}\) and \(N=70\) loops between two magnets which generate a radial magnetic field of \(B=8\text{mT}\) . When a current runs through the coil, it generates a torque which is opposed by a spring with a torsional spring constant of \(\kappa = 1.5\times 10^{-8}\text{Nmrad}^{-1}\) . If the deflection of the galvanometer’s needle is \(0.7\), what is the current running through the coil?

First, we will determine the magnetic dipole moment of the square coil:

\[\begin{aligned} \mu &= NIA\\[4pt] \mu &=NIa^2\end{aligned}\]

Now that we have the magnetic dipole moment, we can calculate the torque on the square coil that is produced by the magnetic field. Note that, in a galvanometer, the magnetic field is configured such that it is radial and always perpendicular to the magnetic dipole moment of the coil:

\[\begin{aligned} \tau_B &= N\mu B sin(90^{\circ})= NIa^2B\\[4pt]\end{aligned}\]

The deflection, \(\theta\) , for a given current will occur when the torque produced by the wire is equal to the torque produced by the spring. The torque produced by the spring is given by:

\[\begin{aligned} \tau_s =\kappa \theta\end{aligned}\]

where \(\theta\) is measured in radians. The above equation is the rotational equivalent of Hooke’s Law. Equating the torque from the spring and from the magnetic field, we can determine the current:

\[\begin{aligned} \tau_B&=\tau_S\\[4pt] NIa^2B &= \kappa \theta\\[4pt] I &= \frac{\kappa \theta}{Na^2B} = \frac{(1.5\times 10^{-8}\text{Nm(rad)}^{-1}) (0.7\text{rad})}{70(0.025\text{m})^2(8\times 10^{-3}\text{T})}\\[4pt] &= 30\mu\text{A}\end{aligned}\]

Exercise \(\PageIndex{3}\)

Integrate the equation \(d\vec F = Id\vec l \times \vec B\) over a circular path to show that the torque exerted on a circular loop of radius, \(R\) , carrying current, \(I\) , immersed in a uniform magnetic field, \(\vec B\) , has a magnitude given by \(\tau=\mu B\) , where \(\vec \mu\) is the magnetic dipole moment of the loop. You may simplify the problem by modeling the loop when its magnetic moment is perpendicular to the magnetic field.

Figure \(\PageIndex{2}\) illustrates a loop of radius, \(R\) , carrying current, \(I\) . The loop is in the \(x-z\) plane, and there is a magnetic field, \(\vec B\) , in the negative \(x\) direction. By setting the loop up this way, it is easier to visualize some of the three-dimensional aspects.

Figure \(\PageIndex{2}\) : A current-carrying loop in a magnetic field.

Consider an infinitesimal section of the loop, with length, \(dl\) , located on the loop at a position labeled by the angle, \(\theta\) , as illustrated. The vector, \(d\vec l\) , is given by:

\[\begin{aligned} d\vec l = dl (-\sin\theta \hat x + \cos\theta \hat z)\end{aligned}\]

The magnetic force on that element of the loop is given by:

\[\begin{aligned} d\vec F &=Id\vec l \times \vec B\\[4pt] &=Idl(-\sin\theta \hat x + \cos\theta \hat z) \times (-B\hat x)\\[4pt] &=-IBdl\cos\theta (\hat z \times \hat x)\\[4pt] &=-IBdl\cos\theta\hat y\end{aligned}\]

and the force on that element of wire is out of the page (negative \(y\) direction), as illustrated. That infinitesimal force will create an infinitesimal torque:

\[\begin{aligned} d\vec \tau = \vec r \times d\vec F\end{aligned}\]

where \(\vec r\) is the vector from the axis of rotation (through the center of the loop, parallel to the \(z\) axis) to the point where the force is exerted. The length of the vector, \(\vec r\) , is simply \(r=R\cos\theta\) , and the force is perpendicular to the vector \(\vec r\) . Thus, the torque on the infinitesimal element is given by:

\[\begin{aligned} d\vec \tau &= \vec r \times d\vec F= (R\cos\theta \hat x)\times (-IBdl\cos\theta\hat y)\\[4pt] &=-IBR\cos^2\theta dl (\hat x \times \hat y)=-IBR\cos^2\theta dl \hat z\end{aligned}\]

and the torque on that infinitesimal element is in the negative \(z\) direction, as anticipated from the direction of the force. Note that had we considered the loop to be oriented such that the magnetic field is not in the plane of the loop, the vector \(\vec r\) in the torque would have a component in the \(y\) direction.

We can sum the torques on each element of the loop, from \(\theta = 0\) to \(\theta=2\pi\) . We can express the length, \(dl\) , using the infinitesimal angle, \(d\theta\) , that subtends the arc of length, \(dl\) , on the circle of radius, \(R\) :

\[\begin{aligned} dl = Rd\theta\end{aligned}\]

The net torque is then given by:

\[\begin{aligned} \vec \tau &= \int d\vec \tau=\int -IBR\cos^2\theta dl \hat z= (-IBR^2\hat z)\int_0^{2\pi} \cos^2\theta d\theta =(-IBR^2\hat z)\pi\end{aligned}\]

The magnetic moment of the loop is:

\[\begin{aligned} \mu = IA = I\pi R^2\end{aligned}\]

so that the torque is indeed given by \(\tau = \mu B\) . If we had rotated the loop so that the vector, \(\vec r\) , had a \(y\) component, then we would have found the general formula with a cross-product.

Net force word problems

Find here in this lesson some easy and challenging net force word problems.

Problem #1:

What is the net force on the airplane in the figure below?

The airplane is moving with a force of 800 N. However, there are two forces moving in the opposite direction on the airplane.

Just add these two forces: 40 N + 60 N = 100 N

Subtract to get the net force: 800 N - 100 N = 700 N

The net force is 700 N.

The airplane will move with a force of 700 N as a result of air friction and wind.

Net force on an airplane

Problem #2 : You and your brother are pushing a car with a dead battery with forces of 20 N and 25 N in the same direction. What is the net force applied on the car?

Solution:

Since you are pushing the car in the same direction, the forces will be added together.

Net force = 20 N + 25 N

Net force = 45 N.

Problem #3 : A brother is pulling a toy from his sister with a force of 6 N. The sister is pulling back with a force of 8 N.

Who gets the toy?

What is the net force?

The sister gets the toy of course since she is pulling with a stronger force.

Net force = 8 N - 6 N

Net force = 2 N.

More challenging net force word problems

Problem #4 :

4 people are playing a tug of war. Two are pulling on the right side. Two are pulling on the left side. On the right side, one is pulling with a force of 60 N and the other with a force of 70 N. On the left side, one is pulling with a force of 30 N. How much force should the second person on the left apply to keep the rope in equilibrium?

The rope will be in equilibrium is the net force is 0.

The forces on the right is equal to 60 N + 70 N = 130 N

Let x be the force that must be applied by the second person on the left.

30 N + x = 130 N

Since 30 N + 100 N = 130 N, x = 100 N

The other person should pull with a force of 100 N to keep the rope in equilibrium.

Equilibrium

Problem #5 :

Your friend is pulling upward on an object with force of 3 N. You are pulling to the right with a force of 4 N.

Find the net force and the direction the object moves.

Just build the rectangle and find the resultant. The red arrow shows the direction the object will move.

This net force word problem is a little challenging. To find the net force, we need use the Pythagorean Theorem .

What is a net force?

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Magnetic Field: Solved Problems for grade 12 and AP Physics

On this page, some Problems on magnetic fields for high school and colleges are solved. Each section is separated for easier reading.

Magnetic Field Problems: Force on a single Moving Charge

Problem (1): A proton moves with a speed of $2\times 10^6 \,\rm m/s$ at an angle of $30^\circ$ with the direction of a magnetic field of $0.2\,\rm T$ in the negative $y$-direction. (a) What is the magnitude and direction of the magnetic force on the proton? (b) What acceleration does undergo by the proton?

Solution : The magnitude of the magnetic force on a single charge moving at speed of $v$ in a uniform magnetic field of $B$ is determined by the formula $F=qvB\sin \theta$ where $\theta$ is the angle between velocity and magnetic field. $q$ is also the magnitude of the charge, which in this case is $q=1.6\times 10^{-19}\,\rm C$. (a) Substituting all numerical values into the above equation, we have \begin{align*} F&=qvB\sin\theta \\\\ &=(1.6\times 10^{-19})(2\times 10^6)(0.3) \sin 30^\circ \\\\ &=4.8\times 10^{-14}\,\rm N \end{align*} An extremely small force. This was the magnitude of the magnetic force. To find the direction of the magnetic force on a positive moving charge , we should use the right-hand rule.

First, identify the directions of $\vec{B}$ and $\vec{v}$ as illustrated in the figure below. Now, according to this rule, put your right fingers along the direction of the charge's velocity $\vec{v}$, then curl your fingers toward the magnetic field $\vec{B}$ through the smaller angle. As a result, your thumb will point in the direction of the magnetic force $\vec{F}$ exerted on the positive charge.

In this case, the thumb points out of the page $\odot$.

(b) Applying Newton's second law, $F=ma$ and solving for $a$ gives \[a=\frac{F}{m}=\frac{4.8\times 10^{-14}}{1.67\times 10^{-27}}=2.87\times 10^{13}\,\rm \frac{m}{s^2}\] Although the magnetic force exerted on the proton was quite small, it experiences a huge acceleration since its mass is also extremely small.

Problem (2): An electron experiences a force of $3.5\times 10^{-15}\,\rm N$ when moving at an angle of $37^\circ$ with the direction of a magnetic field of $2.5\times 10^{-3}\,\rm T$. How fast was the electron?

Solution : Again, the magnitude of the magnetic force exerted by a uniform magnetic field $B$ on a charged particle moving at $v$ with an angle of $\theta$ relative to the $\vec{B}$ is found to be $F=qvB\sin\theta$. Substituting the numerical values into it and solving for $v$, gives \begin{align*} v&=\frac{F}{qB\sin\theta} \\\\ &=\frac{3.5\times 10^{-15}}{(1.6\times 10^{-19})(2.5\times 10^{-3})\sin 37^\circ} \\\\ &=13.7\times 10^6\quad\rm m/s \end{align*} Note that in this formula, only the magnitude of the electric charge is included not its sign. The sign of charge determines the direction of the magnetic force on it.

Problem (3): In the following diagrams, find the direction of the magnetic force on a negative charge. ($\otimes$ and $\odot$ indicate the directions into and out of the page, respectively.)

A negative charge moves in a uniform magnetic field problem

Solution : Notice that in this magnetic field Problem, the charge is negative, so the right-hand rule must be applied with caution. According to this rule, put outstretched fingers of your right hand along the direction of the velocity $\vec{v}$ and curl them toward the magnetic field vector $\vec{B}$ through the smaller angle between them. In this case, your thumb points in the direction of the force.

This instruction is only for positive charges. If the charge was negative, simply reverse the direction of the force.

By doing so, the correct force direction on each diagram is shown below. The blue dotted arrows show the force direction if the charge had been positive.

In the right lower diagram the angle between $\vec{B}$ and $\vec{v}$ is $\theta=180^\circ$, so $\sin \theta=0$, and no force is applied to the particle, $F=0$.

direction of the magnetic force on a negative charge is shown.

Solution : Place your right-hand fingers in the direction of the velocity $v$ and curl them toward the direction of the magnetic field $B$. The direction of the thumb is in the direction of the force. Note that $\otimes$ and $\odot$ represent vectors pointing into the page and out of the page, respectively.

In the left lower diagram the angle $\theta=180^\circ$, so $\sin 180^\circ=0$, and the force is zero. Pay attention to the sign of the electric charge. For a negative charge, the force direction obtained by the right-hand rule must be reversed.

The direction of the magnetic force is drawn.

Solution : (a) Substituting all known values into the following formula and solve for $F$.\begin{align*} F&=qvB\sin\theta \\ &=(1.6\times 10^{-19})(7.5\times 10^6)(45) \sin 45^\circ \\ &=3.8\times 10^{-11}\,\rm N \end{align*} (b) The right-hand rule gives us the direction of the magnetic force into the plane of the page, $\otimes$. Place the fingers of your right hand along the velocity vector (red arrow) and rotate them toward the magnetic field. As a result, your thumb, which shows the direction of the force, points into the page, $\otimes$.

Problem (6): An alpha particle moving at a speed of $2\times 10^7 \,\rm m/s$ toward the positive $z$-direction perpendicularly enters a uniform magnetic field $\vec{B}$ and experiences an acceleration of $1.0\times 10^{13}\,\rm m/s^2$ in the positive $x$-direction. Find the magnitude and direction of the magnetic field?

Solution : The mass of an alpha particle is four times $m_{\alpha}=4m_p$ and its charge is twice $q_{\alpha}=2e$ that of a proton.

The alpha particle's velocity is given as $\vec{v}=2\times 10^7\, (+\hat{k})\quad \rm m/s$.

In this magnetic field Problem, the magnitude and direction of the acceleration acquired by the alpha particle was given. We can use these information to find the magnitude and direction of the applied force to it as below \begin{align*} F&=m_{\alpha}a \\\\ &=(4)(1.67\times 10^{-27})(1.0\times 10^{13}) \\\\ &=6.68\times 10^{-14}\quad \rm N \end{align*} This force directed toward the positive $x$-direction, i.e., $+\hat{i}$, in the same direction of the acceleration.

As you can see, in the following diagram the angle between $\vec{v}$ and $\vec{B}$ is $90^\circ$, so $\sin \theta=1$.

After collecting all these the given information, use the magnetic force formula, $F=qvB\sin\theta$, and solve for $B$ to find the unknown magnitude of the magnetic field \begin{align*} B&=\frac{F}{qv\sin\theta}\\\\ &=\frac{6.68\times 10^{-14}}{(2)(1.6\times 10^{-19}) (2\times 10^7)} \\\\ &=0.01\quad \rm T\end{align*} Now, apply the right-hand rule for a positive charge to find the direction of $\vec{B}$. Here, we choose a coordinate system in which out of the page indicates the positive $z$-direction.

Hold your right hand so that the four fingers point in the $\vec{v}$ (out of the page) and the thumb points toward the $\vec{F}$ (to the right). In this situation, your palm faces in the direction of the magnetic field (up the page).

An alpha particle in a uniform B field.

Problem (7): An electron with a kinetic energy of $1.5\,\rm keV$ perpendicularly enters a $0.02-\rm T$ magnetic field. Determine the radius of its path as it moves through this uniform field?

Solution : First note that $\rm keV$ is a unit of energy in the subatomic level that converts into the usual units of energy, the joules, as below \begin{align*} \rm 1\,keV &=1\times 10^3 \times (1.6\times 10^{-19}) \,J \\ &=1.6\times 10^{-16}\,J\end{align*} By knowing the kinetic energy $K=\frac 12 mv^2$, we can find the particle's velocity as below \begin{align*} v&=\sqrt{\frac{2K}{m}} \\\\ &=\sqrt{\frac{2(1.5)(1.6\times 10^{-16})}{9.11\times 10^{-31}}} \\\\&=2.3\times 10^7\,\rm m/s \end{align*} When a charged particle enters a region of uniform magnetic field at right angle with a speed of $v$, the magnetic field forces it to move in a circular path of radius $r$. This force, whose magnitude is found by $qvB\sin\theta$, serves as a centripetal force radially toward the center of the circle. As we learned in the circular motion Problems section, the acceleration that the object acquire is $a_c=\frac{mv^2}{r}$. Consequently, we have \begin{gather*} F=ma_c \\\\ qvB\sin 90^\circ=\frac{mv^2}{r} \\\\ \Rightarrow \quad r=\frac{mv}{qB} \end{gather*} It is better to memorize this important formula. It always gives the radius of a charged particle moving through a uniform $\vec{B}$ perpendicularly.

Hence, the value of radius is obtained as follows \begin{align*} r&=\frac{(9.11\times 10^{-31})(2.3\times 10^7)}{(1.6\times 10^{-19})(0.02)} \\\\ &=0.65\times 10^{-2}\,\rm m\end{align*} Therefore, this electron having such energy when enters a $0.02\,\rm T$ field bends around a circle of radius $0.65\,\rm mm$. For more information about this bending, refer to the helical motion in a magnetic field .

Problem (8): An electron undergoes the greatest force of magnitude $3.2\times 10^{-13}\,\rm N$, vertically upward, when it travels northward at a speed of $5\times 10^6 \,\rm m/s$ in a uniform magnetic field of unknown strength. What is the magnitude and direction of the magnetic field?

Solution : in all magnetic field questions, the greatest force occurs when the angle between $\vec{v}$ and $\vec{B}$ is $90^\circ$. In this case, $\sin\theta=1$. By solving the equation $F=qvB\sin\theta$ for $B$ and substituting the numbers into it, we get \begin{align*} B&=\frac{F}{qv} \\\\ &=\frac{3.2\times 10^{-13}}{(1.6\times 10^{-19})(5\times 10^6)} \\\\ &=0.4\,\rm T \end{align*} To find the direction of the field, first of all, set a coordinate system and specify all the directions on it as below. In this standard coordinate system, used when geographical direction included in the Problem, we assume north and south directions as into $\otimes$ and out of the page $\odot$, respectively.

Place the four fingers of your right hand in the direction of velocity $\vec{v}$ so that your thumb points to the force direction. In this case, your palm faces toward the magnetic field. This is the right-hand rule for a positive charge.

Doing so, your palm will be toward the left (or west) but the electric charge of the electron is negative so you must reverse that direction (which is shown as a red dotted arrow) to find the correct direction of the field.

The force on an electron in a uniform magnetic field problem

Problem (9): A proton traveling with a speed of $2.4\times 10^6\,\rm m/s$ through a uniform magnetic field of strength $2.5\,\rm T$ experiences a force of magnitude $4.8\times 10^{-13}\,\rm N$. At what angle does the electron enter the field?

Solution : here, the unknown is the angle between $\vec{v}$ and $\vec{B}$. Applying the magnetic force formula, $F=qvB\sin\theta$, and solving for the unknown angle $\theta$ gives \begin{align*} \sin\theta &=\frac{F}{qvB} \\\\ &=\frac{4.8\times 10^{-13}}{(1.6\times 10^{-19})(2.4\times 10^6)(2.5)} \\\\&= 0.5 \end{align*} Taking the inverse sine of both sides, gives us the desired angle \[\theta=30^\circ\]

Problem (10): The path of a negatively charged particle traveling through a magnetic field is shown in the figure below. What is the direction of the $\vec{B}$ field?

Solution :

Magnetic Field Problems: Force on Electric Current

Problem (11): In each of the following diagrams, a current-carrying wire is shown into a uniform external magnetic field. Find the direction of the magnetic force for each diagram.

Focre on a current carrying wire in a B field.

Solution : Place outstretched fingers of your right hand along the direction of the current. For the left diagram, your fingers must point down the page. Now, curl them toward the magnetic field $B$ which is out of the plane of the page. As a result, your thumbs, which represent the force, points to the left.

For the right diagram, there is an interesting point. If look closely, the current is out of the page and the field is into the page, so the angle between them is $180^\circ$ and we know that $\sin 180^\circ=0$. Consequently, there is no force acting on this wire.

The force on a current-carrying wire is drawn.

Problem (12): Find the magnitude of the force per meter exerted on a straight wire carrying a current of $3.5\,\rm A$ through a magnetic field of $0.85\,\rm T$ when (a) it is placed perpendicular to the $\vec{B}$. (b) it makes an angle of $37^\circ$ with the field.

Solution : When a straight wire of length $\ell$ carrying a current $i$ passes through a uniform external magnetic field $\vec{B}$, it experiences a force of magnitude $F=i\ell B\sin \theta$ where $\theta$ is the angle between the direction of the current and $\vec{B}$.

The force ''per meter'' means the force acted on one meter of the wire, so $\ell=1\,\rm m$.

(a) In this case, we have $\theta=90^\circ$, so $\sin\theta=1$. Thus, the magnitude of the force is found to be \begin{align*} F&=i\ell B\sin\theta \\&=(3.5)(1)(0.85) \\&=2.975\quad \rm N \end{align*} (b) Similarly, we have \begin{align*} F&=i\ell B\sin\theta \\&=(3.5)(1)(0.85) \sin 37^\circ \\&=1.785\quad \rm N \end{align*}

Problem (13): A straight wire carrying a current of $4\,\rm A$ is placed perpendicularly in a uniform magnetic field of strength $0.45\,\rm T$. What is the magnitude of the magnetic force on a $35-\rm cm$-long section of this wire?

Solution : All the required information to substitute into the magnetic force formula is given. Hence, we have \begin{align*} F&=i\ell B\sin\theta \\&=(4)(0.35)(0.45) \sin 90^\circ \\&=0.63\quad \rm N \end{align*}

Problem (14): An straight section of a wire $0.40\,\rm m$ long carrying a steady current of $2.5\,\rm A$ toward the $+x$ direction lies in a region where a uniform magnetic field of strength $B=1.5\,\rm T$ to the $+z$ direction is present. What is the magnitude and direction of the magnetic force on this section of wire?

Solution : In this magnetic field Problem, the angle between $\vec{B}$ and the direction of the electric current is $90^\circ$, so $\sin\theta=1$. The magnitude of the magnetic force is also determined as below \begin{align*} F&=i\ell B\sin \theta \\ &=(2.5)(0.40)(1.5) \\ &=1.5\quad \rm N\end{align*} To find the direction of the force acting on the wire, do the following: Point the fingers of your right hand in the direction of the current $i$ and curl them toward the magnetic field direction. Your thumb, then, points in the direction of the magnetic force. This is the right-hand rule for finding the magnetic force on a current-carrying wire in a uniform $\vec{B}$.

The direction of the magnetic force on a wire in a B field is drawn.

Problem (15): At a location where the Earth's magnetic field is $0.5\,\rm G$ from south to north, there is a straight wire of length $2\,\rm m$ that carries a current of $1.5\,\rm A$. In each of the following situations, find the magnitude and direction of the magnetic force exerted on the wire due to the earth's magnetic field? (a) a current flowing from west to east. (b) a current flowing vertically upward. (c) a current flowing north to south.

Solution : The magnitude of the magnetic force is obtained by the equation $F=i\ell B\sin\theta$ and its direction is also found by the right-hand rule. To solve such magnetic field Problems, first set up a coordinate system and specify each direction on it as below. In this coordinate system south and north directions are shown by the symbols $\odot$ and $\otimes$ means out of and into the page, respectively.

The Earth's magnetic field is from south to north, which we can imagine as an arrow points into the page, i.e., $\otimes$. The gauss, $\rm G$ is another unit for magnetic field and is related to the tesla by $\rm 1\, G=10^{-4} \, T$.

Current flows west to east in the Earth's magnetic field

Problem (16): In a region of space, a magnetic force per unit length of $0.24\,\rm N/m$ toward the positive $z$-axis exerts on a current of $I=10\,\rm A$ directed along the positive $y$-direction and perpendicular to the magnetic field $\vec{B}$. Determine the magnitude and direction of the magnetic field?

Solution : the magnitude of the magnetic force on a wire of length $\ell$ carrying current $i$ in a uniform external magnetic field $\vec{B}$, is determined by the following formula \[F=i\ell B\sin\theta\] where $\theta$ is the angle between the direction of the current and the direction of the magnetic field. When this angle is $90^\circ$, the magnetic force acted on the wire is maximum.

Here, the maximum ($\theta=90^\circ$) force per unit length was given $F/\ell=0.24\,\rm N/m$. Substituting the given data into the above formula and solving for $B$ gives \begin{align*} B&=\frac{F}{i\ell} \\\\ &=\frac{0.24}{10}\\\\ &=0.024\quad \rm T \end{align*} The right-hand rule gives us the direction of $\vec{B}$.

To find the magnetic force direction on a wire carrying current in a uniform magnetic field, the following version of the right-hand rule must be applied.

The direction of the magnetic force on a wire in a leftward B field is drawn.

We were asked to make a situation in which the tension in the wires be zero. This holds when the downward conductor's weight force $mg$ is balanced with an upward magnetic force $i\ell B$. Note that, here, the angle between $\vec{B}$ and the wire is $\theta=90^\circ$, so $\sin \theta=1$.

Balancing the weight and the magnetic force

Magnetic Field Due to a Long Straight Wire

Problem (18): A current of $20\,\rm A$ flows vertically upward in a straight wire. What is the magnitude and direction of the magnetic field due to this long wire at a location $15\,\rm cm$ away and on the left side of the wire?

Solution : The magnitude of the magnetic field due to a long straight wire near to it is found as follows \[B=\frac{\mu_0 I}{2\pi r} \] where $\mu_0=4\pi\times 10^{-7}\,\rm T\cdot m/A$ is called the permeability of free space. Plug in the known values gives \begin{align*} B&=\frac{(4\pi\times 10^{-7})(20)}{2\pi\times 0.15} \\\\ &=2\times 10^{-5}\,\rm T\end{align*} To find the direction of the magnetic field due this long wire, we must use the following version of the right-hand rule:

Grasp the wire in your right hand so that the thumb points in the direction of the current. Now, wrap your four fingers around the wire. This is the direction of the $\vec{B}$.

Doing so, in this case, at the point $P$ on the left side of the wire the fingers will come out of the paper, $\otimes$.

Similarly, on the right side, the four fingers will come into the paper, $\otimes$. These are the directions of the $\vec{B}$ at those points.

Author : Dr. Ali Nemati Published : 3/2/2022

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Forces in Physics, tutorials and Problems with Solutions
The concepts of forces, friction forces, action and reaction forces, free body diagrams, tension of string, inclined planes, etc. are discussed and through examples, questions with solutions and clear and self explanatory diagrams. Questions to practice for the SAT Physics test on forces are also included with their detailed solutions.
Calculating Force
The formula for force is . We are given the mass, but we will need to calculate the acceleration to use in the formula. We know the initial velocity (zero because the box starts from rest), final velocity, and distance traveled. Using these values, we can find the acceleration using the formula .
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Work done by force
Push force (F) = 14 Newton. Friction force (F fric) = 10 Newton. Displacement of object (d) = 8 meters. Wanted: Work (W) Solution : There are two forces that act on an object, push force (F) and friction force (F fric). The push force has the same direction as the displacement of the object so that the push force does a positive work.
Force Problems
Hints And Answers For Force Problems. Hint and answer for Problem # 7. The force of gravity pulling down on the block is F1 = Mg sin θ. The maximum friction force opposing the sliding is F2 = Mg cos θμs. At some angle θ the block will be on the verge of sliding. This is the maximum angle θ and occurs when F1 = F2.
6.1 Solving Problems with Newton's Laws
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Newton's Law Problem Sets
Problem 22: Brandon is the catcher for the Varsity baseball team. He exerts a forward force on the .145-kg baseball to bring it to rest from a speed of 38.2 m/s. During the process, his hand recoils a distance of 0.135 m. Determine the acceleration of the ball and the force which is applied to it by Brandon.
6.3: Solving Problems with Newton's Laws (Part 2)
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F=ma Problem Set . Practice solving for net force, using Newtons second law (F=ma), and relating F=ma to the acceleration equations. In these practice problems we will either use F=ma or our 1D motion acceleration equations to solve force problems. 1. What is the acceleration of the 15 kg box that has 500 N of force applied to the right?
4.9: Problem-Solving Strategies
In practice, intuition develops gradually through problem solving, and with experience it becomes progressively easier to judge whether an answer is reasonable. Another way to check your solution is to check the units. If you are solving for force and end up with units of m/s, then you have made a mistake. Summary
Force Examples
Let us consider the problem: Find the force of an object with mass(m) as 600 kg and acceleration as 50 m/s^2 ... Solution: We can calculate force the using the given formula. Force: Where, m = Mass, a = Acceleration . Substituting the values in the formula, = 600 x 50 = 30000 N Hence, force of the object is 30000 Newtons.
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practice problem 1. A 250 kg motorcycle is driven around a 12 meter tall vertical circular track at a constant speed of 11 m/s. Determine the normal and friction forces at the four points labeled in the diagram below. Determine the minimum coefficient of static friction needed to complete the stunt as planned.
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Buoyant Force Formula with Solved Examples
The buoyant force is the same as long as the object displaces the same amount of fluid. The example below shows what is meant by displaced volume in the buoyancy force equation. Example (2): A rectangular container of ethanol with density \rho_ {eth}=800\, {\rm kg/m^3} ρeth = 800kg/m3 floats in the liquid of density \rho_ {f}=1200\, {\rm kg/m ...
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Problem #5: Your friend is pulling upward on an object with force of 3 N. You are pulling to the right with a force of 4 N. Find the net force and the direction the object moves. Just build the rectangle and find the resultant. The red arrow shows the direction the object will move. This net force word problem is a little challenging.
Magnetic Field: Solved Problems for grade 12 and AP Physics
Problem (12): Find the magnitude of the force per meter exerted on a straight wire carrying a current of 3.5\,\rm A 3.5A through a magnetic field of 0.85\,\rm T 0.85T when. (a) it is placed perpendicular to the \vec {B} B. (b) it makes an angle of 37^\circ 37∘ with the field.
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In this paper, a novel cascaded trajectory tracking control strategy based on the characteristics of kinematics and dynamics is proposed for unmanned underwater vehicles (UUVs) under complex disturbances. A kinematic controller based on an improved model predictive control (MPC) is designed to solve the speed jump problem by considering system constraints. Moreover, an improved whale optimizer ...