SAT Physics Dynamics - Newton's Laws Of Motion

SAT Physics Dynamics - Newton's Laws Of Motion

Sir Isaac Newton deduced three laws of motion to describe the relationships among force, mass, and changes in velocity.

Newton’s First Law
In his first law, Newton simply generalizes Galileo’s principle of inertia. Essentially, the first law states that objects at rest will remain at rest, and objects in motion will remain at constant velocity unless acted upon by a net unbalanced force. This law is often referred to as the law of inertia.
When the word equilibrium is used in physics, the forces acting on an object are balanced (equal and opposite). Equilibrium is a stable condition where the net force is zero, 1-1 = 0, and an object does not accelerate, a = 0. If the object is not accelerating, it must be experienc­ing one of only two types of equilibrium.
  1. Static equilibrium. The object has a constant velocity equal to zero.
  2. Dynamic equilibrium. The object has constant velocity not equal to zero.
Newton’s first law of motion governs objects in equilibrium.

Newton’s Second Law
Newton’s second law addresses changes to the inertia of an object. It states that when an object of mass m is acted upon by a net force of Fnet or 1-1 (the sum of all forces acting together), the object will accelerate. In addition,
  • the acceleration of the object will be in the direction of the net force;
  • the acceleration will be directly proportional to the net force, and;
  • the acceleration will be inversely proportional to the mass of the object.
These details are very concisely summarized in one the most famous equations in physics.
Newton’s second law results in several important consquences.
  1. When all the one-dimensional vector forces sum to a nonzero value, the forces are said to be unbalanced. The result is acceleration, which alters either the magnitude or the direction of velocity. When an object is accelerating, there must be a net, unbalanced force acting on the object.
  2. The direction of acceleration is the same as the direction of the net force.
  3. The direction of velocity is not necessarily in the same direction as the acceleration. An example would be the change in velocity of a car as it slows to a stop. The initial veloc­ity is in one direction, but the acceleration is in the opposite direction, causing the car to slow down.
  4. When vector forces add up to zero, the forces are balanced. There can be no accelera­tion. An object will continue at constant velocity based on the first law of motion. When an object is either at rest or moving at constant velocity, the one-dimensional forces acting on the object must cancel each other. The net force must equal zero.
Newton’s Third Law
Newton’s third law states that when objects interact, an equal and opposite force is always exerted between them. The response forces attributed to normal force and to tension are examples of Newton’s third law.
Interactions consist of action-reaction pairs. The agent creates an action force on the object. The object pushes back with a responding opposite and equal reaction force. Although these forces are always equal, the resulting motions experienced by the agent and the object are not necessarily equal.
For example, when a person jumps upward, he or she creates an equal downward force on Earth. Anyone witnessing this event would observe the person moving upward but would not detect Earth moving at all. Why, when the forces are equal, does Earth not experience an equal and opposite motion? Earth has an enormous mass and inertia; therefore, the planet is extremely difficult to accelerate. The force exerted by the person jumping is enough to lift the person off the ground but does little to the motion of Earth

Newton’s Third Law
An 800-newton person is involved in a tug of war with a 600-newton person.
The 800-newton person pulls on the rope between them with 400 newtons of force.
If the rope is not slipping, how much force does the 600-newton person pull with at the other end of the rope?


Newton’s third law states that whenever two objects interact, there is an equal and opposite force between them. The problem is asking for the force in the rope. The first person pulls on the rope with 400 newtons. Therefore, the second person must also pull on the rope with 400 newtons. The weights of each person,800 and 600 newtons, are distractors.

The following four-step method is a suggested attack plan that breaks difficult force problems into a series of steps you can easily solve.
1. Orient the problem. Identify force vectors and the relevant directions. Then sketch a force diagram.
2. Determine the type of motion. Determine if the problem involves
  • equilibrium, constant velocity, balanced forces, 1-1 = 0;
  • dynamics, acceleration, unbalanced forces, 1-2
3. Sum the force vectors in the relevant direction. Only forces, or components of forces, parallel to the direction of
motion  can change the speed of an object. Sum only the one-dimensional force vectors parallel to the
object’s motion. Any force vectors point­ing in the direction of motion will increase the speed of the object
and should be set as positive. Force vectors opposing the motion of the object should be set as negative.
4. Substitute and solve. In this last step, substitute known equations for specific forces and they substitute
numerical values to find the solution.
The following sections include common example problems solved using this four-step strategy. With practice and experience, you will begin to discover shortcuts and will eventu­ally solve problems without realizing you are actually doing each step.

Solve x-Forces and y-Forces Independently
The direction in which forces act is a key consideration when solving force problems. When adding vectors to determine the net force, solve one-dimensional x- and y-force vectors separately and independently. You must resolve any two-dimensional force vectors into separate one-dimensional x- and y-component vectors. For additional information on resolving vec­tors into other components, see Chapter 2. The following examples will illustrate the impor­tance of force vector direction and demonstrate the problem-solving strategy.
Solve x- and y-directions Independently
A 10-kilogram mass is pulled to the right along a rough horizontal surface by a string. The string has a tension of 20 newtons. Determine the acceleration of the mass if the coefficient of kinetic friction between the mass and the surface is 0.10.


Orient the problem: Sketching a free-body diagram, as shown in the diagram on the left below, clearly reveals the force relationships. The object is pulled to the right by a string. Set right as positive in the x-direction. The object is not moving vertically. Set up as positive in the y-direction. Directional plus and minus signs are shown in the figure on the right below.
Determine the type of motion: The x- and y-motions are different.
  • y-direction: stationary, balanced forces, 1-5
  • x-direction: acceleration, unbalanced forces, 1-6
Sum the force vectors in the relevant direction:
  • 1-7
  • 1-8
Substitute and solve: Substitute ma for 1-9 and zero for 1-10 into the above equa­tions. Then substitute values and solve. The y-direction solves first and determines the normal force. This value is needed in the friction equation.
  • y-direction: 0 = N - mg
                         0 = N - (10 kg)(10 m/s2)
                        N = 100 N
  • x-direction: ma = T - μN
                (10 kg)a = (20 N) - (0.10)(100 N)
                            a = 1.0 m/s2
Components of Force
A 10-kilogram mass is pulled along a smooth horizontal surface by a 20-newton force acting 37° above the horizontal. Determine the acceleration of the mass.


Although the force is inclined at an angle, the mass is accelerating parallel to the surface. You must split the 20-newton force into its x- and y-components.

Orient the problem: The force diagram below left shows all the forces acting on the mass, including the components of the diagonal 20-newton force. The 37° angle indicates a 3-4-5 triangle. So you can easily find the components.
Note: The components of the 20 N force are not included in a formal free-body diagram.
The object is accelerating in the x-direction. Only the x-force vectors matter, as shown in the diagram below right.
Determine the type of motion: The problem asks for acceleration. If the mass is accelerating then the mass obeys Newton’s second law, 1-2

Sum the force vectors in the relevant direction: Since there is only one x-direction force vector, this problem is extremely simple.

Substitute and solve: Substitute ma for 1-9 in the above equation. Then substitute the numerical value for the mass and solve for acceleration.
       ma = 16 N
(10 kg)a = 16 N
a = 1.6 m/s2
Incline problems are a special case where the x- and y-axes are not useful. Instead, it is easier to work with tilted axes that are parallel and perpendicular to the incline, as shown in Figure 5.6(a). The left diagram also reveals that although the normal force lies on one of axes, the force of gravity does not.
Figure 5.6. Solving incline problems

As a result, the force of gravity must be split into components, as shown in Figure 5.6(b). Solving the vector components of the force of gravity results in the following equations:
  • Force of gravity parallel to the incline:  1-16
  • Force of gravity perpendicular to the incline:  1-17
The force of gravity perpendicular to the incline pushes the object into the incline. The incline pushes back with an equal and opposite force, creating the normal force N.
Fincline = 1-17
The component of the force of gravity parallel to the incline is unbalanced. If no other forces are present, this component will cause the object to accelerate down the incline. Whenever an incline is present, there is always a component of force acting to pull an object down and parallel to the incline.
The relationship and difference between the force of gravity, Fg, and the force of gravity parallel to the incline, 1-18, confuses many students. The force of gravity is the actual force act­ing on the object and is always included in free-body diagrams. However, it is not used to solve an incline problem mathematically. Calculating the mathematical solution for motion paral­lel to an incline requires force vectors or components of force vectors that are parallel to the incline. The force of gravity parallel to the incline, 1-18, is a component of force that is always present in an incline problem. Since it is a component of a force vector, 1-18 is never included in the free-body diagram. However, it is always included in calculations of the net force acting parallel to an incline. Figure 5.7(a) shows the free-body diagram on an incline. Figure 5.7(b) shows the vector component needed when solving equations for the same situation.
Figure 5.7. Force on an incline
A 5.0-kilogram mass is positioned on a 30° frictionless incline. It is kept stationary by a string pulling parallel to the incline. Determine the tension in the string.


Work with forces and components that are parallel and perpendicular to the slope. 
Orient the problem:
The force diagram below left shows all forces and the components of the force of gravity parallel and perpendicular to the incline. However, to find the tension T, only forces parallel to the slope are needed. The diagram below right shows only the forces relevant to this problem.
Note: The components of Fg are not included in a formal free-body diagram.

Determine the type of motion: The problem asks for the tension in the string that will keep the mass at rest. The mass obeys Newton’s first law, 1-22

Sum the force vectors in the relevant direction: There are two vectors parallel to the incline. Normally, the direction of motion is set as positive. However, the masses are not moving. So you can set either direction as positive.
substitute numerical values, and solve.
      0 = mg sin θ - T
T = mg sin θ
                          T = (5.0 kg)(10 m/s2) sin 30°
Remember: sin 30° = 0.
T= 25 N

Dynamics and Kinematics
Dynamics and kinematics are explicitly linked. Students are commonly given a set of forces and asked to determine their effect on the motion of an object. Just as time was the key variable shared by both x- and y-components of motion in Chapter 4, acceleration is the key variable shared by both force equations and kinematics equations. Typically, the sum of forces helps students find the acceleration. Then the acceleration is used in the kinematic equations to determine displacement and final velocity

Combining Dynamics and Kinematics
Mass m is initially at rest and then is pushed by force F through a distance, x, reaching a final velocity of v. What will be the new final velocity for mass m if the experiment is repeated with a force of 2F?


To find the effect on velocity of a changing force, first determine how acceleration is altered. The force portion of this problem involves a single force whose direction is not important. In cases such as this, the four-step method is not needed. The lone force is responsible for all the acceleration.
F = ma
Mass remains constant. As a result, force and acceleration are directly propor­tional. Therefore, doubling the force doubles the acceleration.
(2F) = m(2a)
The rest of this problem involves analyzing kinematics. This problem has two important aspects. First, there is no mention of time. Second, the object starts at rest.
From above, doubling the force doubles the acceleration.
Acceleration is under the square root. Doubling it is the same as multiplying the velocity by the square root of 2. The new final velocity is √2vf.

Compound Body Problems
A compound body consists of two or more separate masses experiencing the same motion. The masses may be pressing against one another or be tied together with strings so that they move as a single system. Figures 5.8(a) and 5.8(b) illustrate both types of problems.
Figure 5.8. Compound body problems
Both masses experience the same acceleration. The net force pushes or pulls on the total mass of the system. When solving for acceleration, you can sum the masses into a single system (sys).
If asked to find the force exerted by m1 on mass m2 or vice versa, work with each block individually. Draw the free-body diagram for only one of the blocks, and sum the force vec­tors acting on only that block.
Two key forces act between masses in a compound body. In Figure 5.8(a), the masses push on each other with their surfaces creating normal forces. In Figure 5.8(b), the blocks are attached with a string and the force between them is tension. In compound body problems, internal normal forces and tensions acting between masses add to zero.
Compound bodies can also be oriented vertically, as shown in Figures 5.9(a) and 5.9(b). In these problems, gravity is the external force pulling the masses downward.
Figure 5.9. Vertical compound bodies
If any horizontal external forces act on either mass in Figure 5.9(a), then friction may be present between the two masses. However in beginning physics courses, stacked masses most often remain stationary. When the systems shown in Figure 5.9 remain stationary, the acceleration of the system, the net force acting on the system, and the net force acting on each mass are all zero.
If the sum of forces and acceleration are zero, the forces acting on each mass must be bal­anced. The normal forces in Figure 5.9(a) must be equal and opposite the force of gravity pulling down against each surface. The magnitude of the normal force will be equal to but opposite the force of gravity of all the masses stacked above a surface. In Figure 5.9(b) the tension vectors are determined in a similar manner. The tension in a string is equal to but opposite the force of gravity for all the masses hanging from a string.

Compound Bodies
A force of 18 newtons pushes two masses (m1 = 2 kilograms and m2 = 4 kilograms) horizontally to the right.
(A) Determine the acceleration of mass m2.


Both masses are moving together as a system. The acceleration of mass m2 equals the acceleration of the system.

Orient the problem: Visualize the masses as a single system.
Determine the type of motion: The problem asks for acceleration.

Sum the force vectors in the relevant direction: The 18-newton force in the diagram above is the only external force in the direction of motion that is not canceled. Internal normal forces act horizontally between m1, and m2. However, Newton’s third law dictates that the forces are opposite and equal. Therefore, the forces cancel each other. In addition, any vertical forces would not affect the net force in the horizontal direction. Only the 18-newton force remains.
 Substitute and solve: Combine the above equations and substitute values.
     18 = (m1 + m2)a
18 = (2 + 4)a
a = 3 m/s2
(B) Determine the magnitude of the force between the masses


According to Newton’s third law, the two interacting masses push on each other with opposite and equal force. Therefore, solve for the force on either mass. The forces acting between the blocks are normal forces since the surfaces of the masses push against each other.

Orient the problem: Again, the motion is horizontal and only horizontal force vectors contribute forces for this motion.

Determine the type of motion: Acceleration obeys Newton’s second law.
Work with the masses individually and not as part of a system.
Sum the force vectors in the relevant direction: Force F and the normal force of m2, which is, N2, push on m1 Only the normal force of m1 which is N1 pushes on m2. With only one force acting on it, m2 will be easier to solve. However, the solution for both masses is shown to prove that either mass can be used.
values and solve. The acceleration, a = 3 m/s2, was determined in part (A).
m1a = F - N2                                                                           m2a = N1
(2 kg)(3 m/s2) = (18 N) - N2                            (4 kg)(3 m/s2) = N1
    N2 = 12 N                                                                                   N1 = 12 N
The normal forces pressing between the masses are opposite and equal.

Pulley Problems
Pulley problems are compound body problems where more than one mass is connected together by a string draped over a pulley. Two common pulley problems are shown in the following diagram. Figure 5.10(a) depicts an Atwood machine designed by George Atwood to test constant acceleration and Newton’s laws of motion. Figure 5.10(b) shows a modified Atwood machine.
Figure 5.10. An Atwood machine
 Pulleys are used to change the direction of force. On the SAT Subject Test in Physics, these devices will operate under ideal conditions. The pulleys will be massless and frictionless. The strings will also have zero mass.
Although the masses in pulley problems move in different coordinate directions, they do share one common motion. The masses in pulley problems always move in the same direc­tion as the string. Any force aligned with the motion of the string can influence acceleration. When summing the forces in the relevant direction, set all the forces pointing in the direction of the string’s motion as positive and all the forces opposing the string’s motion as negative.

Atwood Machine
Masses m and 2m are connected by a string, which is draped over a pulley.
The masses are released from rest. Determine the magnitude of acceleration of mass m.


When solving for pulley problems, determine the direction the string is moving. Make this the positive direction, and determine all forces parallel to the string. If solving for acceleration, treat both masses as a single system.

Orient the problem: Although the force of gravity acts on both masses, gravity pulls down the 2m mass. This causes the smaller mass m to move upward.
Forces in the direction of motion increase the speed of an object. They have a positive influence on acceleration and are set as positive. Vectors opposing motion slow down an object and are set as negative.

Determine the type of motion: The forces are unbalanced, resulting in the acceler­ation of the entire system. Applying Newton’s second law of motion to the entire system results in the following:

Sum the force vectors in the relevant direction: Include all the forces parallel to the motion of the string. The sum of the vector forces will be:
For the SAT Subject Test in Physics, the pulleys themselves will always be massless. Under these ideal conditions, the tension in a string will be the same everywhere. The equal and opposite tensions will cancel. As a result, tensions can be ignored when solving for the acceleration of the entire system. (Important: tensions cancel only when summing the forces for an entire system. Do not cancel tensions when summing the forces acting on a single independent mass.)
In this problem, the numerical values associated with each mass are coefficients rather than subscripts. Mass 2m is not a mass with a value of 2 units. Instead, it is a mass that is twice as large as mass m. This problem will not have a numerical answer. Instead, the solution will be a simplified equation using variables.
3ma = 2mg - mg
a = g/3

Modified Atwood Machine
Two masses, m1 = 2 kg and m2 = 3 kg, are connected by a string, which is draped over a pulley as shown above. Mass 1 is positioned on a horizontal surface, while mass 2 hangs freely. The masses are released from rest.
(A) Determine the acceleration of mass 2.


To solve for acceleration, treat the masses as a single system and sum the forces acting on both masses simultaneously. 

Orient the problem: Mass m2 will move downward, dragging mass m, along the surface to the right. Friction is never mentioned or alluded to in the problem, so consider its effect to be negligible.
Determine the type of motion: The forces are unbalanced, resulting in the acceleration of the entire system. Newton’s second law of motion says:

Sum the force vectors in the relevant direction: The tensions cancel, leaving only one force acting to accelerate the entire system.
(m1 + m2)a = (m2)g
           (2 kg + 3 kg)a = (3 kg)(10 m/s2)
                 a = 6 m/s2
(B) Determine the tension in the string.


To solve for the tension in the string, sum the forces for only one of the two masses. You cannot sum the forces for the system since this causes the tension to cancel. We can sum the forces for either mass as both will result in the same answer. Solutions using both masses are shown below. Only one is needed.

Orient the problem: Separate the masses into separate diagrams.
 Determine the type of motion: Both masses will accelerate. Newton’s second law of motion applied to each mass results in:

Sum the force vectors in the relevant direction:

Substitute and solve:
Using either mass arrives at the same answer. The easiest solution usually involves selecting the mass with the least amount of forces acting on it.


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