Lesson Explainer: Newton’s Third Law of Motion Mathematics

In this explainer, we will learn how to solve problems on Newton’s third law.

An understanding of Newton’s first and second laws of motion is important for acquiring an understanding of Newton’s third law of motion.

Let us define Newton’s first law of motion.

Definition: Newton’s First Law of Motion

A body in uniform motion will continue in that uniform motion unless a nonzero net force acts on it.

Newton’s first law tells us that the momentum of a body will not change if the forces acting on the body are in equilibrium.

Let us define Newton’s second law of motion.

Definition: Newton’s Second Law of Motion

When a net force acts on a rigid body, the body changes its velocity in the direction of the force. For a body of constant mass, the magnitude of the acceleration depends on the magnitude of the force and on the mass of the body, according to the formula 𝐹=𝑚𝑎, where 𝑚 is the mass of the body and 𝑎 is the acceleration of the body.

The rate of change of momentum of a body is equal to the net force acting on the body. For a constant-mass body, the rate of change of momentum of the body is directly proportional to the rate of change of velocity of the body.

Let us now define Newton’s third law of motion.

Definition: Newton’s Third Law of Motion

Any force that acts on a body is one of a pair of forces that meet the following conditions:

  1. The forces act on different interacting bodies.
  2. The forces act in opposite directions.
  3. The forces act along the same line.
  4. The forces have the same magnitude.
  5. The forces act for the same time interval.

Bodies must exert forces on each other to interact. Consider the pair of forces 𝐹 and 𝐹 that act on bodies 𝐴 and 𝐵 and the pair of forces 𝐹 and 𝐹 that act on bodies 𝐶 and 𝐷, as shown in the following figure.

Each of the pairs of forces are exerted during collisions between the bodies that the forces act on. The forces 𝐹 and 𝐹 might seem to meet the conditions for a pair of forces defined by Newton’s third law of motion, as might the forces 𝐹 and 𝐹.

Body 𝐴 does not act on body 𝐷 to produce 𝐹, however, and body 𝐵 does not act on body 𝐶 to produce 𝐹. Only the pair of forces 𝐹 and 𝐹 and the pair of forces 𝐹 and 𝐹 are examples of pairs of forces that meet the conditions for a pair of forces defined by Newton’s third law of motion, as only these forces act on different interacting bodies.

Newton’s third law is equivalent to the principle of the conservation of momentum. A force accelerating a body changes the momentum of a body. If when a force acts, an equal-magnitude force acts in the opposite direction for the same time interval on another body, then the sum of the changes of momentum of the bodies due to these forces is zero.

Consider two bodies of equal mass that move at equal speeds toward each other along a smooth horizontal surface, approaching a position at which they will collide, as shown in the following figure.

The sum of the momenta of the bodies is given by 𝑝=𝑚𝑣+𝑚(𝑣)𝑝=𝑚(𝑣𝑣)=0.

After the collision, suppose that the velocities of both of the bodies change sign but retain the same magnitude, as shown in the following figure.

The net momentum after the collision is also zero. The change of the momentum of one of the bodies is equal to the change in the momentum of the other body; hence, the forces exerted on the bodies are equal.

Newton’s third law applies to collisions between moving bodies but can apply to bodies that are at rest. A common example of pairs of forces that satisfy Newton’s third law of motion is the forces that act when a body is in contact with a surface on Earth.

A body that is in contact with a surface on Earth is acted on by the weight of the body. The weight of the body is a force that depends on the mass of the body and the acceleration of the body. On Earth, the value of a due to the gravitational force produced by the gravitational field of Earth is 9.8 m/s2 and is represented by 𝑔. A body in contact with a surface is also acted on by a reaction force normal to the surface.

For a body in contact with a surface on Earth, two pairs of forces that satisfy Newton’s third law of motion exist:

  1. the gravitational force that Earth exerts on the body and the gravitational force that the body exerts on Earth,
  2. the reaction force on the body and the reaction force on Earth due to their mutual contact.

The forces acting on the body and on Earth are shown in the following figure, where the forces are shown acting at the centers of mass of the body and of Earth.

The pair of forces labeled 𝑊 that act to attract Earth and the body are gravitational forces, while the pair of forces labeled 𝑅 that act to repel Earth and the body are reaction forces. Each pair of forces that act on both the body and Earth satisfies Newton’s third law of motion. The magnitudes of the forces in either pair equal the magnitudes of the forces in the other pair.

Questions involving the equilibrium of forces on a body at rest on the surface of Earth, where the equilibrium of the forces is the result of Newton’s third law, do not necessarily consider forces that act on Earth. Such questions may only show the forces that act on the body in contact with Earth, as is the case in the following figure of a body at rest on a surface on Earth.

The force 𝑊 in the diagram is the weight, which is the force exerted on the body due to the masses of Earth and the body. The force 𝑅 is the reaction force exerted on the body by the surface of Earth. The forces are of equal magnitude. The gravitational and reaction forces that act on the body meet all the conditions for a pair of forces that satisfy Newton’s third law of motion except for the condition of acting on different bodies.

Considering only the forces acting on a body can give a misleading impression that Newton’s third law can sometimes not apply. Consider a body that is not in contact with Earth that falls toward Earth. Before the falling body reaches the surface of Earth, there is no reaction force on the body and it seems as if the weight of the body acts to increase the momentum of the body without there being a corresponding change in the momentum of another body.

According to Newton’s third law, however, as the body falls and so accelerates toward Earth, Earth accelerates toward the body, as the gravitational forces on the Earth and on the body have equal magnitude. The gravitational force on Earth increases the momentum of Earth in the opposite direction to the momentum of the body, conserving the momentum of the system consisting of the body and Earth. The change in the velocity of Earth due to its change in the momentum is negligible as the mass of Earth is so much greater than the mass of the body. Earth is treated as effectively being at a fixed position when considering bodies that fall toward Earth, as the acceleration of Earth due to the force acting on it due to the falling body is negligible.

The examples in this explainer deal with bodies that are in contact with surfaces, where the surfaces may accelerate either in or opposite to the direction of the gravitational force on the bodies.

To help us understand how to approach such examples, we can use free-body diagrams. A body at rest on a surface that is at rest is shown in the following figure.

The net force on the body is zero. Suppose that the surface on which the body is at rest accelerates uniformly with an acceleration 𝑎, where vertically upward is positive. The net force on the body must be given by 𝐹=𝑚𝑎.net

The net force on the body is not the only force acting vertically upward on the body. A vertically upward force with a magnitude equal to the weight of the body acts on the body even when the body is at rest. When the body accelerates vertically upward, the upward force on the body is given by 𝐹=𝑚𝑔+𝑚𝑎=𝑚(𝑔+𝑎).

If the surface that the body rests on has acceleration vertically downward, then 𝐹=𝑚𝑎,net and the vertically upward force on the body to give it this acceleration is given by 𝐹=𝑚(𝑔𝑎).

The vertically upward force on the body is the reaction force on the body.

Let us look at an example involving forces acting on an object in contact with a moving surface.

Example 1: Finding the Reaction Force between a Man and an Elevator Moving at a Constant Speed

An elevator is moving vertically upward at a constant speed. A man of mass 90 kg is standing inside. Determine the reaction force of the floor on the man. Take 𝑔=9.8/ms.

Answer

The elevator is moving at a constant speed. The man in the elevator is at rest relative to the elevator and so is also moving at a constant speed. A body moving at a constant speed is not accelerating, and the net force acting on a body with no acceleration must be zero.

As the man is in a region where the force of Earth’s gravitational field accelerates unsupported bodies vertically downward at 9.8 m/s2, there is a vertically downward force acting on the man. This force is the weight of the man. If the vertically downward direction is taken as negative, the force of the man’s weight is given by 𝐹=90×(9.8)=882.N

The net force on the man is zero, so the floor of the elevator must be exerting a vertically upward force on the man and this force must have the same magnitude as the weight of the man, so the reaction force from the elevator floor has a magnitude of 882 N.

The constant-velocity motion of a body does not involve the action of a net force, as Newton’s first law of motion states. Motion of a body where there is a change in velocity, however, does involve the action of a net force.

Let us now look at an example involving the forces acting on a body that has a constant acceleration.

Example 2: Finding the Reaction Force between a Man and an Elevator Accelerating Upward

An elevator is accelerating vertically upward at 3.3 m/s2. A man of mass 80 kg is standing inside. Determine the reaction force of the floor on the man.

Answer

The following figure shows the forces acting on the man.

The elevator is accelerating vertically upward at 3.3 m/s2. The man in the elevator is at rest relative to the elevator, so he is also accelerating vertically upward at 3.3 m/s2. The magnitude of the net force acting on the man to provide this acceleration, taking vertically upward motion as positive, is given by 𝐹=80×3.3=264.netN

The force of the man’s weight, 𝐹, acts vertically downward. and it’s given by 𝐹=80×(9.8)=784,N taking the magnitude of 𝑔 as 9.8 m/s2.

The reaction force on the man from the elevator floor must provide a vertically upward force of this magnitude to keep the man at rest relative to the elevator floor. If the elevator was not accelerating, the reaction force vertically upward would have a magnitude of 784 N.

The elevator is accelerating vertically upward at 3.3 m/s2, however, and because the man is at rest relative to the elevator, the force required to accelerate the man at this rate is the net force on the man, 264 N.

The reaction force must therefore provide a vertically upward force on the man to both keep him at rest relative to a point in the gravitational field of Earth and to move that point away from Earth at 3.3 m/s2. The reaction force, 𝐹, has a magnitude given by 𝐹=264(784)=1048.N

This result can also be obtained using the formula 𝐹=𝑚(𝑔𝑎)𝐹=80(9.8(3.3))=80(13.1)=1048.N

Let us consider how the reaction force would be affected by constant acceleration in the direction of the weight of a man rather than in the opposite direction to that weight.

Example 3: Finding the Reaction Force between a Man and an Elevator Accelerating Downward

An elevator was accelerating vertically downward at 1.7 m/s2. Given that the acceleration due to gravity is 𝑔=9.8/ms, find the reaction force of the floor to a passenger of mass 103 kg.

Answer

The following figure shows the forces acting on the passenger.

If the passenger was falling freely, the only force acting on the passenger would be the force of their weight and they would accelerate vertically downward at 9.8 m/s2. As the passenger is accelerating vertically downward at only 1.7 m/s2, a vertically upward force must be acting. The vertically upward acting force is the reaction force from the elevator floor. The reaction force on the passenger must be less than the force of their weight, as the acceleration of the passenger is downward.

The force of the passenger’s weight, taking vertically downward motion as negative, is given by 𝐹=103×(9.8)=1009.4.N

The net force acting on the passenger to provide their acceleration is given by 𝐹=103×1.7=175.1.netN

The reaction force, 𝐹, is therefore given by 𝐹=175.1(1009.4)=834.3.N

This result can also be obtained using the formula 𝐹=𝑚(𝑔𝑎)𝐹=103(9.81.7)=80(8.1)=843.3.N

The reaction force on a person is what provides a person with the sensation of their body’s weight. The subjective experience of weight is the sensation of a resistance from the ground preventing motion through the ground, which is distinct from mere contact with the ground.

Suppose that a person stands on a platform that is in an aircraft flying high above the ground. While the person is standing on the platform, the aircraft floor below the platform is removed. The platform and the person both accelerate vertically downward at 9.8 m/s2 provided that two assumptions are made.

The first assumption is that resistive forces on the person and on the platform due to the motion of the air are negligible. The second assumption is that the gravitational forces exerted by the person and the platform on each other are negligible. If both assumptions are made, then the net force on the platform is equal to the weight of the platform and the net force on the person is equal to the weight of the person.

The feet of the person remain in contact with the platform, but for both platform and person, the only forces that act on them are their respective weights. Therefore, no reaction forces are exerted on either the person or the platform. The person would experience no sensation of the platform pushing their feet upward and so would not feel as if their body weighed anything and would experience apparent weightlessness.

The apparent weight of a body is, therefore, equal to the reaction force on the body. We have seen that the reaction force on a body is given by 𝑅=𝑚(𝑔𝑎), where 𝑎 is the acceleration of the surface that produces the reaction force. The following figures show how the apparent weight𝑊 depends on the mass of a body and the acceleration of the surface that the body rests on.

We can define the apparent weight of a body as follows.

Definition: The Apparent Weight of a Body

A body with a mass 𝑚 that is supported by a surface that has a uniform acceleration 𝑎, within a uniform gravitational field of strength 𝑔, exerts a reaction force, 𝑅, on the body given by 𝑅=𝑚(𝑔𝑎).

The reaction force on the body is known as the apparent weight of the body.

Let us look at an example involving the apparent weight of a body.

Example 4: Finding the Reading on a Spring Balance in an Accelerating Elevator

A body of mass 45 kg was placed in a spring balance fixed to the ceiling of an elevator. If the elevator was accelerating downward at 105 cm/s2, what would the apparent weight of the body be? Consider the acceleration due to gravity to be 9.8 m/s2 and round your answer to two decimal places.

Answer

A spring balance does not have a surface that supports a body, but rather suspends a body from a spring. The tension of the spring produced by the weight of the body acts equivalently to the reaction force that a surface supporting the body would produce. The only difference between the tension force produced by a spring and the reaction force produced by a surface is the point at which the forces act on the body; a surface reaction force acts at a point below the body and a spring tension acts at a point above the body. This difference is of no importance for the purposes of this question as the body that the force acts on has no stated features that distinguish it from a particle; it simply has a mass.

The following figure shows the forces acting on the body.

The magnitudes of the accelerations are not given in the same units, so one of the accelerations must be converted to have the unit of the other. As 𝑔 is a standard value and is given in metres per second squared, both of the accelerations will be expressed in metres per second squared, m/s2.

The question states that the elevator accelerates downward, so the apparent weight of the body as measured by the spring balance has a magnitude given by 𝑊=45×(9.81.05)=393.75.N

Let us look at another example involving apparent weight.

Example 5: Studying the Motion of an Elevator Using the Weight of a Body in Different Conditions

A man was standing on a set of scales in an elevator, recording the readings as the elevator moved. His first reading was when the elevator was accelerating upward at a rate of 7𝑎. He made another reading when the elevator was accelerating downward at a rate of 8𝑎. Given that the ratio between the two readings was 41, determine 𝑎𝑔 where 𝑔 is the acceleration due to gravity.

Answer

The variable 𝑎 used in representing the acceleration of the elevator should not be confused with the variable 𝑎 used in the expression for the apparent weight 𝑊=𝑚(𝑔𝑎).

The value of 𝑎 in this expression is the acceleration of a body, not of an arbitrary variable. The acceleration of a body can be written in the form acceleration=𝑛𝑎, but 𝑎 is not equal to the acceleration of the body unless 𝑛 equals 1.

The upward acceleration rate is stated to be 7𝑎. This acceleration is upward, hence in the opposite direction to 𝑔. The equation 𝑊=𝑚(𝑔𝑎) is in this case equivalent to 𝑊=𝑚(𝑔(7𝑎)).up

Hence, 𝑊=𝑚(𝑔+7𝑎).up

The downward acceleration rate is stated to be 8𝑎. The acceleration direction is the same as the direction of 𝑔, so 𝑊=𝑚(𝑔8𝑎).down

We are told that 𝑊𝑊=41.updown

We see, therefore, that 𝑚(𝑔+7𝑎)=4𝑚(𝑔8𝑎).

The factor of 𝑚 can be eliminated to give 𝑔+7𝑎=4(𝑔8𝑎).

The bracketed terms can be expanded to give 𝑔+7𝑎=4𝑔32𝑎, which can be rearranged to give 3𝑔=39𝑎,3𝑔=39𝑎,𝑔=13𝑎.

The ratios of the magnitudes of 𝑎 and 𝑔, therefore, is 113.

Key Points

  • Newton’s third law of motion states that any force that acts is one of a pair of equal-magnitude forces acting for the same time interval along the same line in opposite directions and on different bodies.
  • The weight of a body is due to the gravitational force on the body, while the apparent weight of a body is due to the reaction force on the body from a surface with which the body is in contact.
  • If a body and its support move uniformly, the apparent weight of the body is equal to the weight of the body.
  • If a body and its support have an acceleration in the opposite direction to gravitational acceleration, the apparent weight of the body is greater than the weight of the body.
  • If a body and its support have an acceleration in the same direction as gravitational acceleration, the apparent weight of the body is less than the weight of the body.
  • If a body and its support have an acceleration in the same direction as gravitational acceleration equal to or greater than the magnitude of the gravitational acceleration, the apparent weight of the body is zero.

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