5.6 Newton’s third law of motion

NEWTON’S THIRD LAW OF MOTION

Newton’s third law of motion says, to every action, there is always an equal and opposite reaction. That is If a body A exerts a force \(\vec{F}\) on another body B, then B exerts a force \(-\vec{F}\) on A. Thus, the force exerted by A on B and that by B on A are equal in magnitude but opposite in direction. According to the third law,
\(
\vec{F}_{A B}=-\vec{F}_{B A}
\)
(force on A by B ) = – ( force on B by A )

Let us note some important points about the third law, particularly in regard to the usage of the terms: action and reaction.

• The terms action and reaction in the third law mean nothing else but ‘force’. Forces always occur in pairs. Force on a body A by B is equal and opposite to the force on the body B by A.
• The terms action and reaction in the third law may give a wrong impression that action comes before reaction i.e. action is the cause and reaction the effect. There is no cause-effect relation implied in the third law. The force on A by B and the force on B by A act at the same instant. By the same reasoning, any one of them may be called action and the other reaction.
• Action and reaction forces act on different bodies, not on the same body. Consider a pair of bodies A and B. According to the third law,
  \(
  \vec{F}_{A B}=-\vec{F}_{B A}
  \)
  (force on A by B ) = – ( force on B by A )

Working with the Tension in a String

The idea of tension was qualitatively introduced in the previous chapter. Suppose a block of mass \(M\) is hanging through a string from the ceiling (figure below).

Consider a cross-section of the string at \(A\). The cross-section divides the string in two parts, the lower part, and the upper part. The two parts are in physical contact at the cross-section at \(A\). The lower part of the string will exert an electromagnetic force on the upper part and the upper part will exert an electromagnetic force on the lower part. According to the third law, these two forces will have equal magnitude. The lower part pulls down the upper part with a force \(T\) and the upper part pulls up the lower part with equal force \(T\). The common magnitude of the forces exerted by the two parts of the string on each other is called the tension in the string at \(A\). What is the tension in the string at the lower end? The block and the string are in contact at this end and exert electromagnetic forces on each other. The common magnitude of these forces is the tension in the string at the lower end. What is the tension in the string at the upper end? At this end, the string and the ceiling meet. The string pulls the ceiling down and the ceiling pulls the string up. The common magnitude of these forces is the tension in the string at the upper end.

Example 5.7: The mass of the part of the string below A in the figure below is \(m\). Find the tension of the string at the lower end and at \(A\).

Answer: To get the tension at the lower end we need the force exerted by the string on the block.
Take the block as the system. The forces on it are
(a) pull of the string, T, upward,
(b) pull of the earth, Mg, downward,
The free-body diagram for the block is shown in the figure below(a). As the acceleration of the block is zero, these forces should add to zero. Hence the tension at the lower end is \(T=M g\).

To get the tension \(T\) at \(A\) we need the force exerted by the upper part of the string on the lower part of the string. For this, we may write the equation of motion for the lower part of the string. So take the string below \(A\) as the system. The forces acting on this part are
(a) \(T^{\prime}\), upward, by the upper part of the string
(b) \(m g\), downward, by the earth
(c) \(T\), downward, by the block.
Note that in (c) we have written \(T\) for the force by the block on the string. We have already used the symbol \(T\) for the force by the string on the block. We have used Newton’s third law here. The force exerted by the block on the string is equal in magnitude to the force exerted by the string on the block.

The free-body diagram for this part is shown in Figure(b). As the system under consideration (the lower part of the string) is in equilibrium, Newton’s first law gives
\(
T^{\prime}=T+m g
\)
But \(T=M g\) hence, \(T^{\prime}=(M+m) g\).

Example 5.8: The block shown in the figure below has a mass M and descends with an acceleration a. The mass of the string below the point A is m. Find the tension of the string at the point A and at the lower end.

Answer: Consider “the block + the part of the string below \(A^n\) as the system. Let the tension at \(A\) be \(T\). The forces acting on this system are
(a) \((M+m) g\), downward, by the earth
(b) \(T\), upward, by the upper part of the string.
The first is gravitational and the second is electromagnetic. We do not have to write the force by the string on the block. This electromagnetic force is by one part of the system on the other part. Only the forces acting on the system by the objects other than the system are to be included.
The system is descending with an acceleration \(a\). Taking the downward direction as the \(X\)-axis, the total force along the \(X\)-axis is \((M+m) g-T\). Using Newton’s law
\(
\begin{aligned}
(M+m) g-T & =(M+m) a . \\
T & =(M+m)(g-a) \dots(i)
\end{aligned}
\)
We have omitted the free-body diagram. This you can do if you can draw the free body diagram in your mind and write the equations correctly.
To get the tension \(T^{\prime}\) at the lower end we can put \(m=0\) in (i)

Effectively, we take the point \(A\) at the lower end. Thus, we get \(T^{\prime}=M(g-a)\).

Example 5.9: A pendulum is hanging from the ceiling of a car having an acceleration \(a_0\) with respect to the road. Find the angle made by the string with the vertical.

Answer: The situation is shown in Figure (a). Suppose the mass of the bob is \(m\) and the string makes an angle \(\theta\) with the vertical. We shall work from the car frame. This frame is noninertial as it has an acceleration \(\overrightarrow{a_0}\) with respect to an inertial frame (the road). Hence, if we use Newton’s second law we shall have to include a pseudo force.

Take the bob as the system.
The forces are :
(a) \(T\) along the string, by the string
(b) \(m g\) downward, by the earth
(c) \(m a_0\) towards left (pseudo force).
The free-body diagram is shown in figure (b). As the bob is at rest (remember we are discussing the motion with respect to the car) the force in (a), (b) and (c) should add to zero. Take \(X\)-axis along the forward horizontal direction and \(Y\)-axis along the upward vertical direction. The components of the forces along the \(X\)-axis give
\(
T \sin \theta-m a_0=0 \quad \text { or, } \quad T \sin \theta=m a_0 \dots(i)
\)
and the components along the \(Y\)-axis give
\(
T \cos \theta-m g=0 \quad \text { or, } T \cos \theta=m g \dots(ii)
\)
Dividing (i) by (ii) \(\tan \theta=a_0 / g\).
Thus, the string makes an angle \(\tan ^{-1}\left(a_0 / g\right)\) with the vertical.

Example 5.10: Two identical billiard balls strike a rigid wall with the same speed but at different angles, and get reflected without any change in speed, as shown in Fig. 5.6. What is (i) the direction of the force on the wall due to each ball? (ii) the ratio of the magnitudes of impulses imparted to the balls by the wall?

An instinctive answer to (i) might be that the force on the wall in case (a) is normal to the wall, while that in case (b) is inclined at \(30^{\circ}\) to the normal. This answer is wrong. The force on the wall is normal to the wall in both cases.
How to find the force on the wall? The trick is to consider the force (or impulse) on the ball due to the wall using the second law, and then use the third law to answer (i). Let \(u\) be the speed of each ball before and after collision with the wall, and \(\mathrm{m}\) the mass of each ball. Choose the \(x\) and y axes as shown in the figure, and consider the change in momentum of the ball in each case:
Case (a)
\(
\begin{array}{ll}
\left(p_x\right)_{\text {intial }}=m u & \left(p_y\right)_{\text {intital }}=0 \\
\left(p_x\right)_{\text {final }}=-m u & \left(p_y\right)_{\text {final }}=0
\end{array}
\)
Impulse is the change in the momentum vector. Therefore,
\(
\begin{aligned}
& x \text {-component of impulse }=-2 m u \\
& y \text {-component of impulse }=0
\end{aligned}
\)
Impulse and force are in the same direction. Clearly, from above, the force on the ball due to the wall is normal to the wall, along the negative x-direction. Using Newton’s third law of motion, the force on the wall due to the ball is normal to the wall along the positive x-direction. The magnitude of force cannot be ascertained since the small time taken for the collision has not been specified in the problem.
Case (b)
\(
\begin{aligned}
& \left(p_x\right)_{\text {initial }}=m u \cos 30^{\circ},\left(p_y\right)_{\text {initial }}=-m u \sin 30^{\circ} \\
& \left(p_x\right)_{\text {final }}=-m u \cos 30^{\circ},\left(p_y\right)_{\text {final }}=-m u \sin 30^{\circ}
\end{aligned}
\)
Note, while \(p_x\) changes sign after collision, \(p_y\) does not. Therefore,
\(
\begin{aligned}
& x \text {-component of impulse }=-2 m u \cos 30^{\circ} \\
& y \text {-component of impulse }=0
\end{aligned}
\)
The direction of impulse (and force) is the same as in (a) and is normal to the wall along the negative \(x\) direction. As before, using Newton’s third law, the force on the wall due to the ball is normal to the wall along the positive \(x\) direction.
The ratio of the magnitudes of the impulses imparted to the balls in (a) and (b) is
\(
2 m u /\left(2 m u \cos 30^{\circ}\right)=\frac{2}{\sqrt{3}} \approx 1.2
\)