5.9 Common forces in mechanics

In mechanics, we encounter several kinds of forces. The gravitational force is, of course, pervasive. Every object on the earth experiences the force of gravity due to the earth. Gravity also governs the motion of celestial bodies. The gravitational force can act at a distance without the need for any intervening medium.

FRICTION (AS THE COMPONENT OF CONTACT FORCE)

When two bodies are kept in contact, electromagnetic forces act between the charged particles at the surfaces of the bodies. As a result, each body exerts a contact force on the other. The magnitudes of the contact forces acting on the two bodies are equal but their directions are opposite and hence the contact forces obey Newton’s third law. The perpendicular component of the contact force is called the normal contact force or normal force and the parallel component is called friction as shown in Figure 5.4.

Figure 5.4

Example 5.12:A body of mass \(400 \mathrm{~g}\)slides on a rough horizontal surface. If the frictional force is \(3.0 \mathrm{~N}\), find (a) the angle made by the contact force on the body with the vertical and (b) the magnitude of the contact force. Take \(g=10 \mathrm{~m} / \mathrm{s}^{2}\).

Solution: Let the contact force on the block by the surface be \({F}\) which makes an angle \(\theta\) with the vertical (figure 5.5).

Figure 5.5

The component of \({F}\) perpendicular to the contact surface is the normal force \(N\) and the component of \({F}\) parallel to the surface is the friction \(f\). As the surface is horizontal, \({N}\) is vertically upward. For vertical equilibrium,
\(
{N}=M g=(0.400 \mathrm{~kg})\left(10 \mathrm{~m} / \mathrm{s}^{2}\right)=4 \cdot 0 \mathrm{~N} \text {. }
\)
The frictional force is \(f=3.0 \mathrm{~N}\).
(a) \(\quad \tan \theta=\frac{f}{N}=\frac{3}{4}\)
or, \(\quad \theta=\tan ^{-1}(3 / 4)=37^{\circ}\).
(b) The magnitude of the contact force is
\(
\begin{aligned}
F &=\sqrt{N^{2}+f^{2}} \\
&=\sqrt{(4 \cdot 0 \mathrm{~N})^{2}+(3 \cdot 0 \mathrm{~N})^{2}}=5 \cdot 0 \mathrm{~N}
\end{aligned}
\)

STATIC FRICTION

Frictional forces can also act between two bodies which are in contact but are not sliding with respect to each other. The friction in such cases is called static friction. Consider a body shown in Figure 5.6 and assume that a force {latex]F[/latex] is applied horizontally to the body. We know from experience that a small applied force may not be enough to move the body. Clearly, the body remains at rest because some other force comes into play in the horizontal direction and opposes the applied force \(F\), resulting in zero net force on the body. This force \(f_{s}\) parallel to the surface of the body is known as frictional force, or simply friction (Figure 5.10a). The subscript stands for static friction to distinguish it from kinetic friction \(f_k\) that we consider later (Fig. 5.10(b)). Note that static friction does not exist by itself.

Note that static friction does not exist by itself. When there is no applied force, there is no static friction. It comes into play the moment there is an applied force. As the applied force \(F\) increases, \(f_{s}\) also increases, remaining equal and opposite to the applied force (up to a certain limit), keeping the body at rest. Hence, it is called static friction. Static friction opposes impending motion. The term impending motion means motion that would take place (but does not actually take place) under the applied force if friction were absent.

We know from experience that as the applied force exceeds a certain limit, the body begins to move. It is found experimentally that the limiting value of static friction \(\left(f_{s}\right)_{\max }\) is independent of the area of contact and varies with the normal force \((N)\) approximately as:
\(
\left(f_{s}\right)_{\max }=\mu_{\mathrm{s}} N \dots(5.13)
\)

where \(\mu_{s}\) is a constant of proportionality depending only on the nature of the surfaces in contact. The constant \(\mu_{\mathrm{s}}\) is called the coefficient of static friction. The law of static friction may thus be written as
\(
f_{\mathrm{s}} \leq \mu_{\mathrm{s}} N \dots(5.14)
\)
If the applied force \(F\) exceeds \(\left(f_{s}\right)_{\max }\) the body begins to slide on the surface. It is found experimentally that when relative motion has started, the frictional force decreases from the static maximum value \(\left(f_{s}\right)_{\max }\). 

The frictional force that opposes relative motion between surfaces in contact is called kinetic or sliding friction and is denoted by \(\mathbf{f}_{\mathrm{k}}\). Kinetic friction, like static friction, is found to be independent of the area of contact. Further, it is nearly independent of the velocity. It satisfies a law similar to that for static friction:
\(
\mathbf{f}_{\mathrm{k}}=\mu_k \mathbf{N} \dots(5.15)
\)
where \(\mu_{k^{\prime}}\) the coefficient of kinetic friction, depends only on the surfaces in contact. As mentioned above, experiments show that \(\mu_k\) is less than \(\mu_s\). When the relative motion has begun, the acceleration of the body according to the second law is \(\left(F-f_k\right) / m\). For a body moving with constant velocity, \(F=f_k\). If the applied force on the body is removed, its acceleration is \(-f_k / m\) and it eventually comes to a stop.

Example 5.13:A boy 30 kg sitting on his horsewhips it as shown in the figure below. The horse speeds up at an average acceleration of \(2.0 \mathrm{~m} / \mathrm{s}^{2}\). (a) If the boy does not slide back, what is the force of friction exerted by the horse on the boy? (b) If the boy slides back during the acceleration, what can be said about the coefficient of static friction between the horse and the boy? Take \(g=10 \mathrm{~m} / \mathrm{s}^{2}\).

Solution: (a) The forces acting on the boy are
(i) the weight \(m g\).
(ii) the normal contact force \(N\) and
(iii) the static friction \(f_{s}\).

As the boy does not slide back, its acceleration \(a\) is equal to the acceleration of the horse. As friction is the only horizontal force, it must act along the acceleration and its magnitude is given by Newton’s second law
\(
f_{s}=m a=(30 \mathrm{~kg})\left(2 \cdot 0 \mathrm{~m} / \mathrm{s}^{2}\right)=60 \mathrm{~N} .
\)
(b) If the boy slides back, the horse could not exert a friction of \(60 \mathrm{~N}\) on the boy. The maximum force of static friction that the horse may exert on the boy is
\(
\begin{aligned}
f_{s} &=\mu_{s}=N=\mu_{s} m g \\
&=\mu_{s}(30 \mathrm{~kg})\left(10 \mathrm{~m} / \mathrm{s}^{2}\right)=\mu_{s} 300 \mathrm{~N}
\end{aligned}
\)
where \(\mu_{s}\) is the coefficient of static friction. Thus,
\(
\begin{aligned}
&\mu_{s}(300 \mathrm{~N})<60 \mathrm{~N} \\
&\mu_{s}<\frac{60}{300}=0.20 .
\end{aligned}
\)

KINETIC FRICTION

When two bodies in contact move with respect to each other, rubbing the surfaces in contact, the friction between them is called kinetic friction. The directions of the frictional forces are such that the relative slipping is opposed by the friction.

Figure 5.7
Suppose a body A placed in contact with B is moved with respect to it as shown in figure (5.7). The force of friction acting on A due to B will be opposite to the velocity of A with respect to B. In figure (5.7) this force is shown towards the left. The force of friction on B due to A is opposite to the velocity of B with respect to A. In figure (5.7) this force is shown towards the right. The force of kinetic friction opposes the relative motion. 

Kinetic friction is denoted by \({f}_{k}\). Kinetic friction, like static friction, is found to be independent of the area of contact. Further, it is nearly independent of the velocity. It satisfies a law similar to that for static friction:
\(
{f}_{k}=\mu_{k} {N}
\)
where \(\mu_{k}\) the coefficient of kinetic friction, depends only on the surfaces in contact. As mentioned above, experiments show that \(\mu_{k}\) is less than \(\mu_{s}\). When the relative motion has begun, the acceleration of the body according to the second law is \(\left(F-f_{k}\right) / m\). For a body moving with constant velocity, \(F=f_{k}\). If the applied force on the body is removed, its acceleration is \(-f_{k} / m\) and it eventually comes to a stop.

Example 5.14: A heavy box of mass \(20 \mathrm{~kg}\) is pulled on a horizontal surface by applying a horizontal force. If the coefficient of kinetic friction between the box and the horizontal surface is 0.25, find the force of friction exerted by the horizontal surface on the box.

Solution: The situation is shown in Figure 5.8. In the vertical direction there is no acceleration, so
\(
N=m g \text {. }
\)

Figure 5.8
As the box slides on the horizontal surface, the surface exerts kinetic friction on the box. The magnitude of the kinetic friction is
\(
\begin{aligned}
f_{k} &=\mu_{k}=N=\mu_{k} m g \\
&=0.25 \times(20 \mathrm{~kg}) \times\left(9.8 \mathrm{~m} / \mathrm{s}^{2}\right)=49 \mathrm{~N} .
\end{aligned}
\)
This force acts in the direction opposite to the pull.

Example 5.15:Determine the maximum acceleration of the train in which a box lying on its floor will remain stationary, given that the coefficient of static friction between the box and the train’s floor is 0.15.

Solution: Since the acceleration of the box is due to static friction,
\(
\begin{aligned}
m a &=f_{s} \leq \mu_{s} N=\mu_{s} m g \\
\text { i.e. } \quad a & \leq \mu_{s} g \\
\therefore a_{\max } &=\mu_{s} g=0.15 \times 10 \mathrm{~m} \mathrm{~s}^{-2} \\
&=1.5 \mathrm{~m} \mathrm{~s}^{-2}
\end{aligned}
\)

Example 5.16: In Figure 5.9 below a mass of \(4 \mathrm{~kg}\) rests on a horizontal plane. The plane is gradually inclined until at an angle \(\theta=15^{\circ}\) with the horizontal, the mass just begins to slide. What is the coefficient of static friction between the block and the surface?

Figure 5.9

Solution: The forces acting on a block of mass \(m\) at rest on an inclined plane are (i) the weight mg acting vertically downwards (ii) the normal force \(N\) of the plane on the block, and (iii) the static frictional force \(f_{s}\) opposing the impending motion. In equilibrium, the resultant of these forces must be zero. Resolving the weight \(\mathrm{mg}\) along the two directions shown, we have
\(m g \sin \theta=f_{s}, \quad m g \cos \theta=N\)
As \(\theta\) increases, the self-adjusting frictional force \(f_{s}\) increases until at \(\theta=\theta_{\max }, f_{s}\) achieves its maximum value, \(\left(f_{s}\right)_{\max }=\mu_{s} N\).
Therefore,
\(
\tan \theta_{\max }=\mu_{s} \text { or } \theta_{\max }=\tan ^{-1} \mu_{s}
\)
When \(\theta\) becomes just a little more than \(\theta_{m}\) there is a small net force on the block and it begins to slide. Note that \(\theta_{\max }\) depends only on $\mu_{s}$ and is independent of the mass of the block.
For \(\theta_{\max }=15^{\circ}\),
\(
\begin{aligned}
\mu_{\mathrm{s}} &=\tan 15^{\circ} \\
&=0.27
\end{aligned}
\)

Example 5.17: What is the acceleration of the block and trolley system shown in Figure 5.10, if the coefficient of kinetic friction between the trolley and the surface is 0.04? What is the tension in the string? (Take \(g=\) \(10 \mathrm{~m} \mathrm{~s} \mathrm{~s}^{-2}\) ). Neglect the mass of the string.

Figure 5.10

Solution: As the string is inextensible, and the pully is smooth, the \(3 \mathrm{~kg}\) block and the \(20 \mathrm{~kg}\) trolley both have the same magnitude of acceleration. Applying the second law to the motion of the block (Figure. \(5.10(\mathrm{~b})\) ),
\(
30-T=3 a
\)
Apply the second law to the motion of the trolley (Figure. \(5.10(\mathrm{c}))\),
\(
\begin{aligned}
T-f_{\mathrm{k}}=& 20 a . \\
\text { Now } \quad f_{k} &=\mu_{k} N, \\
\text { Here } \quad \mu_{k} &=0.04, \\
N &=20 \times 10 \\
&=200 \mathrm{~N} .
\end{aligned}
\)
Thus the equation for the motion of the trolley is \(T-0.04 \times 200=20 a\) Or \(T-8=20 a .\)

These equations give \(a=\frac{22}{23} \mathrm{~m} \mathrm{~s}^{-2}=0.96 \mathrm{~m} \mathrm{~s}^{-2}\) and \(T=27.1 \mathrm{~N}\).