9.0 Friction

  Laws of Motion

• External friction: External friction is also known as kinetic friction. When there is a relative motion between two different bodies. As shown in the figure, $f_1$ is the external friction.

• Internal friction: Internal friction is also known as viscosity. It occurs between the layers of the fluid in motion.
• Sliding friction: Force which acts when one body slides over another body. As shown in the figure, $f_1$ is the sliding friction.
• Rolling friction: Force which acts when one body rolls over another body. As shown in the figure, $f_2$ is the rolling friction

• Static, Limiting and Kinetic (or dynamic) friction: As shown in figure (a), suppose a body $A$ is resting on the ground. Then a gradually increasing force $F$ is applied on body $A$, starting from zero. We will observe that till the magnitude of $F$ reaches a certain value so that $A$ does not start any motion. This is due to the friction known as static friction.
• Then comes a point when the body $A$ is just on the verge of starting its motion. Frictional force acting at this stage is known as limiting friction. Thereafter, if the force $F$ is further increased, the motion of the block $A$ starts. Friction during motion is known as kinetic or dynamic friction.

Note:

• If a body is at rest and no pulling force is acting on it, the force of friction on it is zero.
• If a force is applied to pull the body and it does not move, the friction acts, which is equal in magnitude and opposite in direction to the applied force i.e., friction is a self-adjusting force. Further, as the body is at rest the friction is called as static friction.
• If the applied force is increased the force of static friction also increases. If the applied force exceeds a certain (maximum) value, the body starts moving. This maximum force of static friction up to which body does not move is called limiting friction. Thus, static friction is a self-adjusting force with an upper limit called limiting friction.
• This limiting force of friction $\left( {{f_L}} \right)$ is found experimentally to depend on normal reaction $(N)$.
• Hence, $$\begin{equation} \begin{aligned} {f_L} \propto N \\ {f_L} = {\mu _S}N \\\end{aligned} \end{equation} $$

Here, ${\mu _S}$ is a dimensionless constant and is known as the coefficient of static friction, which depends on nature of surfaces in contact.0

• If the applied force is further increased, the friction opposing the motion is called kinetic or sliding friction. Experimentally, it is well established that kinetic friction is lesser than limiting friction and is given by, $$\begin{equation} \begin{aligned} {f_k} \propto N \\ {f_k} = {\mu _k}N \\\end{aligned} \end{equation} $$
• where ${\mu _k}$ is a dimensionless constant and is known as the coefficient of kinetic friction. It is less than ${\mu _S}$.

Question 16. Suppose a block of mass $1\ kg$ is placed over a rough surface and a horizontal force $F$ is applied on the block as shown in figure.

S. No.	Force $$(F)$$	Friction $$(f)$$	$${F_{net}} = F - f$$	Acceleration $$\left( {a = \frac{{{F_{net}}}}{m}} \right)$$	Diagram	Comments
1.	$$0$$	$$0$$	$$0$$	$$0$$		Block is at rest as no external force is applied on it.
2.	$$2N$$	$$2N$$	$$0$$	$$0$$		Static friction compensates the external force $(F=2N)$
3.	$$4N$$	$$4N$$	$$0$$	$$0$$		Static friction compensates the external force $(F=4N)$
4.	$$5N$$	$$5N$$	$$0$$	$$0$$		Static friction compensates the external force $(F=5N)$. Static friction has reached its maximum value and beyond this, the motion will start.
5.	$$6N$$	$$4N$$	$$2N$$	$$2m/{s^2}$$		As the motion has started, so only kinetic friction will act for any value of external force.
6.	$$8N$$	$$4N$$	$$4N$$	$$4m/{s^2}$$

Graphically, the above situation can be understood as,

Static friction is a straight line passing through the origin.

Therefore it can be written as, $$y=mx$$

At point $A$, $$F = f = 5N$$

Therefore, slope $(m)=1$

Question 17. A block is placed on an inclined plane as shown in the figure. What must be the frictional force between block and the incline, so that the block cannot slide along the incline when the incline is accelerating to the right at $3m/{s^2}$.

Take $\left( {\sin 37^\circ = \frac{3}{5}} \right)$ & $\left( {g = 10m/{s^2}} \right)$

Solution: From FBD we can write, $$\begin{equation} \begin{aligned} N = mg\cos \theta + 3m\sin \theta \quad ...(i) \\ mg\sin \theta = 3m\cos \theta + f\quad ...(ii) \\\end{aligned} \end{equation} $$

But for calculating friction we need only equation $(ii)$,

$$\begin{equation} \begin{aligned} f = m\left( {10} \right)\left( {\frac{3}{5}} \right) - 3m\left( {\frac{4}{5}} \right) \\ f = \frac{{18m}}{5} \\\end{aligned} \end{equation} $$

Question 18. Find the tensions in the two strings when the coefficient of friction between the table and the block is $\mu = 0.3$ as shown in the figure.

Solution: From FBD we can write, $$\begin{equation} \begin{aligned} 3g - {T_1} = 3a\quad ...(i) \\ {T_2} - 1g = 1a\quad ...(ii) \\ {T_1} - \left( {{T_2} + f} \right) = 1a\quad ...(iii) \\ N = 1g\quad ...(iv) \\ f = \mu N\quad ...(v) \\\end{aligned} \end{equation} $$

From equation $(i)$, $(ii)$, $(iii)$, $(iv)$ & $(v)$ we get,

$$\begin{equation} \begin{aligned} 2g - \left( {0.3} \right)\left( 1 \right)g = 5a \\ a = \frac{{1.7g}}{5} \\ a = 3.4m/{s^2}\quad ...(vi) \\\end{aligned} \end{equation} $$

From equation $(i)$ and $(vi)$ we get, $${T_1} = 19.8N$$

From equation $(ii)$ and $(vii)$ we get, $${T_2} = 13.4N$$