Physics > Laws of Motion > 9.0 Friction
Laws of Motion
1.0 Newton's first law of motion
2.0 Newton's second law of motion
3.0 Newton's third law of motion
4.0 Force
5.0 Normal Force
6.0 Gravitional Force
7.0 Tension Force
8.0 Pseudo force
9.0 Friction
10.0 Centripetal force
9.2 Angle of repose $\left( \alpha \right)$
Suppose a block of mass $m$ is placed on an inclined plane whose inclination $\theta $ can be increased or decreased.
Let, $\mu $ be the coefficient of friction between block and plane.
At a general angle $\theta $,
Normal reaction, $N = mg\cos \theta $
Limiting friction, ${f_L} = \mu N = \mu mg\cos \theta $
Driving force (or pulling force), $F = mg\sin \theta \quad \left( {{\text{down the plane}}} \right)$
From these three equations we see that when $\theta $ is increased from $0^\circ $ to $$90^\circ $$, normal reaction $N$ is decreased, so the limiting friction $f_L$ is also decreased while the driving force $F$ is increased.
There is a critical angle called as angle of repose $\left( \alpha \right)$ at which these two forces are equal.
Now, if $\theta $ is further increased, then the driving force $F$ becomes more than the limiting friction $f_L$ and the block starts sliding.
Thus, $$\begin{equation} \begin{aligned} {f_L} = F\quad {\text{at}}\quad \theta = \alpha \\ \mu mg\cos \alpha = mg\sin \alpha \\ \tan \alpha = \mu \\ \alpha = {\tan ^{ - 1}}\mu \\\end{aligned} \end{equation} $$
We see that the angle of friction $\left( \lambda \right)$ is numerically equal to the angle of repose $\left( \alpha \right)$.
$$\lambda = \alpha $$
Note:
- If $\theta < \alpha $ then $F < {f_L}$, then the block is stationary
- If $\theta = \alpha $ then $F = {f_L}$, then the block is at the verge of sliding
- If $\theta > \alpha $ then $F > {f_L}$, then the block slides down with acceleration $a$ which is given by,
$$\begin{equation} \begin{aligned} a = \left( {\frac{{F - {f_L}}}{m}} \right) \\ a = g\left( {\sin \theta - \mu \cos \theta } \right) \\\end{aligned} \end{equation} $$
Let us see graphically how $N$, $F_L$ and $F$ varies with $\theta $.
As we know, $$\begin{equation} \begin{aligned} N = mg\cos \theta \\ \therefore N \propto \cos \theta \\\end{aligned} \end{equation} $$
Similarly, $$\begin{equation} \begin{aligned} {f_L} = \mu mg\cos \theta \\ \therefore {f_L} \propto \cos \theta \\\end{aligned} \end{equation} $$
and $$\begin{equation} \begin{aligned} F = mg\sin \theta \\ \therefore F \propto \sin \theta \\\end{aligned} \end{equation} $$
Note: If $\mu < 1$, then ${f_L} < N$.
Question 19. A block of mass $m$ rests on rough contact with a plane inclined at $30^\circ $ to the horizontal and is just at the verge of sliding. Find the coefficient of friction between the plane and the block.
Solution: Since we know that when the block is at the verge of sliding on an inclined plane, then the inclination of the plane $\left( \theta \right)$ will be equal to the angle of repose $\left( \alpha \right)$.
Mathematically, $$\begin{equation} \begin{aligned} \theta = \alpha \\ \theta = {\tan ^{ - 1}}\mu \\ u = \tan \theta \\ \mu = \tan 30^\circ \\ \mu = \frac{1}{{\sqrt 3 }} \\\end{aligned} \end{equation} $$
Question 20. A block slides down an inclined surface of inclination $30^\circ $ with the horizontal. Starting from rest it covers $9m$ in the first 3 seconds. Find the coefficient of kinetic fiction between the two.
Solution: As the block covers $9m$ in first 3 seconds, starting from rest. So,
$$S = ut + \frac{1}{2}a{t^2}$$
$u=0$ (Starting from rest), $S=9m$, $t=3s$, $$\begin{equation} \begin{aligned} 9 = \frac{1}{2}a{\left( 3 \right)^2} \\ \therefore a = 2m/{s^2}\quad ...(i) \\\end{aligned} \end{equation} $$
The net force along the inclined surface provides the above acceleration.
$$\begin{equation} \begin{aligned} {F_{net}} = \left( {mg\sin \theta - \mu mg\cos \theta } \right) \\ ma = \left( {mg\sin \theta - \mu mg\cos \theta } \right) \\ a = g\left( {\sin \theta - \mu \cos \theta } \right)\quad ...(ii) \\\end{aligned} \end{equation} $$
From equation $(i)$ and $(ii)$ we get, $$\mu = \frac{{\sqrt 3 }}{5}$$
Question 21. The friction coefficient between the board and the floor shown in figure is $\mu $. Find the maximum force that the man can exert on the rope so that the board does not slip on the floor.
Solution: From the FBD we write equations as follows,
$$\begin{equation} \begin{aligned} T + {N_1} = Mg\quad ...(i) \\ {N_2} = mg\quad ...(ii) \\ T = \mu {N_2}\quad ...(iii) \\\end{aligned} \end{equation} $$
From equation $(i)$, $(ii)$ and $(iii)$ we get, $$\begin{equation} \begin{aligned} T = \mu \left[ {(M + m)g - T} \right] \\ T = \frac{{\mu (M + m)g}}{{1 + \mu }} \\\end{aligned} \end{equation} $$
Question 22. Find the acceleration of the block of mass $M$ in the situation as shown in the figure. The coefficient of friction between the two blocks is ${\mu _1}$ and that between the bigger block and the ground is ${\mu _2}$.
Solution: From the FBD we write equations as follows,
$$\begin{equation} \begin{aligned} 2T - {N_1} - {\mu _2}{N_2} = M{a_2}\quad ...(i) \\ {N_2} = T + Mg\quad ...(ii) \\ {N_1} = m{a_2}\quad ...(iii) \\ mg - T - {\mu _1}{N_1} = m{a_1}\quad ...(iv) \\\end{aligned} \end{equation} $$
The constrained equation by work done method is, $$\begin{equation} \begin{aligned} {a_2}(2T) = {a_1}(T) \\ 2{a_2} = {a_1}\quad ...(v) \\\end{aligned} \end{equation} $$
From equation $(iii)$, $(iv)$ and $(v)$ we get, $$T = mg - m{a_2}(2 + {u_1})\quad ...(vi)$$
From equation $(i)$, $(ii)$ and $(iii)$ we get, $$\left( {2 - {\mu _2}} \right)T - {\mu _2}Mg = (M + m){a_2}\quad ...(vii)$$
From equation$(vi)$ and $(vii)$ we get, $${a_2} = \frac{{\left[ {2m - {\mu _2}\left( {M + m} \right)} \right]g}}{{\left[ {M + m\left( {5 + 2\left( {{\mu _1} - {\mu _2}} \right) - {\mu _1}{\mu _2}} \right)} \right]}}$$
