Friction MCQ Questions & Answers in Basic Physics | Physics

Net work done by the block in going from top to bottom of the inclined plane, must be equal to the work done by frictional force.

The block may be stationary, when
$$\eqalign{ & mg\sin \theta \cdot L = \mu \,mg\cos \theta \frac{L}{2} \cr & {\text{or}}\,\mu = \frac{{mg\,\sin \theta \cdot L}}{{mg\,\cos \theta \cdot \frac{L}{2}}} = 2\frac{{\sin \theta }}{{\cos \theta }} = 2\,\tan \theta \cr & \mu = 2\tan \theta \cr} $$

For the upward motion of the body
$$\eqalign{ & mg\sin \theta + {f_1} = {F_1} \cr & {\text{or,}}\,{F_1} = mg\sin \theta + \mu mg\cos \theta \cr} $$
For the downward motion of the body,
$$\eqalign{ & mg\sin \theta - {f_2} = {F_2} \cr & {\text{or}}\,{F_2} = mg\sin \theta - \mu mg\cos \theta \cr & \therefore \frac{{{F_1}}}{{{F_2}}} = \frac{{\sin \theta + \mu \cos \theta }}{{\sin \theta - \mu \cos \theta }} \cr & \Rightarrow \frac{{\tan \theta + \mu }}{{\tan \theta - \mu }} = \frac{{2\mu + \mu }}{{2\mu - \mu }} = \frac{{3\mu }}{\mu } = 3 \cr} $$

For no slipping between $$m$$ and $$M,$$
$$\eqalign{ & F \leqslant \left( {M + m} \right)\frac{g}{3} \cr & F \leqslant 40\,N \cr} $$
For no toppling of $$m$$ block
$$\eqalign{ & F \leqslant \left( {M + m} \right)\frac{g}{4} \cr & F \leqslant 30\,N \cr & \therefore {F_{\min }} = 30\,N \cr} $$

KEY CONCEPT:
For the maximum possible value of $$\alpha ,\,mg\,\sin \alpha $$   will also be maximum and equal to the frictional force.
In this case $$f$$ is the limiting friction. The two forces acting on the insect are $$mg$$  and $$N.$$ Let us resolve $$mg$$  into two components. $$mg\,\cos \alpha $$ balances $$N.\,mg\,\sin \alpha $$   is balanced by the frictional force.
$$\eqalign{ & \therefore N = mg\cos \alpha \cr & f = mg\sin \alpha \cr} $$
But $$f = \mu N = \mu \,mg\cos \alpha $$
$$\therefore \mu \,mg\cos \alpha = mg\sin \alpha \Rightarrow \cot \alpha = \frac{1}{\mu } \Rightarrow \cot \alpha = 3$$

Angle of repose or angle of sliding is defined as the minimum angle of inclination of a plane with the horizontal, such that a body placed on the plane just begins to slide down.

$$AB$$  is an inclined plane such that a body placed on it just begins to slide down
$$\angle BAC = \theta = {\text{angle of repose}}$$
In equilibrium, $$f = mg\,\sin \theta $$
$$\eqalign{ & {\text{and}}\,R = mg\cos \theta \cr & \therefore \frac{f}{R} = \frac{{mg\sin \theta }}{{mg\cos \theta }} = \tan \theta \cr & {\text{i}}{\text{.e}}{\text{.}}\,\,\mu = \tan \theta \cr} $$
NOTE
Coefficient of kinetic friction between any two surfaces in contact is equal to the tangent of the angle of inclination between them.

Let the mass of the block $$B$$ be $$M.$$

In equilibrium, $$T - Mg = 0 \Rightarrow T = Mg\,.......\left( {\text{i}} \right)$$
If blocks do not move, then $$T = {f_s}$$
where, $${f_s}$$ = frictional force $$ = {\mu _s}R = {\mu _s}mg$$
$$\therefore T = {\mu _s}mg\,......\left( {{\text{ii}}} \right)$$
Thus, from Eqs. (i) and (ii), we have
$$\eqalign{ & Mg = {\mu _s}mg \cr & {\text{or}}\,M = {\mu _s}m \cr & {\text{Given,}}\,{\mu _s} = 0.2,\,m = 2\,kg \cr & \therefore M = 0.2 \times 2 = 0.4\,kg \cr} $$

As $$m$$ would slip in vertically downward direction, then
$$\eqalign{ & mg = \mu N \cr & \Rightarrow N = \frac{{mg}}{\mu } = \frac{{10.0}}{{0.5}} = 200\,Newton \cr} $$
Same normal force would accelerated $$M,$$
thus $${a_M} = \frac{{200}}{{50}} = 4\,m/{s^2}$$
Taking $$m+ M$$   as system
$$F = \left( {m + M} \right)4 = 240\,N$$

The force acting on the block along the incline to shift the block downwards

$$ = mg\sin \theta = 2 \times 9.8\sin {30^ \circ } = 9.8N$$
The limiting frictional force
$${f_l} = {\mu _s}mg\cos \theta = 0.7 \times 2 \times 9.8 \times \frac{{\sqrt 3 }}{2} = 11.8N$$
Note : The frictional force is never greater than the force tending to produce relative motion. Therefore the frictional force is $$9 8N.$$

$$\eqalign{ & mg{\text{ }}\sin \theta {\text{ }} = {\text{ }}{f_s}\left( {{\text{for body to be at rest}}} \right) \cr & \Rightarrow m \times 10 \times \sin 30^\circ {\text{ }} = 10 \cr & \Rightarrow m \times 5 = 10 \Rightarrow m = 2.0kg \cr} $$

Block $$B$$ will come to rest, if force applied to it will vanish due to frictional force acting between block $$B$$ and surface, i.e. frictional force = force applied
$$\eqalign{ & {\text{i}}{\text{.e}}{\text{.}}\,\mu mg = ma \cr & {\text{or}}\,\mu mg = m\left( {\frac{v}{t}} \right) \cr & {\text{or}}\,t = \frac{v}{{\mu g}} \cr} $$