Friction MCQ Questions & Answers in Basic Physics | Physics

When a vehicle goes round a curved road, it requires some centripetal force. While rounding the curve, the wheels of the vehicle have a tendency to leave the curved path and regain the straight line path. Force of friction between the wheels and the road opposes this tendency of the wheels. This force (of friction) therefore, acts towards the centre of the circular track and provides the necessary centripetal force.
If $$v$$ is the velocity of the vehicle while rounding the curve, the centripetal force required $$ = \frac{{m{v^2}}}{r}$$
As, this force is provided only by the force of friction,
$$\eqalign{ & \frac{{m{v^2}}}{r} \leqslant \mu mg \cr & \Rightarrow {v^2} \leqslant \mu rg \cr & \Rightarrow v \leqslant \sqrt {\mu rg} \cr & \therefore {v_{\max }} = \sqrt {\mu rg} \cr} $$
Here, radius of curved road $$r = 30 m,$$
coefficient of friction $$\mu = 0.04$$
$$\eqalign{ & \therefore {v_{\max }} = \sqrt {0.4 \times 30 \times 9.8} \cr & = 10.84\,m/s \cr} $$

Graph (B) correctly depicts the acceleration - time graph of the block.

$$\eqalign{ & mg = \mu N = \mu \,ma \cr & \therefore a = \frac{g}{\mu }. \cr} $$

For the block to slide, the angle of inclination should be equal to the angle of repose, i.e.,
$${\tan ^{ - 1}}\mu = {\tan ^{ - 1}}\sqrt 3 = {60^ \circ }$$
Therefore, option (a) is wrong.
For the block to topple, the condition of the block will be as shown in the figure.

In $$\Delta POM,\tan \theta = \frac{{PM}}{{OM}} = \frac{{5cm{\text{ }}}}{{7.5cm}} = \frac{2}{3}$$
For this, $$\theta < 60°.$$  From this we can conclude that the block will topple at lesser angle of inclination. Thus the block will remain at rest on the plane up to a certain anlgle $$\theta $$ and then it will topple.

Given a plank with a box on its one end is gradually raised about the end having angle of inclination is $${30^ \circ },$$  the box starts to slip and slides down $$4m$$  the plank in $$4s$$ as shown in figure.

The coefficient of static friction,
$${\mu _s} = \tan {30^ \circ } = \frac{1}{{\sqrt 3 }} = 0.6$$
So, distance covered by a plank,
$$\eqalign{ & s = ut + \frac{1}{2}a{t^2} \cr & {\text{Here,}}\,u = 0\,{\text{and}}\,a = g\left( {\sin \theta - \mu \cos \theta } \right) \cr & \therefore \quad 4 = \frac{1}{2}g\left( {\sin {{30}^ \circ } - {\mu _k}\cos {{30}^ \circ }} \right){\left( 4 \right)^2} \cr & \Rightarrow 0.5 = 10 \times \frac{1}{2} - {\mu _K} \times 10 \times \frac{{\sqrt 3 }}{2} \cr & \Rightarrow 5\sqrt 3 {\mu _K} = 4.5 \Rightarrow {\mu _K} = 0.51 \cr} $$
Thus, coefficient of kinetic friction between the box and the plank is 0.51.

First of all consider the forces on the blocks

For the 1^st block $$\left[ {\because {m_1} = {m_2} = {m_3}} \right]$$
$$mg - {T_1} = m \times a\,......\left( {\text{i}} \right)$$
Let us consider 2^nd and 3^rd block as a system.
$${\text{So,}}\,\,{T_1} - 2\mu mg = 2m \times a\,.......\left( {{\text{ii}}} \right)$$
Solving Eqs. (i) and (ii),
$$ \Rightarrow mg - {T_1} = m \times a \Rightarrow {T_1} - 2\mu mg = 2m \times a$$
Addiing Eqs. (i) and (ii)
$$mg\left( {1 - 2\mu } \right) = 3m \times a \Rightarrow a = \frac{g}{3}\left( {1 - 2\mu } \right)$$

$$\sin \theta = \frac{r}{{3r}} = \frac{1}{3},\cos \theta = \sqrt {\frac{8}{9}} $$
If mass of smaller cylinder is $$m,$$ then mass of bigger one will be $$4m.$$  For the equilibrium of upper cylinder,

$$\eqalign{ & 2{N_1}\sin \theta = 4\,mg \cr & \therefore {N_1} = \frac{{2mg}}{{\sin \theta }}\,......\left( {\text{i}} \right) \cr} $$
Now for the equilibrium of lower cylinder,
$$\eqalign{ & {N_2} = {N_1}\sin \theta + mg\,......\left( {{\text{ii}}} \right) \cr & {\text{and}}\,\,f = {N_1}\cos \theta \cr & {\text{or}}\,\,\mu {N_2} = {N_1}\cos \theta \,......\left( {{\text{iii}}} \right) \cr} $$
After solving above equations, we get
$$\mu = \frac{1}{{3\sqrt 2 }}.$$

When brakes are on, the wheels of the cycle will slide on the road instead of rolling there. It means the sliding friction will come into play instead of rolling friction. The value of sliding friction is more than that of rolling friction.

According to question, a car is negotiating a curved road of radius $$R.$$ The road is banked at angle $$\theta $$ and the coefficient of friction between the tyres of car and the road is $${{\mu _s}}.$$  So, this given situation can be drawn as shown in figure below.

Considering the case of vertical equilibrium
$$\eqalign{ & N\cos \theta = mg + {f_l}\sin \theta \cr & \Rightarrow mg = N\cos \theta - {f_l}\sin \theta \,......\left( {\text{i}} \right) \cr} $$
Considering the case of horizontal equilibrium,
$$N\sin \theta + {f_1}\cos \theta = \frac{{m{v^2}}}{R}\,......\left( {{\text{ii}}} \right)$$
Divide eqs. (i) and (ii), we get
$$\eqalign{ & \frac{{{v^2}}}{{Rg}} = \frac{{\sin \theta + {\mu _s}\cos \theta }}{{\cos \theta - {\mu _s}\sin \theta }}\,\,\left[ {{f_l} \propto {\mu _s}} \right] \cr & \Rightarrow v = \sqrt {Rg\left( {\frac{{\sin \theta + {\mu _s}\cos \theta }}{{\cos \theta - {\mu _s}\sin \theta }}} \right)} \cr & \Rightarrow v = \sqrt {Rg\left( {\frac{{\tan \theta + {\mu _s}}}{{1 - {\mu _s}\tan \theta }}} \right)} \cr} $$