Question

The magnitude and direction of the force of friction acting on A are

• A. mg, down the plane
• B. $\frac{mg}{2}$, up the plane
• C. $\frac{mg}{\sqrt{2}}$, up the plane
• D. $\frac{mg}{3\sqrt{2}}$, down the plane

Answer 1

The correct option is D $\frac{mg}{3\sqrt{2}}$, down the plane

Given

${\mu }_{AW}=\frac{2}{3}{\mu }_{BW}=\frac{1}{3}$

Mass of A = m

Mass of B = 2m

The system consists of the two masses. When A

moves upwards, B moves downwards. Force

causes this motion and friction opposes it.

Force $=\frac{2mg}{\sqrt{2}}-\frac{mg}{\sqrt{2}}=\frac{mg}{\sqrt{2}}$

This may cause motion

Force of friction $=f_A+f_B={\mu }_A{N}_A{\mu }_B{N}_B$

$=\left(\frac{2}{3}\right)\left(\frac{mg}{\sqrt{2}}\right)+\left(\frac{1}{3}\right)\left(\frac{2mg}{\sqrt{2}}\right)=\frac{4mg}{3\sqrt{2}}$

Since the magnitude of force of friction is greater

than the force which may cause motion of A and

B, the mass system will not move.

Hence acceleration of the system is zero

Consider equilibrium of B:

Whenonly B is considered, we have

Force causing motion $=\frac{2mg}{\sqrt{2}}$

Force of friction $={\mu }_B{N}_B$

$=\left(\frac{1}{3}\right)\times \left(\frac{2mg}{\sqrt{2}}\right)=\left(\frac{1}{3}\right)\left(\frac{2mg}{\sqrt{2}}\right)$

$\therefore$ T = Difference of above two forces on B

or $T=\frac{2mg}{\sqrt{2}}-\frac{1}{3}\left(\frac{2mg}{\sqrt{2}}\right)=\frac{2}{3}\times \frac{2mg}{\sqrt{2}}=\frac{2\sqrt{2}}{3}mg$

Force of friction on block A

T acts upwards while weight component acts

downwards on block A.

$\therefore$ Force of friction

= T. (weight - Component of A)

$=\frac{2\sqrt{2}mg}{3}-\frac{mg}{\sqrt{2}}=\frac{4mg-3mg}{3\sqrt{2}}=\frac{mg}{3\sqrt{2}}$ down the plane