Question

Block A of mass m and block B of 2m are placed on a fixed triangular wedge by means of a light and inextensible string and a frictionless pulley as shown in Fig. The wedge is inclined at ${45}^0$ to the horizontal on both sides. The coefficient of friction between the block A and the wedge is 2/3 and that between the block B and the wedge is 1/3. If the system of A and B is released from rest, then find,
a. the acceleration of A
b. tension in the string
c. the magnitude and direction of the frictional force acting on A?

Answer 1

$Mass=\frac{2}{3},N_{BW}=\frac{1}{3}$

$\text{Mass of A}=m.$

$\text{Mass of B}=2m.$

(i)—

$F=\frac{2mg}{\sqrt{2}}-\frac{mg}{\sqrt{2}}=\frac{mg}{\sqrt{2}}.$

$\text{force of friction}=f_A+f_B$

$=N_A{N}_A+N_B{N}_B$

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

$=\frac{4mg}{3\sqrt{2}}$.

Magnitude of force of rriction is greater which cause of A and B to move, the mass system will not move

acc. of the system is zero

(ii)—

Consider equilibrium of B

$\text{force causing motion}=\frac{2mg}{\sqrt{2}}$

$\text{force of friction}=N_B{N}_B$

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

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

$T=\text{Diff. of above two force on B}.$

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

$=\frac{2}{3}\times \frac{2mg}{\sqrt{2}}=\frac{2\sqrt{2}}{3}mg$

(iii)—

$\text{Force of friction on block A}$

$\text{Facts upwards while weight component acts downwards on block A}$

$\text{Force of friction}$

$=T-(weigh-componentofA)$

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