Question

Block A of mass m and block B of mass 2m are placed on a fixed triangular wedge by means of a massless, inextensible string and a frictionless pulley as shown in the figure. The wedge is inclined at 45° to the horizontal on both sides. If the coefficient of friction between block A and the wedge is (2)/(3) and that between block B and the wedge is (1)/(3), and both blocks A and B are released from rest, the acceleration of A will be: 

• A. \( -1\,\text{m s}^{-2} \)
• B. \( 1.2\,\text{m s}^{-2} \)
• C. \( 0.2\,\text{m s}^{-2} \)
• D. zero

Hint:

In pulley–incline systems, always compare net driving forces on both sides before assuming motion.

Answer 1

The Correct Option is D

Step 1: Component of weight along incline for A:
\( mg\sin 45^\circ = \dfrac{mg}{\sqrt{2}} \),
\( f_A = \mu_A \, mg\cos 45^\circ = \dfrac{2}{3} \cdot \dfrac{mg}{\sqrt{2}} \) 

Step 2: Net force on A along plane:
\( \dfrac{mg}{\sqrt{2}} - \dfrac{2mg}{3\sqrt{2}} = \dfrac{mg}{3\sqrt{2}} \) 

Step 3: For block B:
\( 2mg\sin 45^\circ - \mu_B (2mg\cos 45^\circ) \)
\( = \dfrac{2mg}{\sqrt{2}} - \dfrac{2mg}{3\sqrt{2}} \)
\( = \dfrac{4mg}{3\sqrt{2}} \) 

Step 4: The forces balance through the string, hence no net acceleration occurs.