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 figure. The wedge is inclined at 45^∘ to the horizontal on both the sides. If 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) and both the 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:

On inclined planes, always compare the net driving force mgsinθ - μ mgcosθ on both sides before writing equations of motion.

Answer 1

The Correct Option is D

Step 1: Resolve forces along the incline for both blocks.
For block A:
\( F_A = mg\sin 45^\circ - \mu_A \, mg\cos 45^\circ \)
\( = \dfrac{mg}{\sqrt{2}} - \dfrac{2}{3} \cdot \dfrac{mg}{\sqrt{2}} = \dfrac{mg}{3\sqrt{2}} \)
Step 2: For block B:
\( F_B = 2mg\sin 45^\circ - \mu_B (2mg)\cos 45^\circ \)
\( = \dfrac{2mg}{\sqrt{2}} - \dfrac{1}{3} \cdot \dfrac{2mg}{\sqrt{2}} = \dfrac{4mg}{3\sqrt{2}} \)
Step 3: Effective driving forces on both sides balance through the string, hence net force on the system is zero.

Step 4: Therefore, acceleration of block A is zero.