Question

Three bodies A, B and C of masses 10 g each are tied to a thread–pulley system as shown in the figure. Assume the masses of pulley and threads are negligible and there is no friction in the pulley. The coefficient of friction between A and B with the horizontal surface is 0.1. Find the acceleration with which body C comes down. (Take \(g = 10\, m\,s^{-2}\))

• A. \( \frac{2}{3}\, m\,s^{-2} \)
• B. \( \frac{8}{3}\, m\,s^{-2} \)
• C. \( \frac{1}{3}\, m\,s^{-2} \)
• D. \( \frac{4}{3}\, m\,s^{-2} \)

Hint:

In multi-block pulley systems, always write separate Newton’s equations for each block and eliminate tensions step by step.

Answer 1

The Correct Option is B

Step 1: System parameters.
Each mass = 10 g = 0.01 kg; block C descends, A and B slide horizontally.

Step 2: Equation for C.
mg - T₁ = m a → 0.1 - T₁ = 0.01a.

Step 3: Equation for B with friction.
f_B = μmg = 0.1×0.01×10 = 0.01 N; T₂ - 0.01 = 0.01a → T₂ = 0.01a + 0.01.

Step 4: Equation for A with friction.
f_A = 0.01 N; T₁ - T₂ - 0.01 = 0.01a. Substitute T₂: T₁ = 0.02a + 0.02.

Step 5: Combining with C's equation.
0.1 - (0.02a + 0.02) = 0.01a → 0.08 = 0.03a.

Step 6: Acceleration.
a = 0.08/0.03 = 8/3 m s⁻².