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 \(g = 10\, m\,s^{-2}\))
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 \(g = 10\, m\,s^{-2}\))
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In multi-block pulley systems, always write separate Newton’s equations for each block and eliminate tensions step by step.
Updated On: Jun 19, 2026
- \( \frac{2}{3}\, m\,s^{-2} \)
- \( \frac{8}{3}\, m\,s^{-2} \)
- \( \frac{1}{3}\, m\,s^{-2} \)
- \( \frac{4}{3}\, m\,s^{-2} \)
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The Correct Option is B
Solution and Explanation
Step 1: Understand the system and choose variables.
Each mass is \(10\,g = 0.01\,kg\). Let acceleration of the system be \(a\), where block C moves downward, and blocks A and B move horizontally towards the pulley.
Step 2: Identify forces on block C.
For block C: \[ mg - T_1 = ma \] \[ 0.01 \times 10 - T_1 = 0.01a \] \[ 0.1 - T_1 = 0.01a \]
Step 3: Forces on block B.
Friction on B: \[ f_B = \mu mg = 0.1 \times 0.01 \times 10 = 0.01\,N \] Equation of motion for B: \[ T_2 - 0.01 = 0.01a \] \[ T_2 = 0.01a + 0.01 \]
Step 4: Forces on block A.
Friction on A: \[ f_A = 0.01\,N \] Equation of motion for A: \[ T_1 - T_2 - 0.01 = 0.01a \] Substitute \(T_2\): \[ T_1 - (0.01a + 0.01) - 0.01 = 0.01a \] \[ T_1 = 0.02a + 0.02 \]
Step 5: Substitute in equation of C.
From C: \[ 0.1 - T_1 = 0.01a \] Substitute \(T_1 = 0.02a + 0.02\): \[ 0.1 - (0.02a + 0.02) = 0.01a \] \[ 0.08 - 0.02a = 0.01a \]
Step 6: Solve for acceleration.
\[ 0.08 = 0.03a \] \[ a = \frac{0.08}{0.03} = \frac{8}{3}\, m\,s^{-2} \]
Final Answer: \[ \boxed{\frac{8}{3}\, m\,s^{-2}} \]
Each mass is \(10\,g = 0.01\,kg\). Let acceleration of the system be \(a\), where block C moves downward, and blocks A and B move horizontally towards the pulley.
Step 2: Identify forces on block C.
For block C: \[ mg - T_1 = ma \] \[ 0.01 \times 10 - T_1 = 0.01a \] \[ 0.1 - T_1 = 0.01a \]
Step 3: Forces on block B.
Friction on B: \[ f_B = \mu mg = 0.1 \times 0.01 \times 10 = 0.01\,N \] Equation of motion for B: \[ T_2 - 0.01 = 0.01a \] \[ T_2 = 0.01a + 0.01 \]
Step 4: Forces on block A.
Friction on A: \[ f_A = 0.01\,N \] Equation of motion for A: \[ T_1 - T_2 - 0.01 = 0.01a \] Substitute \(T_2\): \[ T_1 - (0.01a + 0.01) - 0.01 = 0.01a \] \[ T_1 = 0.02a + 0.02 \]
Step 5: Substitute in equation of C.
From C: \[ 0.1 - T_1 = 0.01a \] Substitute \(T_1 = 0.02a + 0.02\): \[ 0.1 - (0.02a + 0.02) = 0.01a \] \[ 0.08 - 0.02a = 0.01a \]
Step 6: Solve for acceleration.
\[ 0.08 = 0.03a \] \[ a = \frac{0.08}{0.03} = \frac{8}{3}\, m\,s^{-2} \]
Final Answer: \[ \boxed{\frac{8}{3}\, m\,s^{-2}} \]
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