First slide

Two block problem

Question

Two masses A and B of 10 kg and 5 kg respectively are connected with a string passing over a frictionless pulley fixed at the corner of a table as shown in figure. The coefficient of friction of A with the table is 0.2. The minimum mass of C that may be placed on A to prevent it from moving is equal to

Moderate

A

15 kg
B

10 kg
C

5 kg
D

20 kg

Solution

$μ (m_{A} + m_{C}) g = m_{B} g$
$∴ m_{C} = \frac{m_{B}}{μ} - m_{A} = \frac{5}{0.2} - 10 = 15 kg$

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