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Variation of G Due to Height and Depth

Two masses A ...

Question

Two masses $A$ and $B$ of $10 k g$ and $5 k g$ 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 (in $k g$ ) of $C$ that may be placed on $A$ to prevent it from moving is:

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Solution

Given: $m_{A} = 10 k g, m_{B} = 5 k g, μ = 0.2$ (between block $A$ and ground)
Since given system is in equilibrium

From $F B D$ of block $B$
$∴ T = m_{B} g = 5 g = 50 N$

For combination of block $A$ and $C$
Friction force between block $A$ and ground $= μ (m_{A} + m_{C}) g = 0.2 (10 + m_{c}) g$

It should be balancing tension acting on them as shown in $F B D$
$∴ 0.2 (10 + m_{c}) g = 50$
$10 + m_{c} = 25$
$m_{c} = 15 k g$

Final Answer: 15 kg

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Variation of G Due to Height and Depth

Standard XII Physics

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