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Q.

A boy of mass 4 kg is standing on a piece of wood having mass 5 kg. If the coefficient of friction between the wood and the floor is 0.5 , the maximum force that the boy can exert on the rope so that the piece of wood does not move from its place is………N.
(Round off to the Nearest integer) $[T a k e g = 10 {ms}^{- 2}]$

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Detailed Solution

(47) From FBD shown, $T = M g - N$
$R = m g + N = (μ + m) g - T$
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For no movement of block,
$\begin{array}{r} T \leq μ R \Rightarrow T \leq μ [(M + m) g - T] \\ \Rightarrow T \leq \frac{μ (M + m) g}{1 + μ} \Rightarrow T = \frac{(0.5) (10 + 4) \times 10}{1 + 0.5} = \frac{70}{1.5} \\ T_{max} = 46.67 V \end{array}$

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A boy of mass 4 kg is standing on a piece of wood having mass 5 kg. If the coefficient of friction between the wood and the floor is 0.5 , the maximum force that the boy can exert on the rope so that the piece of wood does not move from its place is………N. (Round off to the Nearest integer) [Take g=10 ms−2]