On a fixed smooth incline plane of inclination 37^o , two blocks A and B $(m_{A} = 2 kg and m_{B} = 3 kg)$ are placed as shown in figure. When a force F = 55N is applied on block B, it is found that block A just begins to slide on B. The value of coefficient of friction between the blocks is $μ = \frac{4}{k}$ then k is

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answer is 13.

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Detailed Solution

$\begin{array}{l} 55 - 50 \sin 37^{\circ} = (2 + 3) a \\ 55 - 50 \times 0 .6 = 5 a \Rightarrow a = 5 m / s^{2} \end{array}$

$a_{y} = 5 \times 0 .6 = 3 m / s^{2}$

$a_{x} = 5 \times 0 .8 = 4 m / s^{2}$

$N_{1} - 20 = 2 a_{y} = 2 \times 3 N = 26 N$

$\Rightarrow: f = {μN}_{1} = 2 a_{x} μ \times 26 = 2 \times 4 \Rightarrow μ = \frac{4}{13}$

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