As shown in the figure, a block of mass 3 kg is kept on a horizontal rough surface of coefficient of friction 1 3 3 . The critical force to be applied on the vertical surface as shown at an angle 60 °

Question

As shown in the figure, a block of mass $\sqrt{3} kg$ is kept on a horizontal rough surface of coefficient of friction $\frac{1}{3 \sqrt{3}}$ . The critical force to be applied on the vertical surface as shown at an angle $60 °$ with horizontal such that it does not move, will be $3 x$ . The value of $x$ will

$[g = 10 m s^{- 2}; \sin 60 ° = \frac{\sqrt{3}}{2}; \cos 60 ° = \frac{1}{2}]$

TestHub · Accepted Answer

F cos 60 ° = μ N or F 2 = 1 3 3 N … 1 & N = sin 60 ° + 3 g … 2 From equation 1 & 2 F 2 = 1 3 3 F 3 2 + 3 g ⇒ F = g = 10 Newton = 3 x So x = 10 3 = 3 . 33

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