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In the figure, a block of weight 60 N is placed on a rough surface. The coefficient of friction between the block and the surfaces is 0.5. What should be the weight W such that the block does not slip on the surface?

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30 N

$\frac{60}{\sqrt{2}} N$

$\frac{30}{\sqrt{2}} N$

60 N

answer is C.

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Detailed Solution

Frictional force:

$f = μN = 0.5 \times mg = 0.5 \times 60 = 30 N$

Now $F = T_{1} = T_{2} \cos 45^{0}$

or 30 = $T_{2} \cos 45^{0}$

and W = $T_{2} \sin 45^{0}$

Solving them we get W = 30 N

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