A block of mass 15 kg is resting on a rough inclined plane as shown in figure. The block is tied up by a horizontal string which has a tension of 50 N. The coefficient of friction between the surfaces of contact may be (g = 10 m/s 2 )
Text Solution
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(a, b, c): The free body diagram of the block is
N is the normal reaction exerted by inclined plane on the block.
Applying Newton’s second law to the block along and normal to the incline.
mg sin 45° = T cos 45° + µN..............
N = mg cos 45° + T sin 45°...............
On solving we get
µ = 1/2
so, any value of µ greater than 0.5 is answer
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