A solid cube of mass 5 kg is placed on a rough horizontal surface, in xy-plane as shown. The friction coefficient between the surface and the cube is 0.4. An external force 
N is applied on the cube. (use g = 10 m/s 2 )
N.
Text Solution
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(b, c, d)
N = 50 – 20 = 30 N
Limiting friction force = µN = 12 N and applied force in horizontal direction is less than the limiting
friction force, therefore the block will not slide.
For equilibrium in horizontal direction, friction force must be equal to 10 N.
From the top view, it is clear that θ = 37° i.e. 127° from the x-axis that is the direction of the friction
force. It is opposite to the applied force.
Contact force = 
= 
N
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