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Question

# 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 →F=3^i+4^j+20^k N is applied on the cube. (use g=10m/s2)

A
The block starts slipping over the surface
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B
The friction force on the cube by the surface is 5 N.
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C
The friction force acts in xy-plane at angle 1270 with the positive x-axis in clockwise direction.
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D
The contact force exerted by the surface on the cube is 925N.
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Solution

## The correct options are B The friction force acts in xy-plane at angle 1270 with the positive x-axis in clockwise direction. C The contact force exerted by the surface on the cube is √925N. D The friction force on the cube by the surface is 5 N.N=50−20=30NLimiting friction force =μN=0.4×30=12N and appliedforce 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 5 N.From the top view, it is clear that θ=370i.e1270 from the x-axis that is the direction of the friction force. It is opposite tothe applied force. Contact force =√N2+f2=√925N

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