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 $\underset{F}{\to }=3\hat{i}+4\hat{j}+20\hat{k}$ N is applied on the cube. $(useg=10m/s^2)$

• A. The block starts slipping over the surface
• B. The friction force on the cube by the surface is 5 N.
• C. The friction force acts in xy-plane at angle ${127}^0$ with the positive x-axis in clockwise direction.
• D. The contact force exerted by the surface on the cube is $\sqrt{925}N.$

Answer 1

Answer: (B), (C), (D)

The correct options are
B The friction force acts in xy-plane at angle ${127}^0$ with the positive x-axis in clockwise direction.
C The contact force exerted by the surface on the cube is $\sqrt{925}N.$
D The friction force on the cube by the surface is 5 N.

$N=50-20=30N$
Limiting friction force $=\mu N=0.4\times 30=12N$ 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 5 N.
From the top view, it is clear that $\theta ={37}^{0}i.e{127}^0$ from the x-axis that is the direction of the friction force. It is opposite to
the applied force.
Contact force $=\sqrt{N^2+f^2}=\sqrt{925}N$