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 `vecF=6hati+8hatj+20hatkN` is applied on the cube. (use `g=10(m)/(s^2)`)

Options

A. The block start silpping over the surface
B. the friction force on the cube by the surface is 10 N.
C. The friction force acts in xy-plane at angle `127^@` with the positive x-axis in clockwise direction.
D. The contact force exerted by the surface on the cube is `10sqrt(10)N`.

Answer

B, C, D

Explanation

`N=50-20=30N`
Limiting friction force`=muN=12N` and applied force in forizontal 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 `theta=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`=sqrt(N^2+f^2)=10sqrt(10)N`