A solid cube of mass `2 kg` is placed on rough horizontal surface, in `xy-`plane as shown. The friction coefficient between the surface and the cube is `0.2`. An external force `3hat(i)+4hat(j)+5hat(k) N` is applied on the cube. (use `g = 10 m//s^(2)`)

Options

A. The block starts slipping over the surface
B. The friction force on the cube by the surface is `3N`.
C. The frication 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 `15 N`.

Answer

A, B, C

Explanation

`mg = 20 N`
`N = 20 - 5 = 15 N`
Limiting friction force `= muN =m 3 N` and applied force in horizomntal direction is more than the frication force, therefore the block will slide.
So, frication force must be equal to `3N`.

From the top view, it is clear that `theta = 53^(@)` i.e., `127^(@)` from the `x-`axis that is the direction of lthe frication force. If is opposite to the applied force
Contact F=force ` = sqrt(N^(2) - f(2)) = sqrt((15)^(2) + (3)^(2))`
`= sqrt(225 + 9) = sqrt(234) N`