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

• 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 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 on the cube by the surface is 5 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 $\sqrt{925}N$

First of all, lets calculate the maximum value of friction force on the cube

$f_{max}=\mu mg$

$=$ 0.4x5x10

$=$ 20 N

The friction force is acting in the xy-plane because the solid cube is resting on the same plane.

Now, calculate the external force acting in the xy-plane

$F_{xy}=\sqrt{(}{3}^2+4^2)$

$=$ 5 N

The applied external force ($F_{xy}$) is less than the maximum value of friction force.

Therefore, friction force acting on the cube will be equal to the external force.

$f=F_{xy}$

$=$ 5 N

The external force is making ${53}^0$ with x-axis, and we know that the friction force acts in the opposite direction of it.

So, the angle made by the friction force will be ${180}^0-{53}^0={127}^0$ with the positive x-axis and in clockwise direction.

Furthermore, the contact surface will exert two forces on the cube.

(a) Friction force = 5 N in the xy-plane

(b) $mg-20$ in the positive z direction

$=5x10-20$

$=30N$

Resultant contact force exerted by the surface

$=\sqrt{(}frictionforc{e}^2+(mg-20{)}^2)$

$=\sqrt{(}{5}^2+{30}^2)$

$=\sqrt{(}925)$ N

Option B, C, and D are the correct choices.