Question

A body of mass 2 kg is on an inclined plane of inclination 30° and coefficient of friction is $ \frac{1}{\sqrt{3}} $. The minimum force required to move the body up the inclined plane is: \[ \text{(Acceleration due to gravity = 10 ms}^{-2}\text{)} \] Options:

• A. \( 5.77 \, \text{N} \)
• B. \( 10 \, \text{N} \)
• C. \( 20 \, \text{N} \)
• D. \( 15 \, \text{N} \)

Hint:

When a body is moving up an inclined plane, the minimum force required to overcome gravity and friction is the sum of the gravitational force component along the incline and the frictional force.

Answer 1

The Correct Option is C

Given:
Mass of the body, \( m = 2 \, \text{kg} \)
Angle of inclination, \( \theta = 30^\circ \)
Coefficient of friction, \( \mu = \frac{1}{\sqrt{3}} \)
Acceleration due to gravity, \( g = 10 \, \text{ms}^{-2} \)
We need to find the minimum force required to move the body up the inclined plane.
Identify the forces acting on the body.
1. Gravitational force:
The component of the gravitational force acting down the incline is: \[ F_g = mg \sin \theta \] Substituting the values: \[ F_g = 2 \times 10 \times \sin 30^\circ = 20 \times \frac{1}{2} = 10 \, \text{N} \] 2. Frictional force:
The normal force is: \[ F_{\text{normal}} = mg \cos \theta \] Substituting the values: \[ F_{\text{normal}} = 2 \times 10 \times \cos 30^\circ = 20 \times \frac{\sqrt{3}}{2} = 10 \sqrt{3} \, \text{N} \] The frictional force is given by: \[ F_{\text{friction}} = \mu F_{\text{normal}} = \frac{1}{\sqrt{3}} \times 10 \sqrt{3} = 10 \, \text{N} \] 3. Total force required:
The minimum force \( F_{\text{min}} \) required to move the body up the plane is the sum of the gravitational force and the frictional force: \[ F_{\text{min}} = F_g + F_{\text{friction}} = 10 + 10 = 20 \, \text{N} \] Final Answer: \[ \boxed{20 \, \text{N}} \]