Question

A block of mass 5 kg is placed on a frictionless inclined plane. The angle of inclination of the plane is 30 degrees. Calculate the force required to keep the block in equilibrium.

• A. 25.5 N
• B. 2(4)5 N
• C. 26.5 N
• D. 27.5 N

Hint:

Inclined Plane Equilibrium. The component of gravitational force acting parallel down the incline is \(mg \sin\theta\). On a frictionless plane, an equal and opposite force must be applied along the incline to maintain equilibrium.

Answer 1

The Correct Option is B

When a block of mass \(m\) is placed on an inclined plane with angle of inclination \(\theta\), the gravitational force (\(mg\)) acting vertically downwards can be resolved into two components:
(1) Perpendicular to the plane: \(mg \cos\theta\) (balanced by the normal force).

(2) Parallel to the plane, downwards: \(mg \sin\theta\).
To keep the block in equilibrium (i.
e.
, prevent it from sliding down) on a frictionless incline, an external force must be applied upwards along the plane that exactly balances the downward component of gravity.
Force required (\(F_{req}\)) = Component of gravity parallel to the plane $$ F_{req} = mg \sin\theta $$ Given: Mass \(m = 5\) kg Angle \(\theta = 30^\circ\) Acceleration due to gravity \(g \approx 9.
8\) m/s\(^2\) (standard value if not given) \(\sin 30^\circ = 0.
5\) Substitute the values: $$ F_{req} = (5 \, \text{kg}) \times (9.
8 \, \text{m/s}^2) \times \sin(30^\circ) $$ $$ F_{req} = 5 \times 9.
8 \times 0.
5 \, \text{N} $$ $$ F_{req} = 5 \times (4)9 \, \text{N} = 2(4)5 \, \text{N} $$ The force required to keep the block in equilibrium is 2(4)5 N.