Solution: Friction on an Inclined Plane

Question

A block of mass 2 kg is placed on a rough inclined plane inclined at 30° to the horizontal. If the coefficient of static friction between the block and the inclined plane is 0.7, find the magnitude of the frictional force acting on the block.

Given:

• Mass of the block, m = 2 kg
• Angle of inclination, θ = 30°
• Coefficient of static friction, μ_s = 0.7
• Acceleration due to gravity, g = 9.8 m/s²

Step 1: Calculate the component of weight along the inclined plane

The component of gravitational force acting down the inclined plane is:

F = mg sin θ

Substituting the given values:

F = 2 × 9.8 × sin 30°

Since sin 30° = 0.5,

F = 2 × 9.8 × 0.5

F = 9.8 N

Step 2: Calculate the normal reaction

The normal reaction exerted by the inclined plane on the block is:

N = mg cos θ

Substituting the values:

N = 2 × 9.8 × cos 30°

Since cos 30° ≈ 0.866,

N = 19.6 × 0.866

N ≈ 16.97 N

Step 3: Calculate the maximum static friction

The maximum static friction is given by:

f_max = μ_sN

Substituting the values:

f_max = 0.7 × 16.97

f_max ≈ 11.88 N

Step 4: Calculate the actual frictional force

The component of the block’s weight pulling it down the plane is 9.8 N, while the maximum static friction is approximately 11.88 N. Since 9.8 N is less than 11.88 N, static friction is sufficient to prevent the block from sliding. Therefore, the actual static friction balances the component of weight acting down the plane:

f = mg sin θ

f = 2 × 9.8 × 0.5

Important Note

The actual static friction is 9.8 N, not 11.88 N. The value 11.88 N represents the maximum possible static friction. Static friction adjusts itself to balance the applied force, up to its maximum limit.