Question 10

A block rests on a rough inclined plane making an angle of $$30^\circ$$ with the horizontal. The coefficient of static friction between the block and the plane is $$0.8$$. If the frictional force on the block is $$10$$ N, the mass of the block (in kg) is (take $$g = 10 \text{ m/s}^2$$)

Solution

Solution & Explanation

1. Determine the Nature of the Friction Force

When an object rests on an inclined surface, it experiences a downward gravitational force component parallel to the incline ($$mg \cdot \sin\theta$$). For the block to remain stationary ("rests"), a static friction force ($$f_s$$) must act upward along the incline to perfectly balance this component:

$$f_s = mg \cdot \sin\theta$$

Before using this equilibrium equation, we must verify whether the block is actually capable of remaining at rest by checking the maximum possible limit of static friction ($$f_{\text{max}} = \mu_s \cdot mg \cdot \cos\theta$$). The critical angle of repose ($$\theta_r$$) where sliding begins is given by:

$$\tan\theta_r = \mu_s = 0.8$$

$$\theta_r = \tan^{-1}(0.8) \approx 38.66^\circ$$

Since the actual angle of the incline ($$\theta = 30^\circ$$) is strictly less than the angle of repose ($$38.66^\circ$$), the block will not slide down. It remains completely stationary, and the static friction force acting on it is a self-adjusting force equal to the parallel component of gravity.

2. Set Up the Force Equilibrium Equation

Equating the given static frictional force to the component of gravity pulling the block down the incline:

$$f_s = m \cdot g \cdot \sin\theta$$

We are given the following parameters from the problem statement:

Frictional force: $$f_s = 10 \,\, \text{N}$$
Angle of inclination: $$\theta = 30^\circ$$
Acceleration due to gravity: $$g = 10 \,\, \text{m/s}^2$$

Substituting these values into the equilibrium condition:

$$10 = m \cdot 10 \cdot \sin(30^\circ)$$

3. Calculate the Mass ($$m$$)

Since $$\sin(30^\circ) = \frac{1}{2}$$, substitute this into our equation:

$$10 = m \cdot 10 \cdot \left(\frac{1}{2}\right)$$

$$10 = m \cdot 5$$

Isolating the mass variable ($$m$$):

$$m = \frac{10}{5} = 2.0 \,\, \text{kg}$$

Concept Check: The coefficient of static friction ($$\mu_s = 0.8$$) represents the maximum threshold of grip available ($$f_{\text{max}} = 0.8 \cdot 2 \cdot 10 \cdot \cos 30^\circ \approx 13.86 \,\, \text{N}$$). Because the actual required keeping force ($$10 \,\, \text{N}$$) is lower than this maximum capability, the static friction perfectly limits itself to $$10 \,\, \text{N}$$ to keep the $$2.0 \,\, \text{kg}$$ mass from moving.