Static Friction on Inclined Plane Analysis

This problem involves calculating the coefficient of static friction ($\mu_s$) for a mass resting on a horizontal plane that is gradually inclined. The key information is that the mass just begins to slide when the angle of inclination reaches $\theta = 14^\circ$. This angle is also known as the angle of repose.

Understanding Forces on an Inclined Plane

When an object is placed on an inclined plane, the force of gravity ($mg$) acting on it can be resolved into two components:

• A component parallel to the incline, pulling the object downwards: $F_{\parallel} = mg \sin(\theta)$
• A component perpendicular to the incline, pushing the object into the plane: $F_{\perp} = mg \cos(\theta)$

The inclined plane exerts a normal force ($N$) on the object, which is equal in magnitude and opposite in direction to the perpendicular component of gravity. Thus:

$$ N = mg \cos(\theta) $$

The force of static friction ($f_s$) opposes the motion or tendency of motion. It acts parallel to the plane, upwards in this case. The maximum static friction ($f_{s,max}$) is the maximum force that friction can exert before the object starts to move.

The relationship between maximum static friction, the coefficient of static friction, and the normal force is given by:

$$ f_{s,max} = \mu_s N $$

Condition for Sliding

The problem states that the mass *just begins to slide* at $\theta = 14^\circ$. This means that the parallel component of gravity has reached the maximum static friction force:

$$ F_{\parallel} = f_{s,max} $$

Substituting the expressions for these forces:

$$ mg \sin(\theta) = \mu_s N $$

Now substitute the expression for the normal force ($N$):

$$ mg \sin(\theta) = \mu_s (mg \cos(\theta)) $$

Calculating the Coefficient of Static Friction

We can simplify the equation by canceling out $mg$ from both sides:

$$ \sin(\theta) = \mu_s \cos(\theta) $$

To find $\mu_s$, we rearrange the equation:

$$ \mu_s = \frac{\sin(\theta)}{\cos(\theta)} $$

This is equivalent to the tangent of the angle of inclination:

$$ \mu_s = \tan(\theta) $$

Given the angle $\theta = 14^\circ$, and the values:

• $\sin(14^\circ) = 0.2419$
• $\cos(14^\circ) = 0.97$

Substitute these values into the formula:

$$ \mu_s = \frac{0.2419}{0.97} $$

Performing the division:

$$ \mu_s \approx 0.24938 $$

Matching with Options

The calculated value for the coefficient of static friction is approximately 0.24938. Comparing this value with the given options:

• Option 1: 0.15
• Option 2: 0.25
• Option 3: 0.31
• Option 4: 0.42

The calculated value is closest to 0.25.