If an object just begins to slide on a surface inclined at 30º to the horizontal,the coefficient of friction is?

1. Analyze the question

The question asks for the coefficient of friction ($\mu$) when an object is placed on an inclined plane and is just about to slide. This is a classic mechanics problem involving equilibrium on an inclined plane.

2. Explain the concept

When an object is placed on an inclined plane at an angle $\theta$, two main forces act on it relative to the surface:

• The component of weight acting down the plane: $mg \sin\theta$
• The normal reaction ($R$): $mg \cos\theta$
• The frictional force ($F$): $\mu R$

When the object is "just about to slide," it is in a state of limiting equilibrium. At this point, the force pulling the object down the slope is exactly balanced by the maximum static frictional force:

Since $F = \mu R$ and $R = mg \cos\theta$, we have:

By dividing both sides by $mg \cos\theta$, we get the relationship:

3. Show the solution

Given:

• Angle of inclination ($\theta$) = $30^\circ$
• Formula: $\mu = \tan\theta$

Substitute the value of $\theta$:

From trigonometric ratios:

Therefore, the coefficient of friction is $\frac{1}{\sqrt{3}}$.

4. Identify the answer

Comparing our result to the options provided: A. $\sqrt{3}$ B. $\frac{\sqrt{3}}{2}$ C. $\frac{1}{\sqrt{2}}$ D. $\frac{1}{\sqrt{3}}$

The correct option is D.

5. Tip

Always remember the "Angle of Repose" rule: The coefficient of static friction is equal to the tangent of the angle of inclination at which an object just begins to slide. If you forget the formula, just draw a quick free-body diagram and resolve the forces—it will always lead you back to $\tan\theta$!