A box is lying on an inclined plane what is the coefficient of static friction if the box starts sliding when an angle of inclination is 60°

The correct answer is 1.732.

When a box rests on an inclined plane and is on the verge of sliding, the angle of inclination at that critical moment is called the angle of repose. At this angle, the component of gravity pulling the box down the slope is exactly balanced by the maximum static friction force acting up the slope.

Step 1: Identify the forces acting on the box

Consider a box of mass m on a plane inclined at angle θ. Two key force components act along and perpendicular to the surface:

- Component of weight along the plane (tending to cause sliding): $mgsin\theta$
- Normal force perpendicular to the plane: $N=mgcos\theta$

Step 2: Apply the condition at the point of sliding

Just at the moment the box begins to slide, the gravitational component along the plane equals the maximum static friction force:

$mgsin\theta ={\mu }_s·N$

Substituting $N=mgcos\theta$:

$mgsin\theta ={\mu }_s·mgcos\theta$

Step 3: Solve for the coefficient of static friction

Dividing both sides by $mgcos\theta$:

${\mu }_s=\frac{sin\theta }{cos\theta }=tan\theta$

This is the key result: the coefficient of static friction equals the tangent of the angle of repose.

Step 4: Substitute θ = 60°

${\mu }_s=tan\left(\mathrm{60°}\right)=\sqrt{3}\approx 1.732$

Conclusion: The coefficient of static friction between the box and the inclined plane is ${\mu }_s=\sqrt{3}\approx$1.732.