Q.A rectangular box lies on a rough inclined surface. The co-efficient of friction between the surface and the box is $\mu$. Let the mass of the box be $m$.

Solution

Concept: Static friction opposes impending motion up to its maximum value $f_\text{max} = \mu N$; kinetic friction opposes actual motion with $f_k = \mu N$. On an incline, resolve weight into components parallel and perpendicular to the surface.

Setup

The box of mass $m$ sits on a plane inclined at angle $\theta$ (or $\alpha$) to the horizontal. The normal force is $N = mg\cos\theta$, and the component of weight down the plane is $mg\sin\theta$.

(a) Angle at which the box just starts to slide

At the threshold of sliding, static friction reaches its maximum: $f_s = \mu N = \mu mg\cos\theta$. This must balance the down-plane component of weight: …

Why static friction matters on an incline

When a box rests on a rough inclined plane, gravity tries to pull it down while friction resists. The component of weight parallel to the plane is $mg\sin\theta$, and the normal force is $mg\cos\theta$. Static friction can supply up to $\mu N = \mu mg\cos\theta$ to prevent motion. The box begins to slide precisely when the downhill component equals this maximum friction.

Once the box is moving (or about to move), we must account for kinetic friction opposing the motion. The direction of friction always opposes relative motion or impending motion, so when pushing the box upward, friction acts downward, and vice versa.

(a) Angle at which the box just starts to slide

The box is on the verge of sliding when static friction reaches its maximum value.

1. Resolve forces perpendicular to the plane: The normal force balances the perpendicular component of weight:

2. Resolve forces parallel to the plane:

   The component of weight down the plane is $mg\sin\theta$. Maximum static friction up the plane is $f_{\text{max}} = \mu N = \mu mg\cos\theta$.

3. Condition for impending motion:

   The box just starts to slide when these forces are equal:

Dividing both sides by $mg\cos\theta$:

Therefore:

(b) Net force when the angle is increased to $\alpha > \theta$

When the inclination increases beyond the critical angle, the box accelerates down the plane. Friction now acts upward (opposing the motion) as kinetic friction.

1. Normal force at angle $\alpha$:

2. Kinetic friction (up the plane):

3. Component of weight (down the plane):

4. Net force down the plane: The net force is the difference between the gravitational component and friction:

(c) Force to keep the box stationary or moving up with uniform speed

To move the box upward at constant velocity (or keep it stationary on a plane inclined at $\alpha$), the applied force must balance both the downhill component of weight and friction, which now acts downward.

1. Forces opposing upward motion:

   • Component of weight down the plane: $mg\sin\alpha$
   • Kinetic friction down the plane (opposing upward motion): $\mu mg\cos\alpha$

2. Applied force $F$ up the plane:

   For equilibrium (zero acceleration):

   $F = mg\sin\alpha + \mu mg\cos\alpha = mg(\sin\alpha + \mu\cos\alpha)$ …

Concept: Friction on an Inclined Plane — Threshold, Sliding, and Applied-Force Cases

Static friction resists impending motion up to $\mu N$; once sliding, kinetic friction $\mu N$ acts opposite to the direction of (impending) motion. Resolve forces along and perpendicular to the incline in each case.

Step 1: (a) Angle at which sliding just begins

At the verge of sliding, $mg\sin\theta=\mu\,mg\cos\theta \implies \tan\theta=\mu \implies \theta=\tan^{-1}\mu$.

Step 2: (b) Net force down the plane at α > θ

Kinetic friction $\mu mg\cos\alpha$ now acts up the plane, opposing the sliding:

Step 3: (c) Force to hold the box stationary or move it up at uniform speed

Pushing up the plane, friction acts down the plane (opposing the push), adding to the weight component: …