JEE Challenger
More from Laws of Motion
PhyMedAdv 2012 • Paper 1

Equilibrium of Block on Inclined Plane under Horizontal Force

Consider a small block having a mass of 0.1 kg0.1\text{ kg} positioned on a stationary inclined plane PQPQ making an angle θ\theta with the horizontal surface. A horizontal force with magnitude 1 N1\text{ N} acts on the block through its center of mass as depicted. Assuming g=10 m/s2g = 10\text{ m/s}^2, the block stays in static equilibrium if:

Question Diagram 1

Options

A

θ=45∘\theta = 45^\circ

Correct
B

θ>45∘\theta > 45^\circ and the force of friction acts on the block directed towards PP.

C

θ>45∘\theta > 45^\circ and the force of friction acts on the block directed towards QQ.

Correct
D

θ<45∘\theta < 45^\circ and the force of friction acts on the block directed towards QQ.

Topics & Concepts

Laws of MotionFrictionNewton's Laws

Step-by-Step Solution

To determine the condition for static equilibrium of the block on the inclined plane PQPQ, let us analyze all the forces acting on the block of mass m=0.1 kgm = 0.1\text{ kg}.

1. Forces Acting on the Block

  1. Weight (WW): Acts vertically downwards through the center of mass. W=mg=(0.1 kg)×(10 m/s2)=1 NW = mg = (0.1\text{ kg}) \times (10\text{ m/s}^2) = 1\text{ N}
  2. Applied Horizontal Force (FF): Acts horizontally to the left. F=1 NF = 1\text{ N}
  3. Normal Force (NN): Perpendicular to the inclined plane PQPQ. N=mgcos⁡θ+Fsin⁡θ=cos⁡θ+sin⁡θN = mg \cos\theta + F \sin\theta = \cos\theta + \sin\theta
  4. Frictional Force (ff): Acts parallel to the inclined plane PQPQ.

2. Force Components Parallel to the Incline

  • Component of weight down the incline (directed towards PP): Fg=mgsin⁡θ=1⋅sin⁡θ=sin⁡θF_g = mg \sin\theta = 1 \cdot \sin\theta = \sin\theta

  • Component of horizontal force up the incline (directed towards QQ): FH=Fcos⁡θ=1⋅cos⁡θ=cos⁡θF_H = F \cos\theta = 1 \cdot \cos\theta = \cos\theta

The net tendency of motion parallel to the incline is determined by comparing FgF_g and FHF_H:

ΔF=Fg−FH=sin⁡θ−cos⁡θ\Delta F = F_g - F_H = \sin\theta - \cos\theta

3. Analysis of Options

  1. Case 1: When θ=45∘\theta = 45^\circ sin⁡45∘=cos⁡45∘=12\sin 45^\circ = \cos 45^\circ = \frac{1}{\sqrt{2}} Fg=FH=12 NF_g = F_H = \frac{1}{\sqrt{2}}\text{ N} Since the parallel components balance each other, no frictional force is required (f=0f = 0). The block naturally stays in static equilibrium. (Option A is correct)

  2. Case 2: When θ>45∘\theta > 45^\circ For θ∈(45∘,90∘)\theta \in (45^\circ, 90^\circ), we have sin⁡θ>cos⁡θ\sin\theta > \cos\theta. Fg>FH  ⟹  sin⁡θ>cos⁡θF_g > F_H \implies \sin\theta > \cos\theta The downward force component towards PP is greater than the upward force component towards QQ. Hence, the block tends to slide down towards PP.

    To prevent sliding and maintain static equilibrium, the static friction force ff must oppose this tendency and act up the incline, directed towards QQ. (Option C is correct)

  3. Case 3: When θ<45∘\theta < 45^\circ For θ∈(0∘,45∘)\theta \in (0^\circ, 45^\circ), we have cos⁡θ>sin⁡θ\cos\theta > \sin\theta. FH>Fg  ⟹  cos⁡θ>sin⁡θF_H > F_g \implies \cos\theta > \sin\theta The upward force component towards QQ is greater, so the block tends to slide up towards QQ. Friction must act down the incline, directed towards PP. (Options B and D are incorrect)

Conclusion

The correct statements are A and C.

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