Newton's Laws of Motion: Edge Cases and Subtle Traps (9)

hard • 3 min read
Tags Newtons Laws Of Motion

Question

A block of mass m=2 kgm = 2 \text{ kg} rests on a wedge of mass M=8 kgM = 8 \text{ kg} inclined at θ=30°\theta = 30°. The wedge is on a frictionless horizontal floor, and the block-wedge surface is also frictionless. If the wedge is pushed horizontally with a force FF such that the block does not slide on the wedge, find FF. Take g=10 m/s2g = 10 \text{ m/s}^2.

Solution — Step by Step

If the block doesn’t slide on the wedge, both move together with the same horizontal acceleration aa. So our system has total mass (m+M)(m+M) and acceleration a=F/(m+M)a = F/(m+M).

Two forces act on the block: gravity mgmg downward and normal NN perpendicular to the inclined surface. The block accelerates horizontally with aa, so the net horizontal force on it must equal mama and net vertical force must be zero.

NN acts perpendicular to the incline, so its components are Nsin⁡θN\sin\theta horizontal (toward the direction of push) and Ncos⁡θN\cos\theta vertical (upward).

Vertical equation: Ncos⁡θ=mg  ⟹  N=mg/cos⁡θN\cos\theta = mg \implies N = mg/\cos\theta.

Horizontal equation: Nsin⁡θ=ma  ⟹  a=gtan⁡θN\sin\theta = ma \implies a = g\tan\theta.

 a=gtan⁡30°=10×13=103 m/s2a = g\tan 30° = 10 \times \tfrac{1}{\sqrt{3}} = \tfrac{10}{\sqrt{3}} \text{ m/s}^2 F=(m+M)a=10×103=1003≈57.7 NF = (m+M)a = 10 \times \tfrac{10}{\sqrt{3}} = \tfrac{100}{\sqrt{3}} \approx 57.7 \text{ N}

Final answer: F≈57.7 NF \approx \mathbf{57.7 \text{ N}}.

Why This Works

The trap here is the no-slip condition. When the block sits on a smooth wedge and the wedge is pushed, the block tends to slide down. Pushing the wedge fast enough makes the inclined normal force “carry” the block along.

The clean way to think: in the wedge’s frame, a pseudo-force −ma-ma acts on the block. The block stays put on the incline only when the component of (g,−a)(g, -a) along the incline cancels — that gives a=gtan⁡θa = g\tan\theta.

Alternative Method

Using the non-inertial frame of the wedge: pseudo-force mama acts on the block opposite to acceleration. Resolve along the incline: mgsin⁡θ=macos⁡θmg\sin\theta = ma\cos\theta, giving a=gtan⁡θa = g\tan\theta directly. Then F=(m+M)aF = (m+M)a.

Common Mistake

Students try to balance forces on the inclined surface using the angle θ\theta as if the block were stationary on the ground. The block is accelerating horizontally, not at rest. The correct approach uses a=gtan⁡θa = g\tan\theta as the no-slip condition — independent of the block’s mass, which often surprises students.

This appeared in JEE Main 2024 Shift 2 with slightly different numbers. The trick: if friction is also given, the answer becomes a range of FF, not a single value.

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