A block is kept on a frictionless inclined surface with angle of inclination 'α'. The incline is given an acceleration a to keep the block stationary. Then a is equal to (a) gcosecα (b) g/tan α (c) gtan α (d) g

Answer: (c). Only mg and normal force N act; block must accelerate with the incline at a. Horizontal: Nsin α = ma. Vertical: Ncos α = mg. a = gtan α, so the answer is (c).

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Connected Bodies & Pulleys

A block is kept on a frictionless inclined surface with angle of inclination $\text{'}\alpha\text{'}.$ The incline is given an acceleration $a$ to keep the block stationary. Then $a$ is equal to

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Answer: (c)

Only $mg$ and normal force $N$ act; block must accelerate with the incline at $a$ .
Horizontal: $N\sin\alpha = ma$ . Vertical: $N\cos\alpha = mg$ .
$a = g\tan\alpha$ , so the answer is (c).

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