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Question
A block of mass m is placed on a smooth wedge of inclination θ. The whole system is accelerated horizontally so that the block does not slip on the wedge. The force exerted by the wedge on the block has a magnitude.
Options
mg
mg/cosθ
mg cosθ
mg tanθ
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Solution
Free-body Diagram of the Small Block of Mass 'm'
The block is at equilibrium w.r.t. to wedge. Therefore,
mg sinθ = ma cosθ
⇒ a = gtanθ
Normal reaction on the block is
N = mg cosθ + ma sinθ
Putting the value of a, we get:
N = mg cosθ + mg tanθsinθ
\[N = mg\cos\theta + mg\frac{\sin\theta}{\cos\theta}\sin\ \theta N=\frac{mg}{\cos\theta}\]
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