A block of mass m is placed on a smooth wedge of inclination \(\theta\) . 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 (g is acceleration due to gravity) will be
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The wedge is given an acceleration to the left.
\(\cdot\) The block has a pseudo acceleration to the right, pressing against the wedge because of which the block is not moving.
\(\therefore m g \sin \theta = m a \cos \theta\)
\(\text{or } a = \frac{g \sin \theta}{\cos \theta} \\ \text{Total reaction of the wedge on the block is} \\ \text{N} = mg \cos \theta + ma \sin \theta \\ \text{or N} = mg \cos \theta + \frac{mg \sin \theta \cdot \sin \theta}{\cos \theta}. \\ \text{N} = \frac{mg (\cos^2 \theta + \sin^2 \theta)}{\cos \theta} = \frac{mg}{\cos \theta}\)
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