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Question: A block of mass m is placed on a smooth wedge of inclination \(\theta\) . The whole system is accel...

A block of mass m is placed on a smooth wedge of inclination θ\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

A

mgcos⁡θm g \cos \theta

B

mgsin⁡θm g \sin \theta

C

mgm g

D

mg/cos⁡θm g / \cos \theta

Answer

mg/cos⁡θm g / \cos \theta

Explanation

Solution

When the whole system is accelerated towards left then pseudo force (ma) works on a block towards right.

For the condition of equilibrium

mgsin⁡θ=macos⁡θm g \sin \theta = m a \cos \theta ⇒ a=gsin⁡θcos⁡θa = \frac { g \sin \theta } { \cos \theta }

∴\therefore Force exerted by the wedge on the block

R=mgcos⁡θ+masin⁡θR = m g \cos \theta + m a \sin \theta

R =mgcos⁡θ+m(gsin⁡θcos⁡θ)sin⁡θ= m g \cos \theta + m \left( \frac { g \sin \theta } { \cos \theta } \right) \sin \theta =mg(cos⁡2θ+sin⁡2θ)cos⁡θ= \frac { m g \left( \cos ^ { 2 } \theta + \sin ^ { 2 } \theta \right) } { \cos \theta }

R =mgcos⁡θ= \frac { m g } { \cos \theta }

mgsin⁡θ=macos⁡θm g \sin \theta = m a \cos \theta ⇒ a=gsin⁡θcos⁡θa = \frac { g \sin \theta } { \cos \theta }

R=mgcos⁡θ+masin⁡θR = m g \cos \theta + m a \sin \theta

R =mgcos⁡θ+m(gsin⁡θcos⁡θ)sin⁡θ= m g \cos \theta + m \left( \frac { g \sin \theta } { \cos \theta } \right) \sin \theta =mg(cos⁡2θ+sin⁡2θ)cos⁡θ= \frac { m g \left( \cos ^ { 2 } \theta + \sin ^ { 2 } \theta \right) } { \cos \theta }

R =mgcos⁡θ= \frac { m g } { \cos \theta } )