Question

In the adjoining figure, a wedge is fixed to an elevator moving upwards with an acceleration $a$. A block of mass $m$ is placed over the wedge. Find the acceleration of the block with respect to wedge. Neglect friction.

• A. $(g+a)\sin \theta$ down the wedge
• B. $g\sin \theta$ down the wedge
• C. $(g-a)\sin \theta$ down the wedge
• D. $(a-g)\sin \theta$ up the wedge

Answer 1

The correct option is A $(g+a)\sin \theta$ down the wedge

Free Body Diagram of block with respect to wedge is shown in Figure.
The acceleration would had been $gsin\theta$ (down the plane ). If the lift were stationary or when only weight (i.e., $mg$) acts downwards.
Here, downward force is $m(g+a)$

Therefore acceleration of the block with respect to wedge will be $(g+a)sin\theta$ down the plane.