Question 222: Laws of Motion

Question

A block of mass m is placed on a smooth inclined
wedge ABC of inclination q as shown in the
figure. The wedge is given an acceleration ‘a’
towards the right. The relation between a and q
for the block to remain stationary on the wedge
is
![](https://neetcare-image.cdn.examgoal.net/1qpqahkka2jpaun/9d951315-09dc-4b39-92c6-1d612f82df1f/9c784bf0-938e-11ea-aedc-f1c8d8e0194b/file-1qpqahkka2jpauo-375w.png)

Accepted Answer

At equilibrium
ma cos $	heta $ = mg sin $	heta $
$ \Rightarrow $ a = g tan $	heta $
![](https://neetcare-image.cdn.examgoal.net/1qpqahkka2k9g94/2197ea46-1810-4600-91b6-4ac94112a189/ccd36080-9390-11ea-aedc-f1c8d8e0194b/file-1qpqahkka2k9g95-436w.png)

Answer

$a = {g \over {\cos ec	heta }}$

Answer

$a = {g \over {\cos ec	heta }}$

Answer

$a = g\cos 	heta $

Answer

$a = g	an 	heta $

Laws of Motion

A block of mass m is placed on a smooth inclined wedge ABC of inclination q as shown in the figure. The wedge is given an acceleration ‘a’ towards the right. The relation between a and q for the block to remain stationary on the wedge is

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