# Dynamics wedge problem

1. Apr 7, 2016

### vaibhav garg

In the adjoining figure if acceleration of M with respect to ground is a, then
A) Acceleration of m with respect to M is a
B) Acceleration of m with respect to ground is asin(α/2)
C) Acceleration of m with respect to ground is a
D) Acceleration of m with respect to ground is atan(α)

The 2nd question in the image below

The acceleration of m in the vertical direction should be atan(α) by using length constant relation and in the horizontal direction it should be same as the wedge that is a. therefore the resultant should be asec(α), what am i doing wrong here ?

Last edited: Apr 7, 2016
2. Apr 7, 2016

### ehild

Are you sure? The wedge moves horizontally and the block slides down along it, so it moves both horizontally and vertically with respect to the ground.

3. Apr 7, 2016

### vaibhav garg

ok I get what you are saying, so if we take A to be the acceleration of the block wrt to M. therefore it's acceleration with respect to the ground will be
(a - Acosα) horizontally and Asinα vertically. But now how do I find a relation between A and a

4. Apr 7, 2016

### ehild

If the wedge moves by x to the left, the length of the rope between the wall and pulley becomes shorter by x.So the other piece along the wedge becomes longer by the same length x. What does it mean for the accelerations a and A?

5. Apr 7, 2016

### vaibhav garg

that they are both equal. so when we add the the two components it would give the answer to be B. Thanks ehlid :)

6. Apr 7, 2016

### ehild

I got a bit different result, but it might have been my error, or a typo in the problem text. Show your work, please.

7. Apr 7, 2016

### vaibhav garg

the horizontal component would be (a-acosα) which is equal to 2asin2α/2 the vertical component 2asin(α/2)cos(α/2) the resultant would be 2asin(α/2)... I am sorry :P but this dosen't matches any of the options

8. Apr 7, 2016

### ehild

I got the same result, and I do not see any flaw. Sometimes there are typos in the written texts.