# Newton's laws -- Block sliding on a Prism

1. Apr 18, 2016

### Prathamesh

1. The problem statement, all variables and given/known data
All the surfaces shown in fig. are assumed to be frictionless. The block of mass m slides on the prism of mass M which in turn slides backwards on horizontal surface with acceleration of ao. Find the acceleration of smaller block with respect to prism.

2. Relevant equations

3. The attempt at a solution
As we are working from the frame of reference of prism which is non inertial
we have to include a pseudo force in opp. direction of acc of prism.
Now , forces on the block are
1)mg downwards
2)Normal force N
3)pseudo force (mao)
When I resolved these forces using force body diagram , it gave me 2 equations
maocosθ+mgsinθ=ma
i.e aocosθ+gsinθ=a
and
N+maosinθ=mgcosθ

I m stuck here...
How to eliminate N and ao ??
Please help
Ans is
(M+m)gsinθ / M + msin2θ

Last edited by a moderator: Apr 18, 2016
2. Apr 18, 2016

### drvrm

how the prism M moves Ma0 = some force, think about it
i feel a component of N must be responsible for its motion.

3. Apr 20, 2016

### Prathamesh

Is it Nsinθ??

4. Apr 20, 2016

### drvrm

please draw a free body diagram

i think you are correct - M.acceleration of the prism=Nsin (theta)

5. Apr 23, 2016

### atom jana

I am not sure, but here's my two cents:
The angle between the line of acceleration of the prism and the line of acceleration of the block is equal to 90+theta degrees. Use vectors to work out the acceleration of the block with respect to the prism.

6. Apr 23, 2016

### drvrm

so it can be written
-M* a (0) = N sin ( theta)

7. Apr 24, 2016

### ehild

No, M*a0=Nsin(theta).

The block and the prism interact with the normal force of magnitude N. See picture. If $\vec N$ is the normal force acting on the block, then it is $-\vec N$ acting on the prism.
The normal force exerted on the prism has a backward component, accelerating the prism horizontally, to the left. was right.

#### Attached Files:

• ###### upload_2016-4-24_7-38-20.png
File size:
50.8 KB
Views:
40
8. Apr 24, 2016

### drvrm

if a0 is taken as +ve directed on the right side , then naturally
-Ma0 can be seen as vector going to the left.

9. Apr 24, 2016

### ehild

The force causing this acceleration of the block is -N. -Mao=-Nsin(theta), and you can eliminate the minuses.

10. Apr 25, 2016

### Anjum S Khan

Try to do without pseudo forces, which will enhance your understanding further.

11. Apr 25, 2016

### Anjum S Khan

12. Apr 25, 2016

### ehild

You missed the parentheses from the answer. Correctly it is (M+m)gsinθ / ( M + msin2θ)

Together with the equation for the block,
Mao=Nsin(θ)
you have three equations with three unknowns. Just go ahead.

13. Apr 25, 2016

### ehild

The problem wants the acceleration of the block with respect to the prism. The OP's method is correct and the solution is easy that way.

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted