# 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 ??
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.

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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.