# Motion of Wedge

1. Aug 25, 2011

### athrun200

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

It seems I should do part c before part a because part a requires the time.

And for part d, the distance that wedge moves is less than before, it is correct?

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2. Aug 25, 2011

### kuruman

Check your answer to part (a). It is not dimensionally correct and should depend on the sliding mass too, not just the mass of the wedge. Hint: How far does the CM of the wedge plus particle system travel when the particle reaches bottom?

3. Aug 27, 2011

### athrun200

The center of mass doesn't move, and I obtain the answer
mhcot(phi)/(m+M)

But now I am not understand what is the acceleration.
It seems the only acceleration of the wedge is caused by the particle.
So my method to obtain acceleration should be correct.
But why the acceleration I obtain is not the true acceleration?

4. Aug 27, 2011

### kuruman

Correct.
Correct.
It is.
What do you mean by "true" acceleration? What makes you think that if you did the experiment, the acceleration would be something else?

5. Aug 28, 2011

### athrun200

True acceleration

Sorry, I forget to upload new photos.

My tutor gives me the answer of the and I try to use this answer to find acceleration.

But the result doesn't match my work.
So the 'true' acceleration I said refers to the acceleration obtain via answer.

So my question is

Why my work is wrong?

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6. Aug 28, 2011

### kuruman

To find where you went wrong, we need to backtrack and answer the questions in the order that the problem asks them.

You have correctly found the answer to part (a)

$x=\frac{mhcos\phi}{(m+M)sin\phi}$

What is you answer to part (b), the speed of the wedge when the particle reaches bottom and how did you get it? You need the speed to find the time.

7. Aug 28, 2011

### athrun200

By using K.E.=P.E. and conservation of p I obtain this. (The steps are too long and I am sure it is correct because I have the ans.)

So what's wrong with my work?

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8. Aug 28, 2011

### kuruman

OK. Now what equation did you use to find the time and what is that time?

9. Aug 28, 2011

### athrun200

My new attempt to obtain the time.

Can you tell me why the acceleration is wrong?

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10. Aug 28, 2011

### Staff: Mentor

It's a bit hard to follow what you are doing. What does F = mg sinΦ signify? And Fx = mg sinΦ cosΦ? Is that supposed to be the x-component of the normal force? (It's not.)

The only force the mass and wedge exert on each other is the normal force. But note that since the wedge accelerates, you'll have to solve for that normal force.

11. Aug 28, 2011

### kuruman

You are trying to get the acceleration from a free body diagram. You should in principle get it that way if you draw separate FBDs for the wedge and particle and not assume that they have the same acceleration. Your equation (M+m)a = mg cosφ sinφ assumes that. An easier way to find the time (and that is why the questions are sequenced that way) is to use
$\Delta x=\frac{1}{2}(v-v_0)t$ because you already know the displacement and final velocity from parts (a) and (b).

Also, I don't seem to get the factor (M+m) in the denominator under the radical for the speed expression. I just get M. Are you sure it is in the answer that you have?

12. Aug 28, 2011

### athrun200

Oh! Thanks a lot!
I understand now.

13. Aug 28, 2011

### athrun200

My tutor send me a full solution.
Maybe you can take a look.

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14. Aug 28, 2011

### kuruman

Yes, of course. I forgot the relative velocity.

15. Aug 28, 2011

### athrun200

Although you have given me a method which can avoid the value of acceleration.

I still want to know how to obtain the correct acceleration from FBD.

I can also obtain Fx = mg sinΦ cosΦ from the normal force

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16. Aug 28, 2011

### Staff: Mentor

The normal force is not mg cosΦ. That would be true if the wedge were fixed, but not true if it is allowed to accelerate.

17. Aug 28, 2011

### kuruman

I was afraid of that.
I will elaborate on what Doc Al suggested. The normal force is no longer mgcosφ when the wedge accelerates. It is mgcosφ when the wedge is at rest with respect to the Earth, i.e. in the limit when the wedge's mass goes to infinity. As I said, you need to draw two separate free body diagrams, with the two masses having different accelerations, and then find what the normal force ought to be in order to be consistent with these accelerations.

18. Aug 28, 2011

### athrun200

Do you thing it is possible to find acceleration first?

I want to know it because I want to satisfy my curiosity.

I try it all day long but I finally fail.... I feel bad

19. Aug 28, 2011

### kuruman

Curiosity is good.

Yes, you can find the acceleration first. Draw two free body diagrams, one for the particle and one for the wedge, then apply Newton's Second law. You should get three equations that count, two are for the x and y motion of the particle and one for the x motion of the wedge. There are three unknowns, the acceleration of the wedge, the acceleration of the particle and the normal force between the wedge and the particle. There is a fourth equation that will give you the normal force exerted by the surface on the wedge, but that does not relate to the questions asked by the problem.

If you really want to do this, you will have to post your diagrams and equations so that I can check them and help you along.

20. Aug 28, 2011

### ehild

The normal force can be obtained by using the constraint that the particle moves along the slope. Let be its relative position with respect to the bottom point of the wedge (xr,y). Then y/xr=tan(φ) must hold. In the frame of reference in rest, x is the position of the particle and Xw is that of the wedge. xr=x-Xw and x-Xw=y/tan(φ). The same equations holds for the accelerations: ax, ay(particle) Aw(wedge): ax-Aw=ay/tan(φ). Write the three equations for the acceleration components in terms of the normal force N. You have four equations with four unknowns.

ehild