Solving Homework Equations: Part A, B, C, and D

[athrun200]

Homework Statement

Homework Equations

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?

 

[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?

 

[athrun200]

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?

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?

 

[kuruman]

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

Correct.

But now I am not understand what is the acceleration.
It seems the only acceleration of the wedge is caused by the particle.

Correct.

So my method to obtain acceleration should be correct.

It is.

But why the acceleration I obtain is not the true acceleration?

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

 

[athrun200]

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?

Answer

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?

 

[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.

 

[athrun200]

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.

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?

 

[kuruman]

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

 

[athrun200]

My new attempt to obtain the time.

Can you tell me why the acceleration is wrong?

 

[Doc Al]

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.

 

[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?

 

[athrun200]

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.

Oh! Thanks a lot!
I understand now.

 

[athrun200]

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?

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

 

[kuruman]

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

Yes, of course. I forgot the relative velocity.

 

[athrun200]

Yes, of course. I forgot the relative velocity.

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.

It's a bit hard to follow what you are doing. What does F = mg sinΦ signify? And 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.

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

 

[Doc Al]

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.

 

[kuruman]

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 was afraid of that.

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

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.

 

[athrun200]

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.

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

 

[kuruman]

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

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.

 

[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 (x_r,y). Then y/x_r=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. x_r=x-Xw and x-Xw=y/tan(φ). The same equations holds for the accelerations: a_x, a_y(particle) Aw(wedge): a_x-Aw=a_y/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

 

[kyrdmenchou]

im having a problem in solving the given project to us.. about motion, falling object, and projectile motion..

 

[ehild]

Have you given up finding the acceleration in the wedge problem?
If you want help, show what you think and what you have done in order to solve the problem. Finding the accelerations in this problem is difficult. If you have problems generally, solve simple things first.

ehild

 

[athrun200]

Have you given up finding the acceleration in the wedge problem?
If you want help, show what you think and what you have done in order to solve the problem. Finding the accelerations in this problem is difficult. If you have problems generally, solve simple things first.

ehild

I have just finished this problem, 4 unknowns, 4 equations and I solve them correctly.
Thanks for everyone's help.

Now I am wondering how to do the final part, part d.
The answer said the since it is interal force, it won't affect the result.
But I don't understand why.

 

[ehild]

Do not try to solve this one by solving for the accelerations first. Friction consumes energy, its slows down the motions and increases the time for the particle to reach the bottom of the slope. But the CM of the whole system does not accelerate, as no external force acts in the horizontal direction.

ehild

 

[kuruman]

I have just finished this problem, 4 unknowns, 4 equations and I solve them correctly.
Thanks for everyone's help.

Now I am wondering how to do the final part, part d.
The answer said the since it is interal force, it won't affect the result.
But I don't understand why.

You found the answer to part (a) by conserving momentum in the horizontal direction and saying that the CM does not move. Does the CM move when you turn friction on? If so, what is the net horizontal force that causes it to move?

 

[athrun200]

Thanks everyone! I finally complete question 1 !