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athrun200
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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 said: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?
Correct.athrun200 said:The center of mass doesn't move, and I obtain the answer
mhcot(phi)/(m+M)
Correct.athrun200 said:But now I am not understand what is the acceleration.
It seems the only acceleration of the wedge is caused by the particle.
It is.athrun200 said:So my method to obtain acceleration should be correct.
What do you mean by "true" acceleration? What makes you think that if you did the experiment, the acceleration would be something else?athrun200 said:But why the acceleration I obtain is not the true acceleration?
Answerkuruman said: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?
kuruman said: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)
[itex]x=\frac{mhcos\phi}{(m+M)sin\phi}[/itex]
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.
Doc Al said: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 said: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
[itex]\Delta x=\frac{1}{2}(v-v_0)t[/itex] 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?
Yes, of course. I forgot the relative velocity.athrun200 said:My tutor send me a full solution.
Maybe you can take a look.
kuruman said:Yes, of course. I forgot the relative velocity.
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 was afraid of that.athrun200 said: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 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 said:I can also obtain Fx = mg sinΦ cosΦ from the normal force
kuruman said: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.
Curiosity is good.athrun200 said: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
ehild said: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
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 said: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.
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