Two boxes on two slopes

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  • #1
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Homework Statement


http://img15.imageshack.us/img15/407/draw.th.png [Broken]

Initially m and m' are at rest and at the same height. The planes are perpendicular. Find the minimal distance that separates m and m'.


Homework Equations




The Attempt at a Solution


Well the distance that separates them is the length of the vector that describes m 's position relative to m'. But then I fail to find any idea on how to get to that vector.

Thanks for your help.
 
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Answers and Replies

  • #2
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square root of (a cos (alpha)2 + a sin ( alpha) 2)
 
  • #3
AEM
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Homework Statement


http://img15.imageshack.us/img15/407/draw.th.png [Broken]

Initially m and m' are at rest and at the same height. The planes are perpendicular. Find the minimal distance that separates m and m'.


Homework Equations




The Attempt at a Solution


Well the distance that separates them is the length of the vector that describes m 's position relative to m'. But then I fail to find any idea on how to get to that vector.

Thanks for your help.
Make you origin at the intersection of the two planes; choose the y axis along the line from O to m' and the x axis along the line from O to m. Then m is located at
[tex] - a cos(\alpha) \mathbf{i} [/tex]

and m' is located at

[tex] a sin(\alpha) \mathbf{j} [/tex]

Then you can figure out the vector from m to m'.

I'm not sure how the previous post helps you since the a factors out and

[tex] sin^2(\alpha) + cos^2(\alpha) = 1 [/tex]

leaving you with something you already knew.
 
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  • #4
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Thanks for the replies.
I'm sorry, but I think I didn't explain myself correctly.
The thing is the boxes move down the slopes, so at some point either
- m hits O
- m' hits O
(whichever happens first)
So at that moment mm' has a certain length, say d. What I'm looking for is that length which probably is a function of α and a.
 
  • #5
rl.bhat
Homework Helper
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First of all find the component g along the inclined planes, and the lengths of the inclined planes. From the figure it is clear that m hits O first. From the component of g and the
length of the inclined plane, find time taken by m to reach O. Using this time find how much m' moves downward. Knowing the length of the inclined plane containing m', you can find the distance mm'.
 
  • #6
Gokul43201
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From the figure it is clear that m hits O first. From the component of g and the
length of the inclined plane, find time taken by m to reach O.
This is not the right way to solve this problem. First of all, one does not know which block hits O first since the angle \alpha is not specified.

To springo:

1. It does not matter which block hits O first.

2. Can you write and equation that describes the position of m (and hence, also m') as a function of time, t?

3. If you have expressions for the positions of m and m', then you should also be able to deduce an expression for the distance between m and m' as a function of t.

4. What is the last thing you need to do?
 
  • #7
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Well, I guess we could find the expressions of OM and OM' as a function of time using the acceleration or the speed and integrating. Let (O, i, j) be (O, uOM ,uOM')
OM = (a·cos(α) - (1/2)·g·sin(α)·t2) j
OM' = (a·sin(α) - (1/2)·g·cos(α)·t2) i
Is that what you suggested?

Thanks.
 
  • #8
Gokul43201
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Well, I guess we could find the expressions of OM and OM' as a function of time using the acceleration or the speed and integrating. Let (O, i, j) be (O, uOM ,uOM')
OM = (a·cos(α) - (1/2)·g·sin(α)·t2) j
OM' = (a·sin(α) - (1/2)·g·cos(α)·t2) i
Is that what you suggested?

Thanks.
Good. That's exactly correct so far.

Now, isn't there a very simple geometric rule you can use to find the distance (MM') between the two blocks?

PS: Alternatively, if you wish to stick with vectors, you can arrive at this same result by finding the vector MM' out of OM and OM' and then computing its magnitude.
 
  • #9
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Yeah, sure:
d2 = (a·sin(α) - (1/2)·g·cos(α)·t2)2 + (a·cos(α) - (1/2)·g·sin(α)·t2)2
d2 = a2 - a·g·sin(α)·cos(α)·t2 + (1/4)·g2·t4
I had written that down before, but since it wasn't leading me anywhere I thought something was wrong.
 
  • #10
Gokul43201
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I have an extra factor of 2 on the middle term - can you check that?
 
  • #11
Gokul43201
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Next up, you want to find the minimal value of d^2. The way you tackle that depends on whether or not you've had differential calculus. Have you?

If you have, what do you think you should do?
 
  • #12
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d2 = a2 - a·g·sin(2α)·t2 + (1/4)·g2·t4
OK, so since d > 0, the minimum value for d is also the minimum value for d2. So we could take the derivative of that.
d(d2)/dt = 2·a·g·sin(2α)·t + g2·t3
But then I get three roots, the interesting one leaves us with:
gt2 = 4·a·sin(α)·cos(α)
We replace in the first equation and simplify:
d2 = a2·cos2(2α)

d = a·cos(2α)

Is that it? Thanks!
 
  • #13
Gokul43201
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Looks good.

But just for good form you may need to do one final thing to demonstrate that this is indeed a minimum and not a maximum (well, it's obviously not a maximum since acos(2alpha) < a, which was the initial separation, but it's good to develop the practice of identifying the type of optimal value). Do you know how you would do this?
 
  • #14
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The next derivative is f'' (so 2nd, which is even) and it's negative, so it's a maximum. Is that correct?
Thanks again!
 

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