Mechanics problem Watch

Gaz031
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#1
Report Thread starter 13 years ago
#1
Any help would be greatly appreciated.

A small box of mass m_{1} is sitting on a board of mass m_{2} and length L. The board rests on a frictionless horizontal surface. The coefficient of static friction between the board and the box is \mu_{s} and the coefficient of kinetic friction between the board and the box is, as usual, less than \mu_{s}
Find F_{min}, the constant force with the least magnitude that must be applied to the board in order to pull the board out from under the box.

And a minor unrelated check, am I right in thinking that the sum of the angles of a geodesic n-gon on the sphere of radius R is (Area of n-gon)/R² + (n-2)π?

Thank you.
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dvs
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For the first question, do you mean the smallest force that can be applied without causing the box to slip?

And for the second question, I'm not entirely sure, but I think you're right.
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Gaz031
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Report Thread starter 12 years ago
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(Original post by dvs)
For the first question, do you mean the smallest force that can be applied without causing the box to slip?
No - such a force would be 0 and so not very useful.

And for the second question, I'm not entirely sure, but I think you're right.
Thanks

As usual, reputation is on offer.
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dvs
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I meant the force required to pull the board out without causing the box to slip. Anyway, I think I know what you mean now.

Try to think of it as if the box is moving while the board remains at rest. Then we find that the acceleration of the board is a = (u_s m1 g)/m1 = u_s g. This is the acceleration required for the box to slide off. So now we apply N2L, and note that kinetic friction is equal to static friction for the minimum force:
F_{min} - f = u_s m2 g
F_{min} = (m2 + m1)g u_s
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Gaz031
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Report Thread starter 12 years ago
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Thanks for that, I think I understand now.
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