Classical Mechanics: Forces on two cylinders

  • Thread starter Saennir
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  • #1
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Homework Statement



Two identical, uniform and rigid cylinders, each of radius a and mass m, are laid horizontally at rest inside a rigid box of width w. There is no friction acting at any of the four contacts.

i) Draw a diagram for each cylinder showing the forces acting on it alone, using labels to show which forces, if any, have the same magnitudes. Justify your assignments.

ii) Show that the force R acting upon the uppermost cylinder at the contact point the cylinders is given by [itex]R=\frac{mg}{\sqrt{(2(w/2a)-(w/2a)^2)}}[/itex]

The problem + diagram are shown in the following imgur link. (Parts a and b were unrelated)

http://imgur.com/HCzNA

Homework Equations



[itex]R=\frac{mg}{\sqrt{(2(w/2a)-(w/2a)^2)}}[/itex]

The Attempt at a Solution



The first section of the problem seemed relatively straightforward: the left and right cylinder both have the same downwards force acting on them due to gravity, as well as contact forces between them equal in magnitude but opposite in direction, pointing towards the centre of each cylinder, given in part 2 as R. The cylinder on the left also has reaction forces from the bottom and side of the box, and the cylinder on the right has a reaction force from the side on the right.

I mainly had trouble with the second part of the problem: I understand that for the uppermost cylinder the vertical component of R is equal to the force acting on the cylinder due to gravity, so Rsinθ = mg, and thus R = mg/sinθ. However, i'm unsure of how to find the angle/determine sinθ using the dimensions of the box, w and a.

I thought I could possibly solve it using certain situations, ie when w = 2a, R = mg, and when w = 4a, R = 0, but I wasn't sure if that was the right approach or how to continue.

Thanks for any help.
 

Answers and Replies

  • #2
ehild
Homework Helper
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Hi Saennir, welcome to PF.

Consider the geometry of the problem in the figure attached. You certainly can find how the angle is related to the dimensions of the box.

ehild
 

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Last edited:
  • #3
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Ah, I see. Yeah, I solved it using that, thanks a lot.

Thanks for the welcome!
 

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