Equilibrium of a rigid body

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Homework Statement:
We have two masses (M and m) , which are conncected with a massless rod with a length L and placed on a frictionless slope. Find the equilibrium with respect to Φ(horizontal angle).
Relevant Equations:
τ=rxF
Components of gravitional force on M: normal force:M*g*cos(90-α)M*g*sin(α)
Downhill force: M*g*sin(90-α)=M*g*cos(α)

On m: normal force: m*g*cos(α)
Downhill force: m*g*sin(α)
 

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  • #2
anuttarasammyak
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I think the third force, i.e. tension between M and m through rod should be also considered.
 
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Steve4Physics
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Homework Statement:: We have two masses (M and m) , which are conncected with a massless rod with a length L and placed on a frictionless slope. Find the equilibrium with respect to Φ(horizontal angle).
Relevant Equations:: τ=rxF

Components of gravitional force on M: normal force:M*g*cos(90-L)=M*g*sin(L)
Downhill force: M*g*sin(90-L)=M*g*cos(L)

On m: normal force: m*g*cos(L)
Downhill force: m*g*sin(L)
Can you write terms such as cos(L) and sin(L) when 'L' (on your diagram) is a length?

To emphasise what @anuttarasammyak said, does the rod exert a force on each mass?

Have you drawn the free-body diagrams for each mass ?
 
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Can you write terms such as cos(L) and sin(L) when 'L' (on your diagram) is a length?

To emphasise what @anuttarasammyak said, does the rod exert a force on each mass?

Have you drawn the free-body diagrams for each mass ?
Thank you for pointing out the mistake. Yes, the rod does exert force on the masses. I drew one, but I only included the components the gravitional force.
 
  • #5
wrobel
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find critical points of the potential energy ##V=V(\phi),\quad V'(\phi)=0##
 
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Steve4Physics
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I drew one, but I only included the components the gravitional force.
I'd draw *two* free body diagrams (not "one") - one for each mass. Each diagram should show all (3) forces (not components) acting on the mass, or it's not a free body diagram. It will help you if you include the angles between each each force and the plane.

Once you have your two diagrams, resolve parallel to the plane for each mass. This gives you two equations. If you post your diagrams and/or equations we can check them for you.

From the two equations it's not difficult to get Φ.

You can also do it as suggested by @wrobel using potential energy. But post some working for us to see or we can't help.
 
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I came up with this I hope you can read it. As far as the one with potential is concerned, I'm still thinking about how to start that one.
 

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  • #8
haruspex
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I hope you can read it
A forlorn hope, I fear.
Try again, using a darker pen.
Just draw one mass and label the forces on it. Use labels that index that mass, like ##F_{g1}## for gravitational force on mass 1. That way you can later write equations indexing the other mass and we will all know what you mean without needing another diagram.
Don't write expressions for the forces (m1g cos etc.) on the diagram. Type the equations for those into the post, like ##F_{g1}=...##.

See if you can go beyond that, writing equations that relate the forces.
 
  • #9
Lnewqban
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I came up with this I hope you can read it. As far as the one with potential is concerned, I'm still thinking about how to start that one.
Based on the relevant equation shown in the OP, the forces that you should be concerned with are those perpendicular to the massless rod.

If the rod is in horizontal position, a clockwise moment will exist, due to the perpendicular force Mg being greater than force mg.

Therefore, the rod will naturally adopt a unique angle at which both perpendicular forces become equal and a balance of moments is achieved.
 

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