Equilibrium of a Rigid Body: Balancing Forces and Moments

In summary, the forces acting on the masses are: normal force (M*g*cos(90-L)) and downhill force (M*g*sin(L)). The equilibrium state is achieved when the perpendicular forces on the masses become equal.
  • #1
bbal
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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 gravitational 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
I think the third force, i.e. tension between M and m through rod should be also considered.
 
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  • #3
bbal said:
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 gravitational 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 ?
 
  • #4
Steve4Physics said:
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 gravitational force.
 
  • #5
find critical points of the potential energy ##V=V(\phi),\quad V'(\phi)=0##
 
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  • #6
bbal said:
I drew one, but I only included the components the gravitational 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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  • #7
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
bbal said:
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
bbal said:
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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1. What is meant by equilibrium of a rigid body?

Equilibrium of a rigid body refers to the state in which a body is at rest or moving with a constant velocity, with no unbalanced forces acting on it. This means that the body is not accelerating and all the forces acting on it are balanced.

2. How is equilibrium of a rigid body achieved?

Equilibrium of a rigid body is achieved when the sum of all the forces acting on the body is equal to zero and the sum of all the torques (rotational forces) acting on the body is also equal to zero. This means that the forces and torques acting on the body are balanced and there is no net force or net torque acting on it.

3. What is the difference between static and dynamic equilibrium?

Static equilibrium refers to the state in which a body is at rest or moving with a constant velocity, with no unbalanced forces acting on it. On the other hand, dynamic equilibrium refers to the state in which a body is moving at a constant velocity, with no unbalanced forces or torques acting on it. In both cases, the body is in equilibrium, but in dynamic equilibrium, the body is also in motion.

4. How does the center of gravity affect equilibrium of a rigid body?

The center of gravity of a rigid body is the point at which the entire weight of the body can be considered to act. In order for a rigid body to be in equilibrium, the center of gravity must be directly above the point of support. If the center of gravity is not directly above the point of support, the body will be unstable and will topple over.

5. What are some real-life examples of equilibrium of a rigid body?

Some real-life examples of equilibrium of a rigid body include a book resting on a table, a person standing still, and a see-saw with equal weight on both sides. In all of these examples, the forces and torques acting on the body are balanced, resulting in equilibrium.

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