# Static equilibrium minimum cylinder mass problem

1. Jul 5, 2015

### J-dizzal

1. The problem statement, all variables and given/known data
Determine the minimum cylinder mass m1 required to cause loss of contact at A.

2. Relevant equations
ΣF=0, ΣM=0

3. The attempt at a solution
Im not sure where to go with this one. the solution would be in terms of mass(m)

Last edited: Jul 5, 2015
2. Jul 5, 2015

### Dr. Courtney

I suspect there is a figure missing from the problem statement.

3. Jul 5, 2015

4. Jul 5, 2015

### HelaCopter

I have the same question if anyone made progress on it.

5. Jul 5, 2015

### J-dizzal

what do you have for your equilibrium equations?

6. Jul 5, 2015

### Dr. Courtney

Classic equilibrium problem. The sum of the forces is set equal to zero. The sum of the torques is set equal to zero.

I would focus first on the torques on the angled piece. Compute them all. Add and set them equal to zero.

7. Jul 5, 2015

### J-dizzal

Thats what i tried in my solution, im not sure what is wrong with it

8. Jul 5, 2015

### HelaCopter

I have the same moment equation.

9. Jul 5, 2015

### J-dizzal

i think the weight of the object is equal to m. therefore it follows that the weight of the longer bean is 2/3w and the shorter beam has weight w/3.

this might be used to get the ∑F=0 eqation. if that is even needed.

10. Jul 5, 2015

### Nathanael

Hi J-dizzal.

The problem should've put a 90-degree angle between the L shaped part :\

Is what you call Ax the normal force from the wall? What will this be when contact is lost?
(P.S. it would be perpendicular to the wall, I think I can faintly see you drew it perpendicular and then erased it.)

Is what you call M0 is supposed to be the torque about O due to the weight of L-shaped part? You should be able to find an expression for this (assume the density is uniform).

You also need to find an expression for what you call Tm.

Then you should be good.

edit:
Yes this will come into play in the expression for M0
(Also, this assumes the density is uniform, which you have to assume, but it's good to recognize your assumptions.)

Last edited: Jul 5, 2015
11. Jul 5, 2015

### J-dizzal

yes, Ax=0,
edit
Hi Nathanael,
M0 is the reactant force about point 0. was i supposed to draw that reactant force, i think reactant forces are only drawn for the axis they are restricted?
An expression for Tm; maybe in terms of just mg

Last edited: Jul 5, 2015
12. Jul 5, 2015

### Nathanael

This is the advantage of choosing to take the torques about point O, the force of the hinge produces no torque.
(If you calculated it you would multiply a force by zero distance which=0)
In that case you need to include the torques due to the weight of the L shaped object in your equation

Yes it will be in terms of m1

13. Jul 5, 2015

### J-dizzal

Is Ay correctly shown here? ΣM0= Ax2/3 l + m1/2 cos(34) l/3 + m/3 cos(34) 1/9 l - 2/3 m cos(56) = 0
this seems like it has too many variables

14. Jul 5, 2015

### Nathanael

I'm not sure what you mean, what is Ay?

The dimensions are wrong on all these terms. (May be a typo.)

Like you said Ax is zero so just drop this term.

Can you explain why you used L/9?

You haven't multiplied this term by any distance.

The mass of the cylinder will depend on the the mass of the L-shaped thing, but that should be the only variable left in the end.

15. Jul 6, 2015

### J-dizzal

∑M= m1/2 cos(34) l/3 + 3.3cos(34) l/6 - 6.5cos(56) l/3 =0

3.3N= 1/3m(9.8m/s/s)
6.5N= 2/3m(9.8m/s/s)

16. Jul 6, 2015

### Nathanael

mg/3=3.3m not 3.3 but don't plug in g=9.8 anyway, just put g (you will be able to divide the g out of the equation).

It's looking better. The m1 term should also have a factor of g, and for some reason you left "m" out of the equation. Other than that it is good.