Static equilibrium minimum cylinder mass problem

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Homework Statement


Determine the minimum cylinder mass m1 required to cause loss of contact at A.

20150705_173930_zpstbaku2nz.jpg

Homework Equations


ΣF=0, ΣM=0

The Attempt at a Solution


Im not sure where to go with this one. the solution would be in terms of mass(m)
20150705_170558_zpslutfdwqc.jpg
 
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Answers and Replies

  • #2
Dr. Courtney
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I suspect there is a figure missing from the problem statement.
 
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  • #3
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I suspect there is a figure missing from the problem statement.
oops, uploading now
 
  • #4
I have the same question if anyone made progress on it.
 
  • #5
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I suspect there is a figure missing from the problem statement.
I have the same question if anyone made progress on it.
what do you have for your equilibrium equations?
 
  • #6
Dr. Courtney
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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
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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.
Thats what i tried in my solution, im not sure what is wrong with it
 
  • #8
Thats what i tried in my solution, im not sure what is wrong with it
I have the same moment equation.
 
  • #9
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I have the same moment equation.
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
Nathanael
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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:
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.
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.)
 
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Hi J-dizzal

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 assume the density is uniform, which you have to assume, but it's good to recognize your assumptions.)
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
 
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  • #12
Nathanael
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yes, Ax=0,
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?
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

An expression for Tm; maybe in terms of just m1
Yes it will be in terms of m1
 
  • #13
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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
Nathanael
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Is Ay correctly shown here?
I'm not sure what you mean, what is Ay?

m1/2 cos(34) l/3 + m/3 cos(34) 1/9 l - 2/3 m cos(56)
The dimensions are wrong on all these terms. (May be a typo.)

Ax2/3 l
Like you said Ax is zero so just drop this term.

m/3 cos(34) 1/9 l
Can you explain why you used L/9?

- 2/3 m cos(56)
You haven't multiplied this term by any distance.

this seems like it has too many variables
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
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∑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
Nathanael
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∑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)
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.
 
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