Torque equilibrium problem with an angled uniform beam

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Homework Statement



A uniform slender beam of mass 10kg and length 1.2m is resting on a corner which is knife edged. The vertical wall on the left is smooth. What is the angle for equilibrium? a = .5m

Homework Equations



I can't figure out this problem for the life of me. Any help in setting up formulas would be great.

The Attempt at a Solution



I thought of using the knife-edge as the pivot point. This would give me a torque for one side of the beam and an opposite and equal torque for the other side. I gave the left side a mass m1 and the right a mass m2 for a total mass of the 10kg. if the length for the portion of the beam with mass m1 was x then the length for portion m2 is 1.2-x. so with it being in equilibrium i get a torque formula: 10(m1)(x/2)sin(theta)-10(m2)((1.2-x)/2)sin(theta)=0. I also know that cos(theta)=.5/x and sin(theta)=(sqrt(x^2-.5^2) but none of this really helps me come up with a system of equations i can solve. I am not sure if i am going in the right direction or not, so any heklp would be absolutely amazing!
 
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Answers and Replies

  • #2
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Sry i meant sin(theta)=(sqrt(x^2-.5^2)/x)
 
  • #3
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Bump cause im desperate lol... and yet not even a reply :(
 
  • #4
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dont break it in 2 parts ... instead find distance of COM using angle
 
  • #5
I like Serena
Homework Helper
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Since really sharp angles don't exist in nature, the edge must be slightly rounded.
In effect the force the edge exerts on the rod will have to be perpendicular to the rod.

Furthermore you have a relation between m1 and m2 that you did not mention yet, because the rod is uniform. And you did not use the total weight of 10 kg yet.

[EDIT]Hold on, I just saw that your torque equation is not correct yet. I'm missing the normal force of the wall.
And which angle did you mean by theta?[/EDIT]

This gives you enough information to solve the torque equation.
 
  • #6
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the angle labeled as theta is theta... between the table top and the beam
 
  • #7
I like Serena
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the angle labeled as theta is theta... between the table top and the beam
Then you mixed up sin and cos.

And perhaps you could call it phi? Phi is closer to the symbol you used in your drawing.
phi is [tex]\phi[/tex] or [tex]\varphi[/tex] and theta is [tex]\theta[/tex] or [tex]\vartheta[/tex]
[edit]Never mind, I see now that you intended the first form of theta[/edit]
 
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  • #8
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Still havent figured out how to get the answer... any1 have like a formula that will work that i can solve. I seriously have no clue as how to solve this still.
 
  • #9
I like Serena
Homework Helper
6,577
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Still havent figured out how to get the answer... any1 have like a formula that will work that i can solve. I seriously have no clue as how to solve this still.
Step one: consider which forces are in play and draw them.

Step two: choose your coordinates

Step three: make a drawing of the system at rest, make a second drawing of the system in movement (one way or the other), and try to deduce what the important things are.

Step four: set up the equations for equilibrium, that is: sum of vertical forces is zero, sum of horizontal forces is zero, and sum of moments is zero.

I'll give you a couple of clues.
1. The wall exerts a normal force perpendicular to the wall.
2. The edge exerts a normal force perpendicular to the rod.
3. The requested solution asks for an angle, so choose the angle for your coordinate system, and eliminate all other variables.

What are the vertical forces and what size do they have (particularly the normal force at the edge)?
What are the horizontal forces and what size do they have (particularly the normal force at the wall)?
What are the moments and what size and sign do they have?

Your moment-formula uses sin theta twice, which should be cos theta.
The other formulas about theta are correct.

Your moment-formula should contain the moment of the normal force exerted by the wall, which is:
+ force . x sin theta

Your moment-formula contains x, which should be eliminated in favour of theta.
 

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