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Homework Help: Static equilibrium of a hanging, unbalanced beam.

  1. Nov 12, 2011 #1
    1. The problem statement, all variables and given/known data
    a. A uniform beam is hanging from a poing 1.00 m to the right of its center. The beam weights 140 N and makes an angle of 30° with the vertical. At the right-hand end of the beam, a 100-N weight is hung; an unknown weight w hangs at the left end. If the system is in equilibrium, what is w? b. If the beam makes, instead, an angle of 45° with the vertical, what is w?

    2. Relevant equations
    ετ = 0, T = IR

    3. The attempt at a solution
    w(4)(sin150°) - 100(2)(sin30°), but the w I'm getting is if the beam is balanced. How do I calculate it including an angle, so it's not perfectly horizontal? I can't think of an equation that relates the net torque with the change in angle.

    Thank you for any help.
  2. jcsd
  3. Nov 12, 2011 #2
    This question makes no sense !
    There are 2 unknowns.... the length of the rod and the weight 'w' but there is only one equation connecting these together (clockwise moments = anti-clockwise moments)
    No resultant force also implies that the force on the left hand end of the rod should be UP
  4. Nov 12, 2011 #3
    This question makes no sense !
    There are 2 unknowns.... the length of the rod and the weight 'w' but there is only one equation connecting these together (clockwise moments = anti-clockwise moments)
    No resultant force also implies that the force on the left hand end of the rod should be UP
  5. Nov 12, 2011 #4
    I apologize, the length of the beam is 6m. The 100-N weight is 2m to the right of the hanging point, and the other weight "w" is on the left, 4m from the hanging point.
  6. Nov 12, 2011 #5
  7. Nov 12, 2011 #6
    That is much better !! Do you realise that the angle makes no difference at all!! If it is in equilibrium (balanced) it means it will remain in whatever position you place it !!.
    I calculated the MOMENTS of the forces about the pivot point, I ignored the angle because every distance will be multiplied by Sine30 anyway. If you want to do a perfect answer use Sine 30 in your equations.
    Here is what I did:
    clockwise moment = 100 x 2 = 200Nm
    Anticlockwise moment = (140 x 1) + (w x 5)
    clockwise moment = anticlockwise moment

    200 = 140 + 5w
    200 - 140 = 5w
    60 = 5w gives w =12N

    I hope this is what you will get when you include the extra complication of the angle.
  8. Nov 12, 2011 #7
    I think I have given you a big clue about what will happen if the angle is 45.
    A better question would be to ask you what is the tension in the supporting string..... can you do that?
  9. Nov 12, 2011 #8
    how did you get (140 x 1) + (w x 5)? Is there a reason you put all the weight of the beam to the left?
  10. Nov 12, 2011 #9
    The weight of the beam is 140N, it is a uniform beam which means that ALL of its weight acts through the middle of the beam (used to be called the centre of gravity) this means that when you draw this beam in a diagram you MUST draw its weight (140N) as a down arrow at the middle of the beam. The suspension is 1m to the right of the centre of the beam (your diagram shows it quite well) this means that the 140N weight of the beam is Down and 1m from the suspension point. Because it is to the left of the suspension point it is an anti clockwise turning effect (moment) .
    Hope this helps.... ask again if you are not sure
  11. Nov 12, 2011 #10
    I see, makes sense. But how about the (w x 5)? I thought it would be w x 4, for the distance from w to the hanging point.
  12. Nov 12, 2011 #11
    You are on the ball.... and absolutely correct... that is a sloppy mistake by me
    Now I have the moments to be
    clockwise.... 100 x 2 =200
    anticlockwise = (w x 4 ) + (140 x 1) = 4w + 140
    so 200 = 4w + 140
    60 = 4w
    w = 15N

    Well done, that was a sloppy unforgiveable mistake on my part.
  13. Nov 12, 2011 #12
    No problem at all, I understand it better now thanks to you. The only part I don't understand is how a change in the weight would affect the angle. Since you said the angle doesn't really matter, how would the change in angle matter? How does the 15 degree change to 45 degrees relate to the weight of w?
  14. Nov 12, 2011 #13
    First of all.... very well done, if you can cope with someone elses error then you are doing very well. I am sorry about my mistake and not double checking my calculations.
    In your calculations I hope you found yourself using Sine30 because the beam is at 30.
    But I also hope that when you look at your equations this Sine30 is in every expression so it effectively 'cancels out' This does not sound very rigorous or mathematical but I hope you get it.
    Change Sine30 to Sine45.... I bet there is no difference in the final answer.
    Underlying all of this is what is meant in physics by 'equilibrium' it does not simply mean 'balanced' it means it will stay where it is put. Something for you to think about.
    Stay sharp[
  15. Nov 12, 2011 #14
    It makes sense on paper that the angles will just cancel out, but then what changes need to be made to the system to obtain an angle of 45 degrees, from 30 degrees?
  16. Nov 13, 2011 #15
    There is nothing special about any angle.The beam will be balanced at any angle.
    You could get hold of it and move it to 45 (or any angle) and it would stay at that angle
  17. Nov 13, 2011 #16
    That part I don't understand. It's hard to believe that you can hold it at any angle and have it stay there. Are you saying if I held the beam almost vertical, maybe at an angle 10 degrees to the vertical, it would stay that way? There has to be some relation between the weights on the ends and the angle formed. In another case, if the beam was completely horizontal to begin with, if I added a small amount of weight to one end, it would form a slight angle. I doubt if I increased the angle in that case it would stay that way.
  18. Nov 13, 2011 #17
    Look at your diagram and instead of putting the angle = 30 just call the angle ∅
    In equilibrium the sum of clockwise moments = sum of anticlockwise moments
    Take moments about the point where the string is attached to the beam then you will not have to worry about the tension in the string....it will have no turning effect.
    There is only 1 force producing a clockwise moment so
    clockwise moment = 100 x2Sin∅
    There are 2 forces causing anticlockwise moments.... the unknown w and the weight of the beam acting through its centre of mass (the middle since it is a uniform beam)
    So anticlockwise moment = (140 x 1Sin∅) + (w x 4Sin∅)
    So the eqating of moments looks like this:
    100 x 2Sin∅ = (140 x 1Sin∅) + (w x 4Sin∅)
    Do you see that Sin∅ cancels out which means that the angle does not matter
    The equation is now 200 =140 + 4w
    So w = 15N
    I hope this puts you at ease.
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