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Weight/Torque of a beam

  1. Jul 5, 2015 #1
    1. The problem statement, all variables and given/known data
    I have a 100cm beam with the fulcrum at the 20cm point. At the 0cm point is 300g which puts the beam into balance.

    2. Relevant equations
    How can I find the total weight of the beam and the torque due to the weight of beam using the given information.

    3. The attempt at a solution
    I'm so lost, I've been trying to figure this out for a week, I can't grasp how to solve for the weight. I know the center of gravity plays a part, but I'm just so numb at this point, I've become quite frustrated with myself.
  2. jcsd
  3. Jul 5, 2015 #2


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    Hello, Jackie. Welcome to PF!

    Yes, center of gravity plays a part. The important thing to remember is that the center of gravity of an object is the point where you can consider the weight of the entire object to act for the purpose of finding torques.

    So, in your drawing you need to think about where the center of gravity of the beam is located and then draw the (unknown) weight vector for the beam at that point.
  4. Jul 5, 2015 #3
    the COG should be at 10cm on the left and 60cm on the right, but I still have no clue as to what equation I should be using to determine the weight of the full beam. I understand the right portion of the beam must equal the left portion of the beam plus the 300g. I've gotten as far as knowing that I need to divided the beam into sections, but I just can't come up with an equation.
    Last edited: Jul 5, 2015
  5. Jul 5, 2015 #4


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    You don't need to divide the beam into a left part and a right part. Treat the beam as a whole. The entire beam has one COG point. That's where you want to put the weight of the whole beam.
  6. Jul 5, 2015 #5
    so the cog would be at 50cm, how would I devise an equation to solve for this, and how does the cog play a part in solving for the weight of the beam
  7. Jul 5, 2015 #6


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    Think about why the beam remains stationary. What is the condition that must be met in order for the beam not to rotate about the fulcrum?
  8. Jul 5, 2015 #7
    equal weight on both ends, and a net torque of zero. I just don't know how to show this in equation form.
  9. Jul 5, 2015 #8


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    Not sure what you mean here, but it doesn't sound right. There are no weights on the ends of the beam.

    Yes. That's the equation that you will need to set up.

    First, how would you find the torque about the fulcrum due to just the 300 g mass?[/QUOTE]
    Last edited: Jul 6, 2015
  10. Jul 5, 2015 #9


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    The beam is in equilibrium so the sum of torques about any point is zero. The fulcrum point is a convenient place to calculate torques about because the normal force at the fulcrum does not contribute a torque so can be ignored.

    So you have two torques which are equal and opposite, one due to the hanging weight which is anticlockwise, one due to the self weight of the ruler which is CW.
    T = Fr
    Fweightrweight = Frulerrruler

    The only unknown is Fruler so you can solve it.
    Be aware that while the COG of the ruler is at the mid (50 cm) point of the ruler rruler is not 50 cm..
  11. Jul 5, 2015 #10

    Torque = lever arm x force so just for just the 300g at 0cm to 20cm would be .3kg x .2m = .06, but wouldn't I need to include the 20cm of beam weight as well?
  12. Jul 5, 2015 #11


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    The 20 cm part of the beam will already be included when you treat the beam as a whole. If you really want to, you can break it up into a 20cm part of the beam and an 80cm part of the beam, but this is mathematically equivalent to treating the entire beam as a single object. (It takes more effort, though.) You might as well try it both ways to convince yourself that it doesn't matter :wink:

    Anyway you see that torque from the hanging weight is not zero, thus the torque from the weight of the beam (which depends on it's mass which you want to find) will need to be just right so that the net torque is zero. Can you write this as an equation?
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