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Finding the center of mass of a rod

  1. Aug 4, 2011 #1
    This is a pretty tough question for me. I am trying to learn, so please no rude people or trolls please.
    Here is the problem I was given.

    A 50 dyne steel rod is fixed at a point 20cm. from the right end. The clamp at this point provides a force of 200 dynes upward. The rod is in equilibrium. Where is the center of mass of the rod if the rod is in equilibrium?

    Finding a place to start was difficult. I am taking it that the 20 cms are hanging off of the right side of the clamp and the longer part of the bar is hanging on the left side of the clamp(with an unknown length). I also believe the 50 dynes is the bar itself and not just the force on one side of the bar.

    I have gone in circles trying to figure this problem out.
    Any help would be greatly appreciated.
    Thank you
     
  2. jcsd
  3. Aug 4, 2011 #2
    If the rod is in equilibrium, then the torque about any point will be 0. If you pick the right end as your pivot, then you have the torque provided by the clamp. The gravity through the center of mass must cancel this torque for the rod to be in equilibrium.
     
  4. Aug 5, 2011 #3

    BruceW

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    I agree with Yuqing.
    Although Drew777, this forum usually requires that you make an attempt at solving the question before you ask for help.
    (I have got in trouble for helping someone that hasn't given an attempt at the problem in the past).
     
  5. Aug 5, 2011 #4
    come on, show some effort
     
  6. Aug 5, 2011 #5
    Umm..... I have been working on this problem for 4 days. Isn't that enough effort? When stuck what else is one supposed to do? Also am I incorrect in assuming the clamp is the pivot point?
     
    Last edited: Aug 5, 2011
  7. Aug 5, 2011 #6

    BruceW

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    You should have put some of your work from the last 4 days in the 'attempt at solution' section of the homework question template - that's what its there for. I don't mean to put you off physics forums, but you should stick to the template. If we just put the solution down for you, then it won't help you learn.

    I'd say you're right about the clamp being the pivot point, since the clamp is the only contact force acting on the rod. Now think about the force due to gravity: where must it act to keep the rod in equilibrium?
     
  8. Aug 5, 2011 #7
    I agree with you. I am taking this course to learn after all. If you just gave me the solution I would actually be a little upset. The point of me posting the problem is so I can better understand it. I have done many different kinds of equations that are pretty much unrelated. If you need to see one I will put the one down I believe is closest.

    (L-20/L) x (L-20/2) = 200 + (20/L x 50) x 10
    25(L-20)^2 = 200L + 10,000
    25(L^2-40L + 400) = 200L +10,000
    25L^2 - 1000L = 200L
    25L^2 = 1200L
    25L= 1200
    L= 1200/25
    L=48
    If that is the whole length then I figure the center of mass would be at 24 cm.
    Now this took me awhile to put together and I am not entirely sure if I did this right.
    I was reading my text book and trying to duplicate some things I saw in there.
    If I got off track can someone please put me back on. Solutions do me no good. I really need to understand this problem.
    Thank you.
     
  9. Aug 5, 2011 #8

    tiny-tim

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    Welcome to PF!

    Hi Drew777! Welcome to PF! :smile:
    I don't understand the question. :confused:

    There must be another force downward on the rod (perhaps a load) …

    are you sure that is the full question?​
     
  10. Aug 5, 2011 #9

    BruceW

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    Usually in this type of problem, you use the principle of moments. Generally, you can specify any point on the rod, then calculate all the moments associated. If the total moment is nonzero, the rod will rotate. So for the rod to be in equilibrium, the total moments around any given point on the rod must be zero.
    A moment is the distance from the point in question to where an applied force acts times by the perpendicular component of that force. i.e
    [tex] moment = distance \times F_{perpendicular} [/tex]
    And then you add up all the moments (anticlockwise moments are positive, clockwise are negative), to get the total moment around the point in question.
    As Yuqing said, the torque around any point is zero, since the rod is in equilibrium. And when he said 'pick the right end as your pivot', he simply means calculate the moments around the point at the right end.
    Generally, you can pick any point on the rod to calculate the moments. I'm guessing you've learned this stuff about moments (a.k.a. torque) already?
    In the specific problem you've been set, there are only two forces on the rod: one upwards from the clamp and one downwards due to gravity. The force due to gravity is at the point of the centre of mass, so you need to find where gravity acts to find where the centre of mass is.
    Using the principle of moments, you can calculate where gravity must act, while keeping the rod in equilibrium. You can calculate the moments around any point, but there is one point which makes the problem much simpler.
     
  11. Aug 5, 2011 #10

    BruceW

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    Re: Welcome to PF!

    I think he means that the force of 200 dyne upward from the clamp is the max possible force, not necessarily the force that it is actually providing...
     
  12. Aug 5, 2011 #11

    tiny-tim

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    but any clamp worth the name is also capable of exerting a torque :wink:
     
  13. Aug 5, 2011 #12

    ehild

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    The rod can be in equilibrium only if the net force on it is zero. 50 unit weight is not balanced by 200 unit upward force. If these are the only forces, the rod will fly up.

    ehild
     
  14. Aug 5, 2011 #13
    I suggest that you first go back to wherever you learn about the center of mass and typing the definition and how to calculate it, then I can help or maybe thats that in which you would have already figured out by then.
     
  15. Aug 6, 2011 #14

    BruceW

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    I guess Drew777's teacher has only very weak clamps that break when they exert any torque :)
     
  16. Aug 7, 2011 #15
    LOL. Thanks everybody. The problem is hard to understand. My teacher is of very little help and won't clear up anything for me. We just now got into dealing with torque. I say we have spent one hour on it so far. What we went over was nothing like this problem, I think that is why I am having some trouble in this class.
     
  17. Aug 8, 2011 #16
    I keep thinking something is missing as well. It is nerve racking.

    You seem to know how to do this problem. Is there nothing missing from it?
    E= sum of
    EF= 200-50 (This should equal zero)
    ET= should equal zero (200)(20)-(50)(x)=0
    Best I can come up with is this.
    -50x=(200)(20)
    4000/-50= -80
    Would 80 be the distance from the right end over to the center of gravity?
    Sorry for all the questions.
     
  18. Aug 9, 2011 #17

    BruceW

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    It all depends on what your teacher means by: "The clamp at this point provides a force of 200 dynes upward."
    If he literally means that the force on the rod due to the clamp is 200 dynes, then the rod cannot be in equilibrium (as you have worked out).
    But since the question also says "The rod is in equilibrium", I assume that he meant that the max force the clamp can produce is 200 dynes.
    Therefore, you should solve for EF=0 to get the actual force exerted by the clamp, then you can find out where the force of gravity must act.
     
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