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Cantilever equilibrium problem

  1. Jun 22, 2015 #1
    I am trying to resolve for the forces in the attached diagram. In order for this object to be in rotational equilibrium, the torques of the normal force and the garvitational force would have to be equal. However, this would imply that normal force produced by the fulcrum would be far greater than force of gravity at center of mass. The problem is that for an object to be in mechanical equilibrium, the forces must also be equal. So how can both of these be true? The other issue I am having is that you are allowed to find torques using any axis. If that is true and I set the axis of rotation to be the point of contact with the fulcrum, then there will be a net non zero torque. So how is this situation pssible?
     

    Attached Files:

  2. jcsd
  3. Jun 22, 2015 #2

    Dale

    Staff: Mentor

    So at the attachment of a cantilever beam there is a torque as well as a net force. Without that additional torque, as you mention, there is no static solution. For example, if you attach with a hinge instead of a cantilever then the beam will fall until it hangs below the hinge.
     
  4. Jun 22, 2015 #3
    But then, if a net force is present, how is it possible for the beam to remain in translational equilibrium?
     
  5. Jun 22, 2015 #4

    Dale

    Staff: Mentor

    If it is in equilibrium then there is no net force and no net torque.

    ##F_{normal }=-F_{g}##
     
  6. Jun 22, 2015 #5
    Then that means that there is a net force on the beam. But the beam is in static equilibrium and for that to be the case there cannot be a net torque or net force.
     
  7. Jun 22, 2015 #6

    Dale

    Staff: Mentor

    Sorry, my words are confusing. The net force on the beam is 0. The force at the cantilever attachment is non zero.
     
  8. Jun 22, 2015 #7
    So then if the forces are equivalent, how then can the torques be equal? The normal force has a smaller lever arm than the gravitational force.
     

    Attached Files:

  9. Jun 22, 2015 #8

    Dale

    Staff: Mentor

    A cantilever attachment also supports a torque (aka bending moment). There are other attachments, like a hinge, that do not support a torque. Think of what would happen if you tried to attach a cantilever beam using a hinge. Clearly, that shows that there is a torque at the attachment.
     
  10. Jun 22, 2015 #9
    I am not sure I understand what you mean. But it seems to me that unless a third force is present the conditions for both translational and rotational equilibrium cannot be met. Would this be a more correct diagram?
     

    Attached Files:

  11. Jun 22, 2015 #10

    Dale

    Staff: Mentor

    Usually it is just represented as a torque at the attachment rather than a second force. But yes, the basic idea is that you need an additional degree of freedom in your free body diagram.
     
  12. Jun 22, 2015 #11
    What I don't understand though is that it should require less normal force if the fulcrum were placed at the center of mass. However, for translational equilibrium to hold Fn=-Fg
     
  13. Jun 22, 2015 #12

    Dale

    Staff: Mentor

    It would require the same normal force. It would require less torque.
     
    Last edited: Jun 22, 2015
  14. Jun 22, 2015 #13
    But the torque produced by gravity would have to be constant as both the center of mass and the weight of the object would have to be constant. So if gravity's torque is constant, the torque produced from the normal force would have to be constant too in order for the object to be in rotational equilibrium.
     
  15. Jun 22, 2015 #14

    Dale

    Staff: Mentor

    There is the torque from the normal force, but in addition there is also a torque from the bending moment at the attachment point. You have to account for both. The torque produced from gravity, the torque from the normal force, and the additional torque at the attachment must sum to zero.
     
  16. Jun 22, 2015 #15
    What is the additional torque produced by? How would you describe the additional force?
     
  17. Jun 22, 2015 #16

    Dale

    Staff: Mentor

    The additional torque is produced by the attachment. In order for the beam to qualify as a "cantilever" it must, by definition, have an attachment which supports additional torque.

    You didn't specify the construction of the cantilever, but for example, suppose that the beam is welded in place, then the additional torque would be produced by the metal of the weld.
     
  18. Jun 23, 2015 #17
    I am still having issues with this. When I put in actual numbers I still end up having a net force or net torque. Can you see my work? Screenshot_2015-06-23-18-33-21.png
     
  19. Jun 23, 2015 #18

    Dale

    Staff: Mentor

    So again, I would not treat it as two forces. I would follow the standard practice of treating it as a force and a torque. If you do that then the force is clearly equal to ##-F_g##. The torque can then be calculated by setting the axis at the normal force and calculating the opposite of the torque due to gravity.

    However, if you do choose to represent it as two forces rather than a force and a torque, you can still solve it. First, look at how you have drawn it. Think about the torques produced by each force. Can you see that one of the two forces must point downwards? Can you tell which one?
     
  20. Jun 23, 2015 #19
    Is this how it should be? Screenshot_2015-06-23-21-00-01.png
     
  21. Jun 23, 2015 #20

    Dale

    Staff: Mentor

    Yes. That is the right idea, but I think your numbers are off.
     
    Last edited: Jun 24, 2015
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