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Statics of Rigid Bodies

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  1. Jul 1, 2015 #1
    How can I identify the direction of a force in finding its moment? :rolleyes:
     
  2. jcsd
  3. Jul 1, 2015 #2
    Do you know that magnitude of the moment? The direction of the moment? The moment arm? The magnitude of the force?
     
  4. Jul 1, 2015 #3
    I can hardly elaborate my question. :frown: I mean when you have to find its components and get the moment using its components. It's force x distance right? But the negative and positive thing there. It confuses me.
     
  5. Jul 1, 2015 #4
    Are you familiar with the cross-product and the right hand rule?

    It sounds like you need to think about the basic concept of what a moment is.

    Have you ever tried to push open a door by pressing on the door very near the hinge of the door? Did you find that to be difficult? What happens when you press on the door further away from the hinge?
     
  6. Jul 1, 2015 #5
    Yes we had that discussion in my physics 1 class. I have here an example. I hope this could help us understand each other.
     

    Attached Files:

  7. Jul 1, 2015 #6
    It's 2D.
     
  8. Jul 1, 2015 #7


    So, ##M = \sum(r \times F)## The object tends to rotate about the axis of the moment vector.
     
  9. Jul 1, 2015 #8
    Sir, thank you so much for the help.
     
  10. Jul 1, 2015 #9
  11. Jul 1, 2015 #10
    Thank you. :)
     
  12. Jul 1, 2015 #11




    Hopefully these videos help.
     
  13. Jul 1, 2015 #12
    That is really simple question. Firstly, you should find the moment of point "O" ,that is zero, to find Fx and Fy. Finally, F is sqrt[(Fx)^2 + (Fy)^2].
     
  14. Jul 1, 2015 #13
    Do you know what the moment about O is or what the magnitude of the (3,4,5) force is? Or do you need to find the magnitude of this force so that the moment about O would be equal to 0?

    The total moment about O is the sum of the moments about O for each force. The first step is to find the moment arm between O and the point at which the forces are being applied. Then you do cross-products.
     
  15. Jul 1, 2015 #14
    This was it. I would always have hard time in determining the negative and positive thing when it comes to problems like this one.
     

    Attached Files:

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  16. Jul 2, 2015 #15
    Just take it one step at a time, and make sure each step makes sense.

    In this problem, if you were going to find the moment about A, it would just be the sum of the moments of each force about A. The first thing to do would be to use the right hand rule to make sure you understand the orientation of the moment about A of each individual force. For example, by the right hand rule, you can tell that the force Q produces a counter-clockwise (CCW) moment about A. The upwards 80 lb force would produce a CCW moment about A, but the downwards 80 lb force would produce a CW moment about A. Also, the downwards 80 lb force would produce a moment about A of greater magnitude than the moment produced by the upwards 80 lb force. Can you see why? Can you figure out the orientation (CW or CCW) of the moments produced by the rest of the forces?
     
  17. Jul 2, 2015 #16
    Take a look these below.

    1.JPG 2.JPG 3.JPG 4.JPG 5.JPG 6.JPG
     
  18. Jul 2, 2015 #17
    Thank you tho. :) We had this problem in our class for Mechanics and fortunately I was able to finally get it. :biggrin: Salamat for your help. :)
     
  19. Jul 2, 2015 #18
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