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Rigid body dynamics

  1. Oct 18, 2007 #1
    1. The problem statement, all variables and given/known data
    A 3.0 m long rigid beam with a mass of 130 kg is supported at each end. A 65 kg student stands 2.0 m from support 1. How much upward force does each support exert on the beam?

    http://i241.photobucket.com/albums/ff4/alg5045/p13-56.gif


    2. Relevant equations



    3. The attempt at a solution

    I tried doing Fnet = N1 - W - N2 = 0, but that didn't work.
     
  2. jcsd
  3. Oct 18, 2007 #2
    If you have to determine total upward force on the beam (N1+N2) than you have to equate net force on the beam = 0. In the equation written, you have missed out one of the weights out of the weight of the man and weight of the beam (unless W represents sum of the two weights). However, if you need N1 and N2 separately, then take moments of the forces about two supports and then solve for N1 and N2.
     
  4. Oct 18, 2007 #3
    I was using a similar example in the book, which said that Fnet = n1 - w - n2 = 0. Then it said that n2 = d1w/(d1 + d2). I tried 1m ((130+65)9.8)/1+2, but that was wrong.
     
  5. Oct 19, 2007 #4

    learningphysics

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    The normal forces act upward...

    so N1 + N2 - w = 0

    now take the torque about any point of your choosing... set the torque = 0... that gives a second equation. choosing the right point can make the math a little easier...

    2 equations, 2 unknowns N1 and N2... you can solve for both.
     
  6. Oct 20, 2007 #5
    I still don't understand.
     
  7. Oct 20, 2007 #6

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    What is the torque about the left support in terms of the forces and distances...?
     
  8. Oct 20, 2007 #7
    T = rF, and the F is N1 = d2w/(d1+d2)?
     
  9. Oct 20, 2007 #8

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    How did you get the N1 = d2w/(d1+d2)? I don't know if that's right or not...

    N1 exerts no torque about the left support... because the distance from the support is 0.

    start with the basics.

    Torque about the left support = w*2 - N2*3 (taking clockwise positive counterclockwise negative).

    do you see how I got this?

    set this equal to zero:

    0 = w*2 - N2*3

    you also have the equation

    0 = N1 + N2 - w (sum of forces in the y-direction must be 0)

    solve these 2 equations to get N1 and N2.
     
  10. Oct 20, 2007 #9
    I think I see how you got the equation. Is the weight (mass of beam + mass of student)g?
     
  11. Oct 20, 2007 #10

    learningphysics

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    oh... I forgot the beam... I was using w for the student...

    torque = (rod's mass)*g*1.5 + (student's mass)*g*2 - N2*3

    0 = (rod's mass)*g*1.5 + (student's mass)*g*2 - N2*3
     
  12. Oct 20, 2007 #11
    For N2 I got 1061.667. To find N1 I know I use 0 = N1 + N2 - w, but what is w? Is it the same as above?
     
  13. Oct 20, 2007 #12

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    I messed that up too just like the moment equation... I forgot the weight of the student... but you can fix the equation... we need the sum of the forces in the vertical direction... take up as positive down as negative...

    what equation do you get?
     
  14. Oct 20, 2007 #13
    The sum of the forces in the vertical direction would include the two normal forces, but I'm unsure about how to factor in the weight. I imagine torque is going to be involved.
     
  15. Oct 20, 2007 #14

    learningphysics

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    No torque is only for the torque equation...

    Just simply add up the forces taking up as positive, down as negative... no tricks.
     
  16. Oct 20, 2007 #15
    If there are no tricks, then F = N1 + N2 - (mass of the beam + mass of the student)g.
     
  17. Oct 20, 2007 #16

    learningphysics

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    yes, exactly... this has to add to zero.

    N1 + N2 - (mass of the beam + mass of the student)g = 0.

    In all these statics (no moving parts) problems... you'll generally use 3 equations.

    [tex]\Sigma F_x = 0[/tex] (sum of forces in the x-direction = 0)
    [tex]\Sigma F_y = 0[/tex] (sum of forces in the y-direction = 0)

    and

    net torque about any point = 0. (you are free to choose any point).
     
  18. Oct 20, 2007 #17
    Awesome. See, I can do the forces part....just not the torque.
     
  19. Oct 20, 2007 #18

    learningphysics

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    think force*distance, and clockwise/counterclockwise... you'll get used to it. Did you solve both N1 and N2?
     
  20. Oct 20, 2007 #19
    Yep, N2 = 1061.667 and N1 = 849.333
     
  21. Oct 20, 2007 #20

    learningphysics

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    cool. that's what I get.
     
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