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Homework Help: Question about rotational motion and another about man on a beam

  1. Mar 31, 2005 #1
    :mad: #1. Consider the following mass distribution where the xy coordinates are given in meters: 5 kg at (0.0, 0.0) m, 3 kg at (0.0, 4.0) m, and 4 kg at (3.0, 0.0) m. Where should a fourth object of 8 kg be placed so the center of gravity of the four-object arrangement will be at (0.0, 0.0) m?
    I know that it has to be in the bottom left of the xy coordinate just didnt really know how to start the problem.

    #2. A window washer is standing on a scaffold supported by a vertical rope at each end. The scaffold weighs 200 N and the scaffold is 3.1 m long. What is the tension in each rope when the 710 N worker stands 2.1 meters from one end?
    Smaller tension and larger one?
    I know you have the find the summation of the y forces and then the summation of the torque but I dont know whether to subtract 200 and 710. I just need some direction. Don't need an answer.
    Any help would be appreciated.
    Last edited: Mar 31, 2005
  2. jcsd
  3. Mar 31, 2005 #2
    #1: You can ignore the 5kg on teh center, since this is your center of gravity, any mass on it does not affect the center of gravity of the system.

    You have 3kg 4m to the right, and 4kg 3m to the right. You want to place an 8kg object at a distance x so that the CM = (0,0).

    You can solve this with a torque system. Each object exerts a torque around the center of mass (origin):

    [tex] m_1x_1 + m_2x_2 + m_3x_3 = 0 [/tex]

    [tex] m_1 = 4kg, m_2 = 3kg, m_3 = 8kg, x_1 = 4m, x_2 = 3m, x_3 = ? [/tex]

    [tex] (4)(3)+(3)(4)+(8)(x) = 0 [/tex]

    Solving this for x will get you an appropriate distance for the third mass.
  4. Mar 31, 2005 #3
    The 200N force and 710N force are both in the same direction.
  5. Mar 31, 2005 #4
    So is this (0, -3) or (-3, 0).
    So it is 910 N for the direction downward. Where does the man standing 2.1 m from one side come into effect. This problem is really messing with me.
  6. Mar 31, 2005 #5
    My bad on #1, I didnt realize they were on separate axes, I'll work on that in a sec.

    You know the upward torque is the same as the downward torque, if the man was in the center, then he would apply the same torque on each rope. If he is off center, one rope experiences more torque than the other.
  7. Mar 31, 2005 #6
    Yeah I understand that. I just didnt really know how to get the summation of the y forces. I think I multiply 710 by 1.1 and 200 by 1.55. Is this right? I'm really lost on this problem.
  8. Mar 31, 2005 #7
    System in equilibrium:

    [tex] m_nx_n = 0 [/tex]

    Object one exerts a torque in the y direction of magnitude (3kg)(4m) = 12kgm.
    Object two exerts a torque in the x direction of magnitude 12kgm.
    You want to place object 3 at a point where it will cancel all the torque in the x direction and y direction.

    Object 3's torque msut then be :

    [tex] \tau_x = -12kgm [/tex]

    [tex] \tau_y = -12kgm [/tex]

    Using [tex] \tau = mx [/tex], and m = 8kg, find an x that satisfies the above requirement.
  9. Mar 31, 2005 #8
    Where did you get 1.1 and 1.55?
  10. Mar 31, 2005 #9
    Is (-1.5, -1.5) anywhere in the range?

    I thought that you would take half of the beam and then how far the man is from one side 2.1-1. I am really lost on this problem. I have tried now for 3 days and can't figure it out.
  11. Mar 31, 2005 #10
    You tell me. Are the torques in each direction cancelled?
  12. Mar 31, 2005 #11
    Well I divided -12 by 8 and got 1.5. It seems like they should be farther out than that. We did this experiment in lab but that was around 2 weeks ago. We put different masses at different places and figured out the center of gravity.

    SHould i multiply -12 by 8? That number seems like it would be way too big.
    Last edited: Mar 31, 2005
  13. Mar 31, 2005 #12
    [tex] \tau_x = m_1x_1 + m_3x_3 = 0 [/tex]

    [tex] \tau_x = (3)(4) + (8)(x) = 0 [/tex]

    [tex] \tau_y = m_2y_2 + m_3y_3 = 0 [/tex]

    [tex] \tau_x = (3)(4) + (8)(y) = 0 [/tex]
  14. Mar 31, 2005 #13
    x=-1.5, -1.5
  15. Mar 31, 2005 #14
    I guess you are done with me. Thanks for your help though.
  16. Mar 31, 2005 #15
    Yeah, for some reason I thought 3x4=24 and was expecting [-3,-3].

    Now for the ropes.

    [tex] \sum \tau = 0 [/tex], and [tex] \sum Force = 0 [/tex]

    [tex] \tau_{left} = -\tau_{right} [/tex], and [tex] F_{down} (910) = -F_{up} [/tex]

    On the left rope:

    [tex] \tau = F_{man}x_{left} + F_{scaffold}{x_cm} - F_{right}x_{right}[/tex] You can calculate this. The center of mass is the point equidistant from both ropes since the scaffold is uniform. The force of the right rope is the only force stopping the rotation. It is in the opposite direction which is why it is negative.

    Solving that equation and using what you know about the total upward force, try to find the tension in each rope.
  17. Mar 31, 2005 #16
    So the equation should be 710N(1m)+200N(2.1m)-?
  18. Mar 31, 2005 #17
    [tex] \tau = F_{man}x_{left} + F_{scaffold}{x_{cm}} - F_{right}x_{right}[/tex]

    The center of mass would be at 3.1/2 = 1.55m.
    If you want the man to stand at the left end then thats fine, its a matter of choosing one and sticking to it.

    Knowing the total torque is 0,

    [tex] 0 = F_{man}x_{left} + F_{scaffold}{x_{cm}} - F_{right}x_{right}[/tex]

    [tex] 0 = (710N)(1)+(200N)(1.55m)-(F_{right})(3.1} [/tex]

    Solve that for F right. Remember the total upward force is equal to the downward force, and that the upward force is F right + F left
  19. Mar 31, 2005 #18
    Alright, so the smaller force is 329 N. How do I go about finding the larger one?
  20. Mar 31, 2005 #19
    [tex] F_{right} + F_{left} = F_{up} = -F_{down} [/tex]

    Please read more carefully :)
  21. Mar 31, 2005 #20
    So it is 329+ Fleft=-910? and that would give me....581?
  22. Mar 31, 2005 #21
    Yeah I got it....thanks man for all your help....I really appreciate it...Have an awesome day.
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