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HELP Rotational equilibrium material!

  1. Feb 25, 2005 #1
    HELP!! Rotational equilibrium material!

    HELP!!---Okay, I am having some issues grasping the concepts of rotational equilibrium and dynamics. Specifically the equations:

    T=I*alpha
    L=I*omega
    T*delta t= delta L
    W=T*theta
    P=T* omega

    And there are also problems that I can't understand so I don't really know where to begin or if I'm doing them correctly. Heres an example:

    A 700.0N window washer is standing on a uniform scffold supported by a verticle rope at each end. The scaffold weighs 200.0N and is 3.00m long. What is the force in each rope when the window washer stands 1.00m from one end?

    This is what I've got for work:

    L=3.00m
    Theta=90 degrees
    d=1.00m
    Force of window washer (Fw)= 700.0N
    Force os scaffolding (Fs)= 200.0N
    Force of the ropes=Ft

    T=Ft*L - Fw*(L/2) - Fs*d
    T=Ft*3 - 200*(3/2) - 700*1
    T=3Ft - 300 - 700
    -3Ft= -300-700
    -3Ft/-3= -1000/-3
    Ft = 3.33*10^2N

    I have no idea if any of it is right...I tried to follow th example in the book and that was as far as I got. There are a ton of similar problems in the book and on my test. If someone could please help... :cry: thank

    Sidenote: This is my first Forum post so if the math and such is dificult to understand I'm sorry.
     
  2. jcsd
  3. Feb 25, 2005 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    The problem asked "what is the force in each rope". The whole point is that since the window washer is not standing in the center of the scaffold, the force on the two ropes is NOT the same.

    For simplicity, lets say the window washer is 1 meter from the LEFT end rope and, so, 2 meters from the RIGHT end rope. Call the force on the rope on the left side FL and the force on the rope on the right side FR.

    The scaffold weighs 200 N and the window washer weighs 700 N so there is a total force downward of 900 N. Since the window washer and scaffold are NOT going down (that's the point of "equilibrium") the force upward, FL+ FR= 900.

    Now choose any point as a "reference" point and calculate the total 'torque' around that point- since the scaffold is NOT rotating, that must be 0 (that's the "rotational equilibrium"- it's NOT rotating!). There are four points that are of interest: the left end where the force is FL, the place where the window washer is standing, 1 m away where the force is -700 N, the center of the scaffold where the force is -200 N (since the scaffold is uniform, we can take the force at its center) 1.5 m away from the left end, and the right end, 3 m from the left end, where the force is +FR. Since the torque due to each force is the force times the distance from the reference point, taking the reference point to be one of those, means the distance will be 0 and we will have one less term to worry about.

    For example, taking the left end as reference point, and taking "upward" as positive, the total torque is FL(0)-700(1)- 200(1.5)+ FR(3)= 0. That's easy to solve for FR. Once you know that you can use FL+ FR= 900 to solve for FL.
     
  4. Feb 25, 2005 #3
    I think I think I understand how to do it...but where did the 1.5 come from...FL(0)-700(1)- 200(1.5)+ FR(3)= 0? I just want to make sure that I have this right so I can practice some more problems, take the test and only study one portion of physics instead of this and what my class is doing.
     
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