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Statics: System in equilibrium - determine height (h)

  1. Mar 14, 2012 #1
    1. The problem statement, all variables and given/known data

    Good day to all,
    I have been given a problem in statics, ore specifically a system that is in equilibrium, but a system I am having some trouble with. I have attached (in pdf) the problem's context and the free body diagram I have drawn.

    The data we are give are as follows:
    r = 0.575 m, a=1.05 m, W=4.905 N, Wc = 4.905 N

    2. Relevant equations

    We are asked to find h.

    3. The attempt at a solution

    What I have so far is as follows:

    Using the fact the the hoop is in equilibrium, I obtained the two equations:

    Td cos(θ) - Tccos(α) = 0 (along x-axis: [itex]\rightarrow[/itex])
    Td sin(θ) + Tcsin(α) - W= 0 (along y-axis: [itex]\uparrow[/itex])

    We know W and Tc and so we have two equations with three unknowns so far. Then I thought of using the problem's geometry to obtain a third equation.

    Ans so using the sine law, I obtained:
    [itex]\frac{sin180-α-θ}{a}[/itex] = [itex]\frac{sin(α)}{r}[/itex]

    But when I try solving the system with the values, I get:
    α=90, θ=88.17 and Td = -1* 10-81

    which I am quite certain are wrong. I can't seem to pinoint the third equation to help solve the problem. Any help would be greatly appreciated.

    Thank you.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     

    Attached Files:

  2. jcsd
  3. Mar 14, 2012 #2

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    hi gate13! :wink:

    your method looks ok :confused:

    show us how you got those figures :smile:
     
  4. Mar 14, 2012 #3
    Hello tiny-tim,

    First of all, I wish to thank you for taking time to answer my question. The values I listed above were obtained using a TI Voyage 200 calculator. What I entered (in the calclulator) was the following:
    solve(Td*cos(theta)-Tc*cos(alpha)=0 and Td*sin(theta)+Tc*sin(alpha)-W = 0 and r*(180-alpha-theta) = a*sin(apha), {alpha, theta, Td})
    I just realized I should be entering:
    solve(Td*cos(theta)-Tc*cos(alpha)=0 and Td*sin(theta)+Tc*sin(alpha)-W = 0 and r*sin(180-alpha-theta) = a*sin(apha), {alpha, theta, Td}).
    Upon correcting this, I obtained:
    Td = 5.06 N, alpha = 27.99 degrees, theta = 31.00 degrees.

    Once more I wish to thank you for your help ( and I feel a bit embarassed with respect to my mistake).
     
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