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Hanging masses and tensions

  1. Feb 27, 2008 #1
    [SOLVED] Hanging masses and tensions....

    1. The problem statement, all variables and given/known data
    attached is drawing of diagram...

    two masses m1 and m2 are attached as shown in the figure. If the system is in equilibrium, find the angle labled theta.
    m1 = 35.0 kg
    m2 = 65.0 kg




    2. Relevant equations



    3. The attempt at a solution
    well...to start i drew three FBD one for the mass, one for the intersection, and one for the second mass...
    for the first FBD:
    Tm = m * g

    for the second FBD:
    Fx = T2 cos 53 - T1 cos 37 = 0
    Fy = T2 sin 53 + T1 sin 37 - m1*g = 0

    for the third FBD:
    Fx = T3 cos (theta) - T2 cos 53 = 0
    Fy = T3 sin (theta) - T2 sin 53 - m2*g = 0

    so...

    First:
    Tm = m*g; which equals (35.0kg) ( 9.80 m/s^2) = 343N

    Second:
    .602 T2 - .799 T1 = 0 ; so... T1 = (.602/.799)(T2) so...

    .602 T1 +.799 T2 = 343
    .602[((.602)/(.799))T2 + .799 T2 = 343
    .453T2 + .799T2 = 343
    1.25T2 = 343
    T2 = 274.4N

    Third:
    T3=T2(cos 53/cos (theta))

    [T2(cos 53/cos (theta))] sin(theta) - T2 sin 53 = m2*g
    T2 tan(theta) cos 53 = m2*g + T2 sin 53
    tan(theta) = m2*g + T2 sin 53/T2 cos 53
    theta = tan^-1(m2*g + T2 sin 53/T2 cos 53)
    theta = 79.1?


    uhmm any help? thats what i got correct me if i am wrong
     

    Attached Files:

  2. jcsd
  3. Feb 27, 2008 #2
    anyone?
     
  4. Feb 27, 2008 #3

    Doc Al

    User Avatar

    Staff: Mentor

    Looks good to me. Well done!
     
  5. Feb 27, 2008 #4
    thank you much...i have more to come so sit tight!
     
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