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Dynamics homework

  1. May 13, 2007 #1
    Hey all,

    anyone help me verify the answer for the question

    i got

    Va=-5(root 3)i + 3j ft/s and
    Aa=45i-29j ft/s^2
     

    Attached Files:

  2. jcsd
  3. May 13, 2007 #2
    Why not host your image somewhere until it gets approval?
     
  4. May 14, 2007 #3
    for the velocity, i got the j component to be 5. this would give it a speed of 10 which is equal to the speed of the other end which seems reasonable to me due to the symmetry of the problem when the rod is horizontal.
     
  5. May 15, 2007 #4
    eok20: you mind showin me you workin. I'm assumin you got the same for the 'i' component though?
     
  6. May 15, 2007 #5
    yea, i got the same thing as you for the i component. what i did was find an expression involving the distance of end a (call it s_a) to the vertex and the distance of end b to the vertex (s_b). since the distance between a and b is always 2 and the ends stay on those lines, we know that (s_a*cos30 + s_b*cos30)^2 + (s_a*sin30-s_b*sin30)^2 = 2^2. then i differentiated with respect to time and put in 10 for ds_b/dt and 1/cos30 for s_a and s_b.
     
  7. May 15, 2007 #6
    i understood everything u said clearly apart from " then i differentiated with respect to time and put in 10 for ds_b/dt "
     
  8. May 15, 2007 #7
    i differentiated both sides of the equation (s_a*cos30 + s_b*cos30)^2 + (s_a*sin30-s_b*sin30)^2 = 2^2 with respect to time, keeping in mind that s_a and s_b are both functions of time.
     
  9. May 15, 2007 #8
    i'm sorry, ur losing me a bit here. how can u differentiate with respect to time if you dont have a 't' in the whole thing. it's basically a constact.
     
  10. May 16, 2007 #9
    the distance between each end and the vertex is changing over time-- ds_a/dt and ds_b/dt are both nonzero (in general).
     
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