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Diving board with two supports and torque due to diver

  1. Nov 15, 2008 #1
    1. The problem statement, all variables and given/known data
    A uniform diving board, of length 5.0 m and mass 53 kg, is supported at two points; one support is located 3.4 m from the end of the board and the second is at 4.6 m from the end (see the figure below). What are the forces acting on the board due to the two supports when a diver of mass 64 kg stands at the end of the board over the water? Assume that these forces are vertical. [Hint: In this problem, consider using two different torque equations about different rotation axes. This may help you determine the directions of the two forces.]

    http://www.webassign.net/grr/chapter-08/fig-019.gif

    2. Relevant equations
    formula derived from image and question


    3. The attempt at a solution
    N1 is the left support and N2 is the left support.
    (sigma)(Torque)= N1-N2-We-Wp then since the right support = o since it is the axis of rotation I solve accordingly for the left support force :

    1.2N1=2.1We+4.6Wp
    N1=(2.1(53)(9.8)+4.6(64)(9.8))/1.2
    N1= 3313.21N or 3.313kN

    I have tried taking the left support to be the axis of rotation and solved accordingly. I always end up with the answer that is correct for the other support(I have the answers from webassign but I never understood it the first time through and I am reviewing for a test) The answer ends up being 2.17 kN for the left support downwards and 3.31kN for the right support upwards. What am I doing wrong and why do I get the answer for the other support? Is it just coincidence? After I find the correct force for the one I know I can just take the summation of the forces and subtract to get the force acting on the other support.
     
  2. jcsd
  3. Nov 15, 2008 #2
    Use the left support (the one farthest away from the diver) as the axis of rotation with torques in the clockwise direction due to the weight of the diver and the weight of the board's center of mass. Then, the counter-clockwise torque will be the right support, equal and opposite in magnitude of the sum of the CW torques. Then, change your axis of rotation to the right support, with a CCW torque due to the left support and the same CW torques.
     
  4. Nov 15, 2008 #3
    Awesome. Thanks. I will just have to be more careful when I set up the force diagrams I guess. After following the directions you gave I got the right answer no problem.
     
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