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Calculating forces involved in seesaw motion

  1. Feb 22, 2012 #1
    Let's say i have a plank of wood (seesaw) 2in x 8in x 12ft with approx. mass of 50lbs. The fulcrum is 2.36ft high and is in the center of the board. There is 45lb placed on one end and 22 lb place on the other end. The 22lb end is secured to the ground ready for release. The 45lb-end is 4ft 6in off the ground.

    I know you would have to use the tau=I*α. I also know that i have to get the total tau (torque) by the summation of the moments due to the weights.

    Neglecting friction how would the angular velocity and angular acceleration of the seesaw be determined when it is released? Also how would one calculate time it takes for the 45-end to hit the ground?

    Now lets say that there is a kid with a weight of 45lb instead of the 45lb weight what would be the force which he exerts on the ground to bring him to the height of 4ft 6in?
    Thanks
     
    Last edited: Feb 22, 2012
  2. jcsd
  3. Feb 23, 2012 #2

    OldEngr63

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    Gold Member

    Let's take the real easy question first. At the moment of release, the angular velocity is zero because nothing is moving. Bingo! one item down.

    Presumably you can draw a FBD and sum moments, so we will assume that you can get the left side of the moment sum equation. The moment of inertia has to be considered as the distributed MMOI of the plank about the pivot, plus the MMOI of two concentrated masses, one at each end. You know the weights for each item, so convert these into masses, use the parallel axis theorem, and combine the whole thing to get I. From there get alpha at t=0.

    This should get you started. See if you can't take it from there.
     
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