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    A uniform diving board is 4m long and a weight of 250N.

    It is bolted at one end and has a 1m support underneath and extends 3m over the swimming pool.

    A person weighing 650N stands on the edge.

    a) Calculate the force on the bolts;

    b) Calculate the force on the edge of the swimming pool

    Any help please?

    The answers are 2200N and 3100N
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    If the diving board is uniform, where does its weight act?

    The question is a little unclear. Which edge is the person standing on (two points are both referred to as "the edge" ), and do we need to factor in the reaction force of the ground acting 1m away from the bolts?

    EDIT: I think I understand now. Take the moment around the point where the support ends, 1m away from the bolts. With the bolts acting there, I think the system can be considered to be in equilibrium.
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    (Original post by Iota Null)
    If the diving board is uniform, where does its weight act?

    The question is a little unclear. Which edge is the person standing on (two points are both referred to as "the edge" ), and do we need to factor in the reaction force of the ground acting 1m away from the bolts?

    EDIT: I think I understand now. Take the moment around the point where the support ends, 1m away from the bolts. With the bolts acting there, I think the system can be considered to be in equilibrium.
    Ah right.

    (650 x 3) + 250 (The weight of the board)

    and (650 x 4) + (2 x 250)


    So, for the force on the bolts, you take the moment at the edge and for the force on the edge you take the moments at the bolts?

    :confused:
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    Sorry for the late reply. Yes, that's right.
 
 
 
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