A 75kg adult sits at one end of a 9.0m long board.His 25kg child sits on the other end.(a)Where should the pivot be placed so that the board is balanced,ignoring the board's mass?

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Duane Bryant Profile
Duane Bryant answered
To be balanced, weight x length has to be equal on both sides.  Let X be the unknown distance.  X/9 = 25/75  Multiply by 9 to isolate X then do math on right side.  That is one distance.  Subtract that from 9.  That is the other distance.  Where should the pivot be from the adult?  Multiply X x 75.  Does that = (9 - X) x 25?  If it does then that is where the pivot should go.

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