if 2 children sit on a seesaw and one is twice as heavy as theother, in order to

Question

if 2 children sit on a seesaw and one is twice as heavy as theother, in order to achieve balance the lighter child must sit: a. half as far from the fulcrum b. Twice as far from the fulcrum c. the same distance from the fulcrum d. four times as far from the fulcrum e. None of the above if 2 children sit on a seesaw and one is twice as heavy as theother, in order to achieve balance the lighter child must sit: a. half as far from the fulcrum b. Twice as far from the fulcrum c. the same distance from the fulcrum d. four times as far from the fulcrum e. None of the above

Explanation / Answer

let the first child sit at (0,0) and the second child sit at(l,0).the coordinates of the center of mass of the two childrenis X = (m1 * x1 + m2 * x2/m1 + m2) ----------(1) and Y = (m1 * y1 + m2 * y2/m1 + m2) ---------(2) m1 = 2m2 from (1) and (2) we get X = (2m2 * 0 + m2 * l/2m2 + m2) = (1/3)l and Y = (2m2 * 0 + m2 * 0/2m2 + m2) = 0 or (X,Y) = ((1/3)l,0) In order to achieve balance the lighter child must sit at adistance d given by d = [((l/3) - 0)^2 + (0 - 0)^2]^(1/2) = (l/3) therefore,the lighter child must sit at a distance which is(1/3) times the total length of the seesaw.

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