(a) Calculate the individual torques about the fulcrum produced by the weights of the girl and boy where the total distance of the seasaw is 6m, given that Wgirl = 300 N and Wboy = 580 N. Assume the see-saw is not in motion. girl's torque Nm ---Select--- clockwise counterclockwise
boy's torque Nm ---Select--- clockwise counterclockwise
What is the net torque?
(b) Calculate the distance a 580 N boy should sit from the fulcrum.
(c) Calculate the distance a 300 N girl should sit when the boy weighs 400 N. Assume the boy is located at the distance found in part (b).
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(a) Assuming the girl sits at the left end of the seesaw and the boy sits at the right end of the seesaw. The distance each child is from the fulcrum is L/2, where L is the length of the seesaw. The torque of the system is then
T = (L/2)*Wb - (L/2)*Wg
T = (L/2)*(Wb - Wg)
where Wb is the weight of the boy and Wg is the weight of the girl. The torque due to the boy is clockwise and that from the girl counterclockwise.
(b) To balance the system the net torque should be zero. To achieve this, the boy moves to a distance d from the fulcrum.
0 = d*Wb - (L/2)*Wg
Solving for the new distance
d = (L/2)*Wg/Wb
(c) Taking the new weight of the boy to be Wb2. With this new boy sitting at the distance d found above, the girl needs to move to a distance d2 to balance the torque.
0 = d*Wb2 - d2*Wg
Solving for d2
d2 = d*Wb2/Wg
Substituting the distance d in terms of the original boy's weight.
d2 = (L/2)*(Wg/Wb)*(Wb2/Wg)
d2 = (L/2)*Wb2/Wb