A see – saw of length 50 meter is pivoted and balanced at the centre. Kuhu, who weighs 400 N, sits 20 m from centre on the left. She is balanced by Jiya who weighs 600 N. How far is Jiya from the centre?
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Answer:
ANSWER
If the kid of mass 15kg sit at the end he will produce maximum torque about fulcrum. The seasaw will not be balance for any position of kid of mass 10kg. It is clear that the 10kg kid should sit at the end and the 15kg kid should sit closer to the centre. Suppose his distance from the centre is x. As the kids are in equilibrium, the normal force between a kid and the seesaw equals the weight of that kid. Considering the rotational equilibrium of the seesaw, the torque of the forces acting on it should add to zero. The forces are as follows:
(a) (!5kg)g=150N downwards by the 15kg kid.,
(b) (10kg)g=100N downwards by the 10kg kid,
(c) weight of the seesaw, and
(d) the normal force by the fulcrum
Taking torques about the fulcrum
150×x=100×2.5 or x=1.7m
