A uniform plank of seesaw is 5meter long.a boy weighing40kgf sits 1.5meter from the one end.what will be the distance of a girl weighing 25kgf sit at other end,so as to balance the seesaw.
Distance from each end to the center =
5/2 = 2.5 m.
2.5 - 1.5 = 1.0 m from center.
40*1 = 25*d
d = 1.6 m from center = 0.9 m from the
end.
To balance the seesaw, the torques on each side of the fulcrum need to be equal. Torque is calculated by multiplying the weight of an object by its distance from the fulcrum.
Let's calculate the torque on the side with the boy first:
Torque of the boy = Weight of the boy × Distance of the boy from the fulcrum
= 40 kgf × 1.5 meters
Next, we need to calculate the torque on the other side of the seesaw, where the girl will sit:
Torque of the girl = Weight of the girl × Distance of the girl from the fulcrum
Since the seesaw is balanced, the torques on both sides are equal. Therefore:
Torque of the boy = Torque of the girl
40 kgf × 1.5 meters = 25 kgf × Distance of the girl from the fulcrum
To find the distance of the girl from the fulcrum, we rearrange the equation:
Distance of the girl from the fulcrum = (40 kgf × 1.5 meters) / 25 kgf
Now, plugging in the values:
Distance of the girl from the fulcrum = (40 kgf × 1.5 meters) / 25 kgf
= (60 kgf·m) / 25 kgf
Simplifying:
Distance of the girl from the fulcrum = 2.4 meters
Therefore, the girl needs to sit at a distance of 2.4 meters from the other end of the seesaw in order to balance it.