Two kids are seated on either side of a see-saw. If one of them weighs 32 kg and is seated 3 feet from the fulcrum, how far away from him should the other weighing 30 kg sit to keep the see saw horizontal.
The two kids are sitting on either side of a see-saw. One of them has a mass 32 kg and is seated 3 feet away from the fulcrum. If the see-saw has to remain horizontal, the total torque acting about the fulcrum should be equal to 0.
Torque is the product of force and distance and is given by T = F*d.
The force exerted by the kid with mass 32 kg is 32*9.8 = 313.6 N. As he is seated 3 ft away from the fulcrum the torque exerted is 313.6*3 = 940.8 N*ft. The force exerted by the second kid is 30*9.8 = 294 N. If the distance he sits at from the fulcrum is d ft, the torque created is 294*d N*ft. This should equal 940.8
=> 294*d = 940.8
=> d = 3.2
As the two of them are seated on opposite sides of the fulcrum the second kid should sit 3.2 + 3 = 6.2 ft away from the first kid.
In order for the see-saw to be in equilibrium, the anti-clockwise moment has to be equal to the clockwise moment. Therefore,
32 x 3 = y x 30
So, y = 3.2
The other student should sit 3.2 m away from the fulcrum.
Let another sit d feet away from fulcrum
By principle of weight
32 x 3 =30 x d
d= (32 x 3)/30