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.
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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
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