Question

A 60lb. child and a 150lb. adult are each holding 30lb. weights and balancing ona see saw. If the child sits 6 feet from the pivot point, how far from the pivot point must the adult sit?

Answer 1

Take moments about the pivot point

\[\sum \tau=0\]
\[\sum \tau (clockwise) = \sum \tau (anticlockwise)\]
\[rF (child + weights) = rF(adult + weights)\]

Answer 2

I've never worked with imperial units though... Do you usually convert to metres and kilograms?

Answer 3

If so...

m(child and weights) = 90 lb = 40.8 kg
m(adult and weights) = 180 lb = 81.7 kg
distance of child from pivot point = 6 feet = 1.8 m

Answer 4

Torque = rF
where r = radius or distance from pivot point
F = force applied

Answer 5

In this case the force applied is the weight force of the child and weights on one side of the see saw, and the adult and weights on the other side of the see saw

Back to taking moments around pivot point of see saw...

rF (child + weights) = rF (adult + weights)

1.8 * (40.8 * 9.8) = r * (81.7 * 9.8)

\[r=\frac{ 1.8 \times 40.8 \times 9.8 }{ 81.7 \times 9.8 }\]

r=0.899 m

So the adult must sit 0.899 m from the pivot point to balance the see saw