A father and son sitting on opposite sides of a seesaw balance each other. What is the father's weight if he is seated at 0.5m away from fulcrum and son is seated at 1.5m away from the fulcrum and weighs 200N?

200(1.5) = .5 w

w = 600 Newtons

3.25 To find the father's weight, we need to consider the principle of moments in rotational equilibrium. According to this principle, for an object to be in rotational equilibrium, the sum of the clockwise moments must equal the sum of the counterclockwise moments.

In this case, we have a seesaw with the father and son sitting on opposite sides of the fulcrum. The son weighs 200N and is seated 1.5m away from the fulcrum. Since the seesaw is balanced, the clockwise and counterclockwise moments must be equal.

We can calculate the counterclockwise moment created by the son using the formula:

Counterclockwise moment = Weight of the son * Distance of the son from the fulcrum

= 200N * 1.5m

= 300N·m

To balance this moment, the father's weight must create an equal and opposite clockwise moment. Since the father is seated at 0.5m away from the fulcrum, we can calculate his weight using the formula:

Clockwise moment = Weight of the father * Distance of the father from the fulcrum

300N·m = Weight of the father * 0.5m

Dividing both sides by 0.5m, we get:

Weight of the father = 300N·m / 0.5m

Weight of the father = 600N

Therefore, the father's weight is 600N.