1. Two children balance a see-saw in horizontal equilibrium. One weighs 80 lb. and the other weighs 60 lb. and is sitting 4ft. from the fulcrum. Find the force the fulcrum applies to the beam and the distance to the fulcrum to the 80 lb. child. (Neglect the mass of the see-saw)

See your 10:39am post for solution.

To find the force the fulcrum applies to the beam and the distance to the fulcrum to the 80 lb. child, we can use the principle of moments.

The principle of moments states that for a system in rotational equilibrium, the sum of the moments about any point is equal to zero.

Let's assume the force applied by the fulcrum is F in pounds and the distance from the fulcrum to the 80 lb. child is d feet.

The moment (torque) of the 60 lb. child about the fulcrum is 60 lb. (weight of the 60 lb. child) multiplied by 4 ft. (distance from the fulcrum). This can be written as 60 lb. * 4 ft. = 240 lb.ft.

Similarly, the moment of the 80 lb. child about the fulcrum is 80 lb. multiplied by d ft., which can be written as 80 lb. * d ft. = 80d lb.ft.

Since the see-saw is in horizontal equilibrium, the sum of the moments about the fulcrum should be zero:

240 lb.ft. - 80d lb.ft. = 0

Simplifying the equation, we get:

240 lb.ft. = 80d lb.ft.

To solve for d, we divide both sides of the equation by 80 lb.ft.:

240 lb.ft. / 80 lb.ft. = d

d = 3 ft.

Thus, the distance from the fulcrum to the 80 lb. child is 3 ft.

Now, since we have the distance and weight of the 80 lb. child, we can find the force applied by the fulcrum using the equation:

Force * Distance = Moment

F * 3 ft. = 80 lb. * 4 ft.

Simplifying the equation, we get:

3F ft. = 320 lb.ft.

Dividing both sides by 3 ft., we get:

F = 320 lb.ft. / 3 ft.

F ≈ 106.67 lb.

Therefore, the force the fulcrum applies to the beam is approximately 106.67 lb.