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## Homework Statement

A seesaw with a mass of 5 kg has one block of mass 10 kg two meters to the left of the fulcrum and another block 0.5 m to the right of the fulcrum. If the seesaw is in equilibrium,

A. find the mass of the second block.

B. find the force exerted by the fulcrum

I have trouble finding the force exerted by the fulcrum; I keep setting up the wrong equation. I do not understand why the normal forces for m1 and m2 do not come into play for this problem. Any help would be greatly appreciated.

## Homework Equations

Σtorque = 0

ΣFy = 0

## The Attempt at a Solution

m1 = 10 kg (2 m left of fulcrum)

m2 = 40 kg (0.5 m right of fulcrum)

seesaw mass (M) = 5 kg

F1 = upward force (normal) exerted by fulcrum onto seesaw

Mg = downward force

designated fulcrum as pivot point

up → positive

down → negative

clockwise rotation → positive

counterclockwise rotation →negative

d1 (⊥ distance from the line of action of force of m1g to pivot point) = 2 m

d2 (⊥ distance from the line of action of force of m2g to pivot point) = 0.5 m

A. ∑torque = -m1g (d1) + m2g (d2) = 0

⇒m2 = 40 kg

B. For m1,

∑Fy = n- m1g = 0

⇒ n = m1g

⇒ n = 10 kg (9.8 m/s^2)

For m2,

∑Fy = n- m2g = 0

⇒ n = m2g

⇒ n = 40 kg (9.8 m/s^2)

∑Fy = n-m1g + n-m2g - F1- Mg = 0