A30.0-kg child sits on one end of a long uniform beam having a mass of 20.0 kg, and a 40.0-kg child sits on the other end. the beam balances when a fulcrum is placed below the beam a distance 1.10 m from the 30.0-kg child. how long is the beam?

let the length of the beam be "L"

from the diagram

AD = length of beam = L

AC = CD = AD/2 = L/2

BC = AC - AB = (L/2) - 1.10

BD = AD - AB = L - 1.10

m = mass of beam = 20 kg

m₁ = mass of child on left end = 30 kg

m₂ = mass of child on right end = 40 kg

using equilibrium of torque about B

(m₁ g) (AB) = (mg) (BC) + (m₂ g) (BD)

30 (1.10) = (20) ((L/2) - 1.10) + (40) (L - 1.10)

L = 1.98 m

not too sure thinking it will always be directed towards the centre of the earth. could be wrong not sure

15.3 cos 65 degrees trig