Balancing a beam

let A=angle of fulcrum pivot point to end connection point of the weights on beam $A=(cos (0.1/1) = 84.260°$

let B= angle of displacement from horizontal of the beam do to hanging weights The beam is rotated clock wise about the fulcrum pivot point by the difference in mass on each end of the beam. The resulting angle from horizontal to the connection point on the left side is increasing A+B and decreeing on the right side A-B.

Sum of moments around center of protractor

$0.9 (sin(A+ B)) = 1 (sin(A- B))$ Using the angle sum and difference identities

$0.9 (sin(A) cos(B) + cos(A) sin(B)) = sin(A) cos(B) - cos(A) sin(B)$ $0.9 sin(A) cos(B) + .09 cos(A) sin(B) = sin(A) cos(B) - cos(A) sin(B)$ $1.9 . cos(A) sin(B) = 0.1 sin(A) cos(B)$ divide both sides by cos(B) cos(A) $1.9 sin(B)/cos(B) = 0.1 sin(A)/cos(A)$

Using identity tan(A)= sin(A)/cos(A)

$tan(B) =\frac{0.1 \times tan(A)}{1.9}$ $B = arc tan(\frac{0.1 \times tan(84.260)}{1.9})$

$B = 27.6°$