The bar is in equilibrium with tension forces Tl and Tr on the left and right cords respectively and mass m. The equations for equilibrium are written involving the vertical and horizontal force components and torque. Tl and Tr are eliminated to solve for the displacement x. Using trigonometric identities, x is found to be 2.20 m for the given values of θ = 36.9° and φ = 53.1°.

1. 27. The bar is in equilibrium, so the forces and the torques acting on it each sum to zero. Let Tl be the
tension force of the left-hand cord, Tr be the tension force of the right-hand cord, and m be the mass of
the bar. The equations for equilibrium are:
vertical force components Tl cos θ + Tr cos φ − mg = 0
horizontal force components −Tl sin θ + Tr sin φ = 0
torques mgx − Tr L cos φ = 0 .
The origin was chosen to be at the left end of the bar for purposes of calculating the torque.
The unknown quantities are Tl , Tr , and x. We want to eliminate Tl and Tr , then solve for x. The second
equation yields Tl = Tr sin φ/ sin θ and when this is substituted into the first and solved for Tr the result
is Tr = mg sin θ/(sin φ cos θ + cos φ sin θ). This expression is substituted into the third equation and the
result is solved for x:
sin θ cos φ sin θ cos φ
x=L =L .
sin φ cos θ + cos φ sin θ sin(θ + φ)
The last form was obtained using the trigonometric identity sin(A + B) = sin A cos B + cos A sin B. For
the special case of this problem θ + φ = 90◦ and sin(θ + φ) = 1. Thus,
x = L sin θ cos φ = (6.10 m) sin 36.9◦ cos 53.1◦ = 2.20 m .