A uniform bar of length l and weight w is attached to a wall with a hinge that exerts a horizontal force hx and a vertical force
hy on the bar. the bar (which makes an angle θ with respect to wall) is held by a cord that makes a 90◦ angle with respect to bar. what is the magnitude of the tension t in the cord?
The answer is Hx = ½ Wsin θ cos θ The explanation for this is: Analyzing the torques on the bar, with the hinge at the axis of rotation, the formula would be: ∑T = LT – (L/2 sin θ) W = 0 So, T = 1/2 W sin θ. Analyzing the force on the bar, we have: ∑fx = Hx – T cos θ = 0Then put T into the equation, we get:∑T = LT – (L/2 sin θ) W = 0
