Question

A uniform drawbridge must be held at a 35.0 ∘ angle above the horizontal to allow ships to pass underneath. The drawbridge weighs 4.65×104 N , is 18.5 m long, and pivots about a hinge at its lower end. A cable is connected 3.75 m from the hinge, as measured along the bridge, and pulls horizontally on the bridge to hold it in place.

Tension of cable: 1.64x10⁵ N

Magnitude of the force of the hinge exerted on the bridge: 1.70x10⁵ N

Part C:

Find the direction of the force the hinge exerts on the bridge.

Part D:

If the cable suddenly breaks, what is the initial angular acceleration of the bridge?

Answer 1

A) Sum the moments about the pivot:
ΣM = 0 = T*3.75m*sin35º - 46500N*9.25m*cos35º
tension T = 163808.58 N

B) Horizontally we have Fx = T = 163808.58 N
and vertically Fy = 46500, so
|F| = √(Fx² + Fy²) = 170280.65 N

C) Θ = arctan(46500/163808.58) = 15.848º above horizontal

D) Treat the bridge as a rod pivoting about its end:
I = mL²/3 = (46500N / 9.8m/s²)(18.5m)²/3 = 541313.78 kg·m²

torque τ = mg*(L/2)*cosΘ = 46500N * 9.25m * cos35º =352337.77 N·m
and τ = I*α, so
352337.77 N·m = 541313.78 kg·m² * α
α = 0.651 rad/s²