A uniform drawbridge must be held at a 32.0 angle above the horizontal to allow

Question

A uniform drawbridge must be held at a 32.0 angle above the horizontal to allow ships to pass underneath. The drawbridge weighs 4.00×104 N , is 18.5 m long, and pivots about a hinge at its lower end. A cable is connected 3.10 m from the hinge, as measured along the bridge, and pulls horizontally on the bridge to hold it in place.

What is the tension in the cable?

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

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

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

Explanation / Answer

A) Sum the moments about the pivot:

M = 0 = T*3.1m*sin32º - 40000N*1/2*18.5m*cos32º

tension T = 191007.6696 N

B)

Horizontally we have Fx = T = 191007.6696 N

and vertically Fy = 40000 N, so

|F| = (Fx² + Fy²) = 195151.0437 N

C)

= arctan(Fy/Fx) = 11.82º above horizontal

D)

Treat the bridge as a rod pivoting about its end:

I = mL^2/3 = (40000N / 9.8m/s^2)(18.5m)^2/3 = 465646.25 kg·m^2

torque T = mg*(L/2)*cos = 40000N *1/2*18 .5m * cos32º = 313777.79 N·m

and = I*, so

313777.79 N·m = 465646.25kg·m^2 *

= 0.6738 rad/s^2

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