Question

A uniform drawbridge must be held at a 37 ? angle above the horizontal to allow ships to pass underneath. The drawbridge weighs 45,000 N and is 14.0 m long. A cable is connected 3.5 m from the hinge where the bridge pivots (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 magnitude of the angular acceleration of the drawbridge just after the cable breaks?

What is the angular speed of the drawbridge as it becomes horizontal?

Answer 1

Answer:

Explanation:

a ) Let T be tension in the horizontal cable .

Taking torque about hinge of weight and tension

45000 x 7 x cos37 = 3.5 sin 37 x T

T = 119431 N

b ) If R be the reaction force on the hinge

R sin 37 = 45000 ( vertical force balancing each other )

R = 74775 N

c ) The direction of R will be along the bridge

d ) Moment of inertia of bridge

I = Ml² / 3 = (45000 / 9.8 )  x 14² / 3

= 300000.

When cable breaks , the weight creates a torque about the hinge will creates angular acceleration

Torque by weight about the hinge

= 45000 x 7 x cos 37

= 251570

angular acceleration = torque / moment of inertia

= 251570 / 300000

= .84 radian / s²