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 5.00×104 N , is 10.5 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. Part A What is the tension in the cable? T = 1.36×105 N THIS Answer is correct

Part B Find the magnitude of the force the hinge exerts on the bridge. F = N

Part C Find the direction of the force the hinge exerts on the bridge. = 20.3 above the horizontal THIS ANSWER IS CORRECT

Part D If the cable suddenly breaks, what is the initial angular acceleration of the bridge? = rad/s2

if you do not get A and C correct then B and D wont be correct either so please do all the parts

Explanation / Answer

given,

angle = 32 degree

weight = 5 * 10^4 N

length = 10.5 m

distance where the cable is connected = 3.1

since the bridge is in equilibrium net torque = 0

torque about hinge = 5 * 10^4 * (10.5 / 2) * sin(90 - 32) - T * 3.1 * sin(32)

5 * 10^4 * (10.5 / 2) * sin(90 - 32) - T * 3.1 * sin(32) = 0

tension in the cable T = 135512 N

horizontal force on the hinge = T

horizontal force on the hinge = 135512 N

vertical force on hinge = mg

vertical force on hinge = 5 * 10^4 N

net force = sqrt(135512^2 + (5 * 10^4)^2)

net force on hinge = 144442.03 N

direction = tan^-1(5 * 10^4 / 135512)

direction = 20.25 degree

torque = moment of inertia * acceleration

5 * 10^4 * (10.5 / 2) * sin(90 - 32) = (1 / 3) * 5 * 10^4 * 10.5^2 / 9.8 * angular acceleration

angular acceleration = 1.187 rad/sec^2

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