A loaded cement mixer drives onto an old drawbridge, where it stalls with its ce

Question

A loaded cement mixer drives onto an old drawbridge, where it stalls with its center of gravity three-quarters of the way across the span. The truck driver radios for help, sets the handbrake, and waits. Meanwhile, a boat approaches, so the drawbridge is raised by means of a cable attached to its end opposite the hinge (see figure). The drawbridge span is 40.0 m long and has a mass of 11,900 kg; its center of gravity is at its midpoint. The cement mixer, with driver, has a mass of 28,100 kg. When the drawbridge has been raised to an angle of 30

A loaded cement mixer drives onto an old drawbridge, where it stalls with its center of gravity three-quarters of the way across the span. The truck driver radios for help, sets the handbrake, and waits. Meanwhile, a boat approaches, so the drawbridge is raised by means of a cable attached to its end opposite the hinge (see figure). The drawbridge span is 40.0 m long and has a mass of 11,900 kg; its center of gravity is at its midpoint. The cement mixer, with driver, has a mass of 28,100 kg. When the drawbridge has been raised to an angle of 30 degree above the horizontal, the cable makes an angle of 70 degree with the surface of the bridge. What is the tension T in the cable when the drawbridge is held in this position? What are the horizontal and vertical components of the force the hinge exerts on the span?

Explanation / Answer

Net forces on horizontal direction are

H1 (right) acting at hinge

T cos(70) (left) acting at cable

Net forces on vertical direction

H2 (upwards) acting at hinge

T sin(70) (upwards) acting at cable

Wb = 11900 * 9.8 N downwards(weight of bridge)

Wt = 28100 * 9.8 N downwards(weight of truck)

Balancing forces on horizontal and vertical directions

H1 = T cos70 ..............(eq.1)

H2 + T sin70 = Wb + Wt = 392000 .......................(eq.2)

Torques acting about Hinges are

Wb * (20 cos30)   clockwise (torque due to bridge weight)

Wt * (30 cos30)   clockwise (torque due to truck weight)

T * (40 sin70)   anti-clockwise(torque due to Tension in cable)

Balancing Torques

T * (40 sin70) = Wb * (20 cos30) + Wt (30 cos30) ...........................(eq.3)

From eq.3, we get

T = 244082.4723 N

Putting the above T in eq.1 & eq.2 and solving

H1 = 83481.1221 N   (horizontal component)

H2 = 162637.5019 N (vertical component)

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