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, set the handbrake, and waits. Meanwhile, a boat approaches, so the drawbridge is raised by means of a cable attached to the end opposite the hinge. The drawbridge is 40.0 m long and has a mass of 18000kg; its center of gravity is at its midpoint. The cement mixer, with driver, has mass 30000 kg. When the drawbridge has an angle of 30^o above horizontal, the cable makes an angle of 70^o with the surface of the bridge.

1. What is the tension T in the cable when the drawbridge is held in this position?
2. Find the horizontal and vertical components of the force the hinge exerts on the span?

Short answer

1. The tension in the cable is $2.84\times {10}^5$N.
2. Horizontal component is $2.18\times {10}^5$N to the right and vertical component is $2.88\times {10}^5$N upwards.

Step-by-step solution

The given data

Given that 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, set the handbrake, and waits. Meanwhile, a boat approaches, so the drawbridge is raised by means of a cable attached to the end opposite the hinge. The drawbridge is 40.0 m long and has a mass of 18000kg; its center of gravity is at its midpoint. The cement mixer, with driver, has mass 30000 kg. When the drawbridge has an angle of 30^o above horizontal, the cable makes an angle of 70^o with the surface of the bridge.

Mass of drawbridge, $m_B=18000 \text{kg}$

Mass of cement mixer,$m_T=30000 \text{kg}$

Length of bridge$L=40 \text{m}$

Formula used

Torque$\mathit{\tau }\mathbf{=}\mathit{F}\mathit{l}$

Where Fis force exerted and l is distance.

(a)Step 3: Find tension in the cable

Let T be the tension in the cable.

The moment arm for the force T is $\left(1.4m\right)\cos 10°$.

Applying equilibrium conditions on the bridge,

This gives

$$\begin{gathered} TL\sin {70}^o=m_{T}g\left(\frac{3}{4}L\right)\cos {30}^o+m_{B}g\left(\frac{1}{2}\right)\cos {30}^o \\ T=\frac{\left({\displaystyle \frac{3}{4}}{m}_T+{\displaystyle \frac{1}{2}}{m}_B\right)\left(9.8m/s^2\right)\cos {30}^o}{\sin {70}^o} \\ T=2.84\times {10}^{5}N \end{gathered}$$

Tension in rope is $T=2.84\times {10}^5$N

(b)Step 4: Find horizontal and vertical component of force

Let horizontal component of the force that the hinge exerts on the span is$T_x$and vertical component is $T_y$.

Horizontal component : $T\cos \left(70°-30°\right)=\left(2.84\times {10}^5\right)\cos 40°=2.18\times {10}^5 \text{N}$(to the right)

Vertical Component: $\left(m_T+m_B\right)\left(g\right)T\sin \left(70°-30°\right)=2.88\times {10}^5 \text{N}$(upwards)

Hence horizontal component is $2.18\times {10}^5$N and vertical component is $2.88\times {10}^5$N.