91Ó°ÊÓ

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$L=40 \text{m}$

Torque$\mathit{Ï„}\mathbf{=}\mathit{F}\mathit{l}$

Where Fis force exerted and l is distance.

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

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

Tension in rope is $T=2.84×{10}^5$N

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

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

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

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