91Ó°ÊÓ

Total torque about the point of rotation which is fixed is zero.

The distance of the CG from the left hinge is \({{\rm{r}}_{\rm{l}}}{\rm{ = 1}}{\rm{.5}}\;{\rm{m}}\).

The distance to the opposite shore is \({{\rm{l}}_{\rm{r}}}{\rm{ = 9}}\;{\rm{m}}\).

The mass of the bridge is \({\rm{m = 2500}}\;{\rm{kg}}\).

The diagram is shown below:

\(\begin{array}{c}{{\rm{\tau }}_{{\rm{db}}}}{\rm{ = W \times }}{{\rm{r}}_{\rm{l}}}\\{\rm{ = 2500 \times 9}}{\rm{.8 \times ( - 1}}{\rm{.5)}}\\{\rm{ = - 36750}}\;{\rm{N \times m}}\end{array}\)

The torque due to the weight of the car is(clockwise),

\(\begin{array}{c}{{\rm{\tau }}_{{\rm{car}}}}{\rm{ = }}{{\rm{W}}_{{\rm{car}}}}{\rm{ \times - }}{{\rm{r}}_{{\rm{car}}}}\\{\rm{ = 900 \times 9}}{\rm{.8 \times ( - 4}}{\rm{.5)}}\\{\rm{ = - 39690}}\;{\rm{N \times m}}\end{array}\)

The torque due to tension on the ropes is,

\(\begin{array}{c}{{\rm{\tau }}_{{\rm{rope}}}}{\rm{ = T \times }}{{\rm{r}}_{{\rm{rope}}}}{\rm{ \times sin40^\circ }}\\{\rm{ = T \times 9 \times sin40^\circ }}\\{\rm{ = 5}}{\rm{.79 T}}\end{array}\)

This torque is anticlockwise.

For equilibrium, the net torque is zero. So,

\(\begin{array}{c}{{\rm{\tau }}_{{\rm{db}}}}{\rm{ + }}{{\rm{\tau }}_{{\rm{car}}}}{\rm{ = }}{{\rm{\tau }}_{{\rm{rope}}}}\\{\rm{ - 36750 + ( - 39690) = 5}}{\rm{.79 T}}\\{\rm{T = }}\frac{{36750 + 39690}}{{5.79}}\\{\rm{T = 13213}}\;{\rm{N}}\end{array}\)

Hence, the force is \({\rm{13213}}\;{\rm{N}}\).

The horizontal component is,

\(\begin{array}{c}{\rm{Tcos40^\circ }}\\{\rm{ = 13213cos40^\circ }}\\{\rm{ = 10122}}\;{\rm{N}}\end{array}\)

The vertical component is,

\(\begin{array}{c}{\rm{mg + }}{{\rm{m}}_c}{\rm{g - Tsin40^\circ }}\\{\rm{ = (2500 + 900)}} \times {\rm{9}}{\rm{.8 - 13213sin40^\circ }}\\{\rm{ = 24827}}\;{\rm{N}}\end{array}\)

Due to the force in hinge inclined at\({\rm{\theta }}\)with the horizontal, the horizontal component is,

So,

\(\begin{array}{c}{\rm{tan\theta = }}\frac{{{\rm{24827}}}}{{{\rm{10122}}}}\\{\rm{\theta = 67}}{\rm{.8^\circ }}\end{array}\)

The force is,

\(\begin{array}{c}{\rm{F = }}\sqrt {{\rm{10122}}{}^{\rm{2}}{\rm{ + 2482}}{{\rm{7}}^{\rm{2}}}} \\{\rm{ = 26811}}\;{\rm{N}}\end{array}\)

Hence, the force is \({\rm{26811}}\;{\rm{N}}\) with \({\rm{67}}{\rm{.8^\circ }}\)approaching the opposite shore, with the bridge.