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Q13PE

Expert-verifiedFound in: Page 316

Book edition
1st Edition

Author(s)
Paul Peter Urone

Pages
1272 pages

ISBN
9781938168000

**Suppose a \({\rm{900}}\;{\rm{kg}}\)car is on the bridge in ****Figure **\({\rm{9}}{\rm{.34}}\)** ****with its center of mass halfway between the hinges and the cable attachments. (The bridge is supported by the cables and hinges only.) (a) Find the force in the cables. (b) Find the direction and magnitude of the force exerted by the hinges on the bridge.**

(a) The force in the cables is \({\rm{13213}}\;{\rm{N}}\).

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

**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:

- The torque by the weight of drawbridge is,

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

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 \({\rm{13213}}\;{\rm{N}}\).

- We have the tension from part (a) is \({\rm{T = 13213}}\;{\rm{N}}\).

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.

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