Chapter 9, Problem 13PE

To determine

(a)

The force on the cables.

Answer to Problem 13PE

The force in the cables is $13213.5\text{N}\cdot \text{m}$.

Explanation of Solution

Given:

The mass of the bridge is $M=2500\text{kg}$.

The mass of the car is $m=900\text{kg}$.

The distance of weight of the bridge from point $\text{A}$ is $D_{\text{A}}=1.5\text{m}$.

The distance of weight of the car from point $\text{A}$ is $d=4.5\text{m}$.

The distance of weight of the bridge from point $\text{B}$ is $D_{\text{B}}=7.5\text{m}$.

The length of the bridge is $L=9.0\text{m}$.

The angle of force from point $\text{B}$ is $\beta =40°$.

Formula used:

The relation between the weight and the mass of the bridge is

  $W=Mg$  (1)

The relation between the weight and the mass of the car is

  $w=mg$  (2)

The relation between the torque and weight of bridge is

  ${\tau }_{\text{B}}=W{D}_{\text{A}}$  (3)

The relation between the torque sand weight of car is

  ${\tau }_{\text{C}}=wd$  (4)

The relation between torque of cable is

  $\tau =LT\sin \theta$  (5)

The relation between the torque of car, bridge and cable is

  ${\tau }_{\text{C}}+{\tau }_{\text{B}}+\tau =0$  (6)

  

Figure-(1)

Calculation:

Substituting the given values in equation (1), we get

  $\begin{array}{c}W=\left(2500\text{kg}\right)\left(9.8{\text{m/s}}^{\text{2}}\right)\\ =24500\text{N}\end{array}$

Substituting the given values in equation (2), we get

  $\begin{array}{c}w=\left(900\text{kg}\right)\left(9.8{\text{m/s}}^{\text{2}}\right)\\ =8820\text{N}\end{array}$

Substituting the given values in equation (3), we get

  $\begin{array}{c}{\tau }_{\text{B}}=-\left(24500\text{N}\right)\left(1.5\text{m}\right)\\ =-36750\text{N}\cdot \text{m}\end{array}$

Substituting the given values in equation (4), we get

  $\begin{array}{c}{\tau }_{\text{C}}=-\left(8820\text{N}\right)\left(4.5\text{m}\right)\\ =-39690\text{N}\cdot \text{m}\end{array}$

Substituting the given values in equation (5), we get

  $\begin{array}{c}\tau =T\left(9.0\text{m}\right)\sin 40°\\ =T\left(5.785\text{m}\right)\end{array}$

Substituting the given values in equation (6), we get

  $\begin{array}{c}-39690\text{N}\cdot \text{m}-36750\text{N}\cdot \text{m}+T\left(5.785\text{m}\right)=0\\ T=\frac{76440\text{N}\cdot \text{m}}{5.785\text{m}}\\ =13213.5\text{N}\end{array}$

Conclusion:

Thus, the force in the cables is $13213.5\text{N}\cdot \text{m}$.

To determine

(b)

The direction and magnitude of force exerted by the hinges on the bridge.

Answer to Problem 13PE

The magnitude of force exerted by the hinges on the bridge is $26811.7\text{N}$ and the direction is $68°$ above the horizontal axis towards the other side of the bridge.

Explanation of Solution

Formula used:

The relation between force in the horizontal direction and torque is

  $F_{\text{x}}=T\cos \theta$  (7)

The relation between force in the vertical direction and torque is

  $F_{\text{y}}=W+w-T\sin \theta$  (8)

The relation between the force and vertical and horizontal forces is

  $F=\sqrt{F_{\text{x}}^2+F_{\text{y}}^2}$  (9)

The relation between the angle made by force and vertical and horizontal forces is

  $\beta ={\tan }^{-1}\left(\frac{F_{\text{y}}}{F_{\text{x}}}\right)$  (10)

Calculation:

Substituting the given values in equation (7), we get

  $\begin{array}{c}{F}_{\text{x}}=\left(13213.5\text{N}\right)\cos 40°\\ =10122\text{N}\end{array}$

Substituting the given values in equation (8), we get

  $\begin{array}{c}{F}_{\text{y}}=24500\text{N}+8820\text{N}-\left(13213.5\text{N}\right)\sin 40°\\ =33320\text{N}-8493.5\text{N}\\ \text{=24827}\text{.5}\text{N}\end{array}$

Substituting the given values in equation (9), we get

  $\begin{array}{c}F=\sqrt{{\left(10122\text{N}\right)}^2+{\left(\text{24827}\text{.5}\text{N}\right)}^2}\\ \approx 26811.7\text{N}\end{array}$

Substituting the given values in equation (10), we get

  $\begin{array}{c}\beta ={\tan }^{-1}\left(\frac{\text{24827}\text{.5}\text{N}}{10122\text{N}}\right)\\ \approx 68°\end{array}$

Conclusion:

Thus, the magnitude of force exerted by the hinges on the bridge is $26811.7\text{N}$ and the direction is $68°$ above the horizontal axis towards the other side of the bridge.