Chapter 9, Problem 12PE

To determine

(a)

The fraction of weight supported by the opposite shore.

Answer to Problem 12PE

The fraction of weight supported by the opposite shore is $0.1667$.

Explanation of Solution

Given:

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

Formula used:

The relation between the mass and weight is,

  $w=mg$

The expression for condition of equilibrium is,

  $\begin{array}{l}\sum \tau =0\\ w\left(1.5\text{m}\right)-T\sin {40}^{\text{ο}}\left(9.0\text{m}\right)=0\end{array}$

The relation between the fraction of weight and tension is,

  $w_F=\frac{T\sin {40}^{\text{ο}}}{w}$

The free body diagram for the bridge is shown below.

  

  Figure-(1)

Calculation:

The weight of the bridge is calculated as,

  $\begin{array}{c}w=mg\\ =\left(2500\text{kg}\right)\left(9.8\text{m}/{\text{s}}^{\text{2}}\right)\\ =24525\text{N}\end{array}$

The tension is calculated as,

  $\begin{array}{c}w\left(1.5\text{m}\right)-T\sin 40°\left(9.0\text{m}\right)=0\\ \left(24525\text{N}\right)\left(1.5\text{m}\right)-T\sin 40°\left(9.0\text{m}\right)=0\\ T\sin 40°\left(9.0\text{m}\right)=36787.5\text{N}\cdot \text{m}\end{array}$

Solve further:

  $\begin{array}{c}T=\frac{36787.5\text{N-m}}{\sin {40}^{\text{ο}}\left(9.0\text{m}\right)}\\ T=6359\text{N}\end{array}$

The fraction of weight is calculated as,

  $\begin{array}{c}{w}_F=\frac{T\sin 40°}{w}\\ =\frac{\left(6359\text{N}\right)\sin 40°}{24525\text{N}}\\ =\frac{4087\text{N}}{24525\text{N}}\\ =0.1667\end{array}$

Conclusion:

The fraction of weight supported by the opposite shore is $0.1667$.

To determine

(b)

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

Answer to Problem 12PE

The magnitude of the force is $21009\text{N}$ and the direction of force is $76.59°$ anti-clockwise.

Explanation of Solution

Given:

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

Formula used:

The expression for condition of equilibrium is,

  $\begin{array}{c}F\sin \theta +T\sin 40°=w\\ F\sin \theta =w-T\sin 40°\end{array}$

The expression for force along horizontal is,

  $F\cos \theta =T\cos {40}^{\text{ο}}$

Calculation:

The vertical component of the force is calculated as,

  $\begin{array}{c}F\sin \theta =w-T\sin 40°\\ =24525\text{N}-\left(6359\right)\sin 40°\\ =20437.5\text{N}\end{array}$

The horizontal component of the force is calculated as,

  $\begin{array}{c}F\cos \theta =T\cos 40°\\ =\left(6359\text{N}\right)\cos 40°\\ =4871\text{N}\end{array}$

The magnitude of the force is given as,

  $\begin{array}{c}F=\sqrt{{\left(F\cos \theta \right)}^2+{\left(F\sin \theta \right)}^2}\\ =\sqrt{{\left(4871\text{N}\right)}^2+{\left(20437.5\text{N}\right)}^2}\\ =21009\text{N}\end{array}$

The direction of the force is given as,

  $\begin{array}{c}\tan \theta =\frac{F\sin \theta }{F\cos \theta }\\ =\frac{20437.5\text{N}}{4871\text{N}}\\ \tan \theta =4.195\\ \theta =76.59°\end{array}$

Conclusion:

The magnitude of the force is $21009\text{N}$ and the direction of force is $76.59°$ anti-clockwise.