Chapter 8.5, Problem 24E

a.

To determine

To draw: a sketch of the given situation.

Explanation of Solution

Given information:

Let the forces acting on an object be $\overrightarrow{F_1}\text{,}\overrightarrow{F_2}\text{ and }\overrightarrow{F_3}$ .

  $\begin{array}{c}\left|\overrightarrow{F_1}\right|\text{= 70 pounds}\\ \left|\overrightarrow{F_2}\right|=\text{115 pounds}\\ \left|\overrightarrow{F_3}\right|=\text{135 pounds}\end{array}$

The force $\overrightarrow{F_1}$ is applied along $+x\text{-axis}$ , whereas $\overrightarrow{F_2}$ is applied below $x\text{-axis}$ at $120°$ with $\overrightarrow{F_1}$ .

Angle between $\overrightarrow{F_2}\text{ and }\overrightarrow{F_3}$ is $75°$ .

Angle between $\overrightarrow{F_1}\text{ and }\overrightarrow{F_3}$ is $165°$ .

Calculation:

The sketch of the given situation is shown below.

  

b.

To determine

To state: whether the vectors are in equilibrium, if not then find the magnitude and direction of the resultant force.

Answer to Problem 24E

No equilibrium

Magnitude: 134.5 pounds

Direction: $28.7°$ south of west

Explanation of Solution

Given information:

Let the forces acting on an object be $\overrightarrow{F_1}\text{,}\overrightarrow{F_2}\text{ and }\overrightarrow{F_3}$ .

  $\left|\overrightarrow{F_1}\right|\text{= 70 pounds, }\left|\overrightarrow{F_2}\right|=\text{115 pounds and }\left|\overrightarrow{F_3}\right|=\text{135 pounds}$

The sketch of the given situation is shown below.

  

Calculation:

Use the sketch to resolve the components as shown below.

  $\begin{array}{c}\overrightarrow{F_1}=\left(70\right)\overrightarrow{i}\\ \overrightarrow{F_2}=\left(135\cos 165°\right)\overrightarrow{i}+\left(135\sin 165°\right)\overrightarrow{j}\\ \overrightarrow{F_3}=\left(115\cos 240°\right)\overrightarrow{i}+\left(115\sin 240°\right)\overrightarrow{j}\end{array}$

The magnitude of the resultant force is evaluated as follows:

  $\begin{array}{c}\left|\overrightarrow{F_1}+\overrightarrow{F_2}+\overrightarrow{F_3}\right|=\sqrt{{\left(70+135\cos 165°+115\cos 240°\right)}^2+{\left(135\sin 165°+115\sin 240°\right)}^2}\\ \approx 134.5\text{ pounds}\end{array}$

Since $\left|\overrightarrow{F_1}+\overrightarrow{F_2}+\overrightarrow{F_3}\right|\ne 0$ , so no equilibrium exists.

The direction of the resultant force is evaluated as follows:

  $\begin{array}{c}\tan \theta =\frac{135\sin 165°+115\sin 240°}{70+135\cos 165°+115\cos 240°}\\ \theta ={\tan }^{-1}\left(\frac{135\sin 165°+115\sin 240°}{70+135\cos 165°+115\cos 240°}\right)\\ \approx 208.7°\\ or\\ 28.7°\text{ south of west }\left[\because \text{ west is at 180}°\right]\end{array}$

Thus the magnitude and direction of resultant force is 134.5 pounds and $28.7°$ south of west respectively.