Chapter 2.4, Problem 2.93P

Knowing that the tension is 425 lb in cable AB and 510 lb in cable AC, determine the magnitude and direction of the resultant of the forces exerted at A by the two cables.

Fig. P2.93 and P2.94

To determine

The direction and magnitude of the resultant forces exerted at $A$ by two cables.

Answer to Problem 2.93P

The direction and magnitude of the resultant forces exerted at $A$ by two cables is given by ${\theta }_x=\underset{\_}{48.2°}$, ${\theta }_y=\underset{\_}{116.6°}$, ${\theta }_z=\underset{\_}{53.4°}$, and $R=\underset{\_}{913\text{lb}}$.

Explanation of Solution

The tension in the cable $AB$ is $425\text{lb}$.

The tension in the cable $AC$ is $510\text{lb}$.

Write the equation to find the magnitude of the vector $AB$.

$\left|AB\right|=\sqrt{A{B}_x^2+A{B}_y^2+A{B}_z^2}$ (I)

Write the equation to find the unit vector along $\overrightarrow{AB}$.

$\lambda =\frac{\overrightarrow{AB}}{\left|AB\right|}$ (II)

Write the equation of $F_{AB}$ in terms of its rectangular components.

$F_{AB}=T_{AB}\lambda$ (III)

Here, $T_{AB}$ is the tension in the cable $AB$, and $\lambda$ is the unit vector along $\overrightarrow{AB}$.

Write the equation for force $F_{AB}$ in component form.

$F_{AB}=F_{x}i+F_{y}j+F_{z}j$ (IV)

Write the equation to find the magnitude of the vector $AC$.

$\left|AC\right|=\sqrt{A{C}_x^2+A{C}_y^2+A{C}_z^2}$ (V)

Write the equation to find the unit vector along $\overrightarrow{AC}$.

$\lambda =\frac{\overrightarrow{AC}}{\left|AC\right|}$ (VI)

Write the equation of $F_{AC}$ in terms of its rectangular components.

$F_{AC}=T_{AC}\lambda$ (VII)

Here, $T_{AC}$ is the tension in the cable $AC$, and $\lambda$ is the unit vector along $\overrightarrow{AC}$.

Write the equation for force $F$ in component form.

$F_{AC}=F_{x}i+F_{y}j+F_{z}j$ (VIII)

Refer figure P2.93.

Write the equation to find the resultant of the two vectors.

$R=F_{AB}+F_{AC}$ (IX)

Here, $R$ is the resultant of the two vectors $F_{AB}$ and $F_{AC}$

Write the equation to find the magnitude of the vector $R$.

$\left|R\right|=\sqrt{R_x^2+R_y^2+R_z^2}$ (X)

Write the equation to find the direction of vector $R$.

$\text{cos}{\theta }_n=\frac{R_n}{R}$ (XI)

Here, $R_n$ is the component of the vector corresponding to the direction $n$, $R$ is the magnitude of the vector $R$, and ${\theta }_n$ is the angle corresponding to the direction.

Conclusion:

Refer Fig.P2.93 and calculate the vector coordinates of the vector $AB$.

$\overrightarrow{AB}=\left(40\text{in}\right)i-\left(45\text{in}\right)j+\left(60\text{in}\right)k$

Substitute $40\text{in}$ for $A{B}_x$, $-45\text{in}$ for $A{B}_y$, and $60\text{in}$ for $A{B}_z$ in the equation (I).

$\begin{array}{c}AB=\sqrt{{\left(40\text{in}\right)}^2+{\left(-45\text{in}\right)}^2+{\left(60\text{in}\right)}^2}\\ =85\text{in}\end{array}$

Substitute $\left(40\text{in}\right)i-\left(45\text{in}\right)j+\left(60\text{in}\right)k$ for $\overrightarrow{AB}$, and $85\text{in}$ for $AB$ in equation (II).

$\begin{array}{c}\lambda =\frac{\left(40\text{in}\right)i-\left(45\text{in}\right)j+\left(60\text{in}\right)k}{85\text{in}}\\ =0.47i-0.52j+0.70k\end{array}$

Substitute $425\text{lb}$ for $T_{AB}$, and $0.47i-0.52j+0.70k$ for $\lambda$ in equation (III).

$\begin{array}{c}{F}_{AB}=\left(425\text{lb}\right)\left(0.47i-0.52j+0.70k\right)\\ =\left(200\text{lb}\right)i-\left(225\text{lb}\right)j+\left(300\text{lb}\right)k\end{array}$

Substitute $\left(200\text{lb}\right)i-\left(225\text{lb}\right)j+\left(300\text{lb}\right)k$ for $F_{AB}$ in equation (IV).

$\left(200\text{lb}\right)i-\left(225\text{lb}\right)j+\left(300\text{lb}\right)k=F_{x}i+F_{y}j+F_{z}j$

Refer Fig.P2.93 and calculate the vector coordinates of the vector $AC$.

$\overrightarrow{AC}=\left(100\text{in}\right)i-\left(45\text{in}\right)j+\left(60\text{in}\right)k$

Substitute $100\text{in}$ for $A{C}_x$, $45\text{in}$ for $A{C}_y$, and $60\text{in}$ for $A{C}_z$ in the equation (V).

$\begin{array}{c}AC=\sqrt{{\left(100\text{in}\right)}^2+{\left(45\text{in}\right)}^2+{\left(60\text{in}\right)}^2}\\ =125\text{in}\end{array}$

Substitute $\left(100\text{in}\right)i-\left(45\text{in}\right)j+\left(60\text{in}\right)k$ for $\overrightarrow{AC}$, and $125\text{in}$ for $AC$ in equation (VI).

$\begin{array}{c}\lambda =\frac{\left(100\text{in}\right)i-\left(45\text{in}\right)j+\left(60\text{in}\right)k}{125\text{in}}\\ =0.8i-0.36j+0.48k\end{array}$

Substitute $510\text{lb}$ for $T_{AC}$, and $0.8i-0.36j+0.48k$ for $\lambda$ in equation (VII).

$\begin{array}{c}{F}_{AC}=\left(510\text{lb}\right)\left(0.8i-0.36j+0.48k\right)\\ =\left(408\text{lb}\right)i-\left(183.6\text{lb}\right)j+\left(244.8\text{lb}\right)k\end{array}$

Substitute $\left(408\text{lb}\right)i-\left(183.6\text{lb}\right)j+\left(244.8\text{lb}\right)k$ for $F_{AC}$ in equation (VIII).

$\left(408\text{lb}\right)i-\left(183.6\text{lb}\right)j+\left(244.8\text{lb}\right)k=F_{x}i+F_{y}j+F_{z}j$

Substitute $\left(200\text{lb}\right)i-\left(225\text{lb}\right)j+\left(300\text{lb}\right)k$ for $F_{AB}$ and $\left(408\text{lb}\right)i-\left(183.6\text{lb}\right)j+\left(244.8\text{lb}\right)k$ for $F_{AC}$ in equation (IX).

$\begin{array}{c}R=\left(200\text{lb}\right)i-\left(225\text{lb}\right)j+\left(300\text{lb}\right)k+\left(408\text{lb}\right)i-\left(183.6\text{lb}\right)j+\left(244.8\text{lb}\right)k\\ =\left(608\text{lb}\right)i-\left(408.6\text{lb}\right)j+\left(544.8\text{lb}\right)k\end{array}$

Substitute $608\text{lb}$ for $R_x$, $408.6\text{lb}$ for $R_y$, and $544.8\text{lb}$ for $R_z$ in the equation (X).

$\begin{array}{c}R=\sqrt{{\left(608\text{lb}\right)}^2+{\left(408.6\text{lb}\right)}^2+{\left(544.8\text{lb}\right)}^2}\\ =912.92\text{lb}\end{array}$

Substitute $x$ for $n$, $912.92\text{lb}$ for $R$, and $608\text{lb}$ for $R_x$, in equation (XI).

$\begin{array}{c}\text{cos}{\theta }_x=\frac{R_x}{R}\\ =\frac{608\text{lb}}{912.92\text{lb}}\\ =0.66599\\ {\theta }_x=48.2°\end{array}$

Substitute $y$ for $n$, $912.92\text{lb}$ for $R$, and $408.6\text{lb}$ for $R_y$, in equation (XI).

$\begin{array}{c}\text{cos}{\theta }_y=\frac{R_y}{R}\\ =\frac{408.6\text{lb}}{912.92\text{lb}}\\ =-0.44757\\ {\theta }_y=116.6°\end{array}$

Substitute $z$ for $n$, $912.92\text{lb}$ for $R$, and $544.8\text{lb}$ for $R_z$, in equation (XI).

$\begin{array}{c}\text{cos}{\theta }_z=\frac{R_z}{R}\\ =\frac{544.8\text{lb}}{912.92\text{lb}}\\ =0.59677\\ {\theta }_z=53.4°\end{array}$

Therefore, the direction and magnitude of the resultant forces exerted at $A$ by two cables is given by ${\theta }_x=\underset{\_}{48.2°}$, ${\theta }_y=\underset{\_}{116.6°}$, ${\theta }_z=\underset{\_}{53.4°}$, and $R=\underset{\_}{913\text{lb}}$.