Chapter 2.4, Problem 2.95P

To determine

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

Answer to Problem 2.95P

The direction and the magnitude of the resultant of the forces exerted by the cables at $B$ by two cables is given by ${\theta }_x=\underset{\_}{89.5°}$, ${\theta }_y=\underset{\_}{36.2°}$, ${\theta }_z=\underset{\_}{126.2°}$ and $R=\underset{\_}{1171\text{N}}$.

Explanation of Solution

The following figure gives the sketch of the system given in the problem.

The tension in the cables $BG$ and $BH$ are $T_{BG}$ and $T_{BH}$ respectively.

The tension in the cable $BG$ is $540\text{N}$.

The tension in the cable $BH$ is $750\text{N}$.

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

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

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

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

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

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

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

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

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

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

$\left|BH\right|=\sqrt{B{H}_x^2+B{H}_y^2+B{H}_z^2}$ (V)

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

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

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

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

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

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

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

Refer figure P2.85.

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

$R=F_{BG}+F_{BH}$ (IX)

Here, $R$ is the resultant of the two vectors $F_{BG}$ and $F_{BH}$

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.P2.85 and calculate the vector coordinates of the vector $BG$.

$\overrightarrow{BG}=\left(-0.8\text{m}\right)i+\left(1.48\text{m}\right)j+\left(-0.64\text{m}\right)k$

Substitute $-0.8\text{m}$ for $B{G}_x$, $1.48\text{m}$ for $B{G}_y$, and $-0.64\text{m}$ for $B{G}_z$ in the equation (I).

$\begin{array}{c}BG=\sqrt{{\left(-0.8\text{m}\right)}^2+{\left(1.48\text{m}\right)}^2+{\left(-0.64\text{m}\right)}^2}\\ =1.8\text{m}\end{array}$

Substitute $\left(-0.8\text{m}\right)i+\left(1.48\text{m}\right)j+\left(-0.64\text{m}\right)k$ for $\overrightarrow{BG}$, and $1.8\text{m}$ for $BG$ in equation (II).

$\begin{array}{c}\lambda =\frac{\left(-0.8\text{m}\right)i+\left(1.48\text{m}\right)j+\left(-0.64\text{m}\right)k}{1.8\text{m}}\\ =-0.44i+0.82j-0.35k\end{array}$

Substitute $540\text{N}$ for $T_{BG}$, and $-0.44i+0.82j-0.35k$ for $\lambda$ in equation (III).

$\begin{array}{c}{F}_{BG}=\left(540\text{N}\right)\left(-0.44i+0.82j-0.35k\right)\\ =\left(-240\text{N}\right)i+\left(444\text{N}\right)j-\left(192.0\text{N}\right)k\end{array}$

Substitute $\left(-240\text{N}\right)i+\left(444\text{N}\right)j-\left(192.0\text{N}\right)k$ for $F_{BG}$ in equation (IV).

$\left(-240\text{N}\right)i+\left(444\text{N}\right)j-\left(192.0\text{N}\right)k=F_{x}i+F_{y}j+F_{z}j$

Refer Fig.P2.85 and calculate the vector coordinates of the vector $BH$.

$\overrightarrow{BH}=\left(0.6\text{m}\right)i+\left(1.2\text{m}\right)j-\left(1.2\text{m}\right)k$

Substitute $0.6\text{m}$ for $B{H}_x$, $1.2\text{m}$ for $B{H}_y$, and $-1.2\text{m}$ for $B{H}_z$ in the equation (V).

$\begin{array}{c}BH=\sqrt{{\left(0.6\text{m}\right)}^2+{\left(1.2\text{m}\right)}^2+{\left(-1.2\text{m}\right)}^2}\\ =1.8\text{m}\end{array}$

Substitute $\left(0.6\text{m}\right)i+\left(1.2\text{m}\right)j-\left(1.2\text{m}\right)k$ for $\overrightarrow{BH}$, and $1.8\text{m}$ for $BH$ in equation (VI).

$\begin{array}{c}\lambda =\frac{\left(0.6\text{m}\right)i+\left(1.2\text{m}\right)j-\left(1.2\text{m}\right)k}{1.8\text{m}}\\ =0.33i+0.66j-0.66k\end{array}$

Substitute $750\text{N}$ for $T_{BH}$, and $0.33i+0.66j-0.66k$ for $\lambda$ in equation (VII).

$\begin{array}{c}{F}_{BH}=\left(750\text{N}\right)\left(0.33i+0.66j-0.66k\right)\\ =\left(250\text{N}\right)i+\left(500\text{N}\right)j-\left(500\text{N}\right)k\end{array}$

Substitute $\left(250\text{N}\right)i+\left(500\text{N}\right)j-\left(500\text{N}\right)k$ for $F_{BH}$ in equation (VIII).

$\left(250\text{N}\right)i+\left(500\text{N}\right)j-\left(500\text{N}\right)k=F_{x}i+F_{y}j+F_{z}j$

Substitute $\left(-240\text{N}\right)i+\left(444\text{N}\right)j-\left(192.0\text{N}\right)k$ for $F_{BG}$ and $\left(250\text{N}\right)i+\left(500\text{N}\right)j-\left(500\text{N}\right)k$ for $F_{BH}$ in equation (IX).

$\begin{array}{c}R=\left(-240\text{N}\right)i+\left(444\text{N}\right)j-\left(192.0\text{N}\right)k+\left(250\text{N}\right)i+\left(500\text{N}\right)j-\left(500\text{N}\right)k\\ =\left(10\text{N}\right)i+\left(944\text{N}\right)j-\left(692\text{N}\right)k\end{array}$

Substitute $10\text{N}$ for $R_x$, $944\text{N}$ for $R_y$, and $-692\text{N}$ for $R_z$ in the equation (X).

$\begin{array}{c}R=\sqrt{{\left(10\text{N}\right)}^2+{\left(944\text{N}\right)}^2+{\left(-692\text{N}\right)}^2}\\ =1170.51\text{N}\end{array}$

Substitute $x$ for $n$, $1170.51\text{N}$ for $R$, and $10\text{N}$ for $R_x$, in equation (XI).

$\begin{array}{c}\text{cos}{\theta }_x=\frac{R_x}{R}\\ =\frac{10\text{N}}{1170.51\text{N}}\\ =0.00854\\ {\theta }_x=89.5°\end{array}$

Substitute $y$ for $n$, $1170.51\text{N}$ for $R$, and $944\text{N}$ for $R_y$, in equation (XI).

$\begin{array}{c}\text{cos}{\theta }_y=\frac{R_y}{R}\\ =\frac{944\text{N}}{1170.51\text{N}}\\ =0.8065\\ {\theta }_y=36.2°\end{array}$

Substitute $z$ for $n$, $1170.51\text{N}$ for $R$, and $-692\text{N}$ for $R_z$, in equation (XI).

$\begin{array}{c}\text{cos}{\theta }_z=\frac{R_z}{R}\\ =\frac{-692\text{N}}{1170.51\text{N}}\\ =-0.5912\\ {\theta }_z=126.2°\end{array}$

Therefore, the direction and the magnitude of the resultant of the forces exerted by the cables at $B$ by two cables is given by ${\theta }_x=\underset{\_}{89.5°}$, ${\theta }_y=\underset{\_}{36.2°}$, ${\theta }_z=\underset{\_}{126.2°}$, and $R=\underset{\_}{1171\text{N}}$.