Chapter 3.4, Problem 3.140P

A flagpole is guyed by three cables. If the tensions in the cables have the same magnitude P, replace the forces exerted on the pole with an equivalent wrench and determine (a) the resultant force R, (b) the pitch of the wrench, (c) the point where the axis of the wrench intersects the xz-plane.

To determine

The resultant force.

Answer to Problem 3.140P

The resultant force is $\underset{\_}{\left(3P/25\right)\left[2\widehat{i}-20\widehat{j}-\widehat{k}\right]}$.

Explanation of Solution

The diagram for the force-couple system is given below:

Refer fig 1.

Write the equation of resultant force.

$\begin{array}{c}\overrightarrow{R}=P{\lambda }_{BA}+P{\lambda }_{DC}+P{\lambda }_{DE}\\ =P\left[{\lambda }_{BA}+{\lambda }_{DC}+{\lambda }_{DE}\right]\end{array}$ (I)

Here, the resultant force is $\overrightarrow{R}$, the magnitude of the forces are $P$, the direction of the force at points is ${\lambda }_{BA},{\lambda }_{DC},{\lambda }_{DE}$

Conclusion:

Substitute, $\left(\frac{4}{5}\widehat{j}-\frac{3}{5}\widehat{k}\right)$ for ${\lambda }_{BA}$, $\left(\frac{3}{5}\widehat{i}-\frac{4}{5}\widehat{j}\right)$ for ${\lambda }_{DC}$, $\left(\frac{3}{5}\widehat{i}-\frac{4}{5}\widehat{j}\right)$ for ${\lambda }_{DE}$ in equation (I)

$\begin{array}{c}\overrightarrow{R}=P\left[\left(\frac{4}{5}\widehat{j}-\frac{3}{5}\widehat{k}\right)+\left(\frac{3}{5}\widehat{i}-\frac{4}{5}\widehat{j}\right)+\left(\frac{3}{5}\widehat{i}-\frac{4}{5}\widehat{j}\right)\right]\\ =\frac{3P}{25}\left[2\widehat{i}-20\widehat{j}-\widehat{k}\right]\end{array}$

The magnitude of the resultant force,

$\begin{array}{c}\left|\overrightarrow{R}\right|=\sqrt{{\left(3P/25\right)}^2\left[{\left(2\right)}^2+{\left(20\right)}^2+{\left(1\right)}^2\right]}\\ =\left(27\sqrt{5}/25\right)P\end{array}$

Thus, the resultant force is $\underset{\_}{\left(3P/25\right)\left[2\widehat{i}-20\widehat{j}-\widehat{k}\right]}$.

To determine

The pitch of the wrench.

Answer to Problem 3.140P

The pitch of the wrench is $\underset{\_}{-0.0988a}$.

Explanation of Solution

Write the equation of pitch of the wrench.

$p=\frac{M_1}{R}$ (II)

Here, the pitch of the wrench is $p$, the momentum along wrench is $M_1$

Since, the $\overrightarrow{R}$ and ${\overrightarrow{M}}_o^R$ are not in same direction, so the form of the wrench is,

${\overrightarrow{M}}_1={\lambda }_{axis}{\overrightarrow{M}}_o^R$ (III)

Here, the constant is ${\lambda }_{axis}$.

Write the expression for the constant is,

${\lambda }_{axis}=\frac{\overrightarrow{R}}{R}$

Write the equation of momentum.

${\overrightarrow{M}}_o^R=\left({\overrightarrow{r}}_1\times {\overrightarrow{F}}_1\right)+\left({\overrightarrow{r}}_2\times {\overrightarrow{F}}_2\right)+\left({\overrightarrow{r}}_3\times {\overrightarrow{F}}_3\right)$ (IV)

Here, the momentum is ${\overrightarrow{M}}_B^R,{\overrightarrow{M}}_A,{\overrightarrow{M}}_B$, the distance is $r_1$.

Rewrite the expression for the momentum of the wrench is,

${\overrightarrow{M}}_1=\left(\frac{\overrightarrow{R}}{R}\right)\left[\left({\overrightarrow{r}}_1\times {\overrightarrow{F}}_1\right)+\left({\overrightarrow{r}}_2\times {\overrightarrow{F}}_2\right)+\left({\overrightarrow{r}}_3\times {\overrightarrow{F}}_3\right)\right]$ (V)

Conclusion:

Substitute, $\left(3P/25\right)\left[2\widehat{i}-20\widehat{j}-\widehat{k}\right]$ for $\overrightarrow{R}$, $\left(27\sqrt{5}/25\right)P$ for $R$, $\left(24a\right)\widehat{j}$ for ${\overrightarrow{r}}_1$, $\left(\frac{-4P}{5}-\frac{3P}{5}\right)\widehat{k}$ for ${\overrightarrow{F}}_1$, $\left(20a\right)\widehat{j}$ for ${\overrightarrow{r}}_2$, $\left(\frac{3P}{5}\widehat{i}-\frac{4P}{5}\widehat{j}\right)$ for ${\overrightarrow{F}}_2$, $\left(20a\right)\widehat{j}$ for ${\overrightarrow{r}}_3$, $\left(-\frac{9P}{25}\widehat{i}-\frac{4P}{5}\widehat{j}+\frac{12P}{25}\widehat{k}\right)$ for ${\overrightarrow{F}}_3$ in equation (V).

$\begin{array}{c}{\overrightarrow{M}}_1=\frac{\left(3P/25\right)\left[2\widehat{i}-20\widehat{j}-\widehat{k}\right]}{\left(27\sqrt{5}/25\right)P}\left[\begin{array}{l}\left\{\left(24a\right)\widehat{j}\times \left(\frac{-4P}{5}-\frac{3P}{5}\right)\widehat{k}\right\}+\left\{\left(20a\right)\widehat{j}\times \left(\frac{3P}{5}\widehat{i}-\frac{4P}{5}\widehat{j}\right)\right\}\\ +\left\{\left(20a\right)\widehat{j}\times \left(-\frac{9P}{25}\widehat{i}-\frac{4P}{5}\widehat{j}+\frac{12P}{25}\widehat{k}\right)\right\}\end{array}\right]\\ =\left(\frac{-8Pa}{675}\right)\left(2\widehat{i}-20\widehat{j}-\widehat{k}\right)\\ =\left(\frac{-8Pa}{15\sqrt{5}}\right)\end{array}$

Substitute, $\left(\frac{-8Pa}{15\sqrt{5}}\right)$ for $M_1$, $\left(27\sqrt{5}/25\right)P$ for $R$ in equation (II)

$\begin{array}{c}p=\frac{\left(-8Pa/15\sqrt{5}\right)}{\left(27\sqrt{5}/25\right)P}\\ =-0.0988a\end{array}$

Thus, the pitch of the wrench is $\underset{\_}{-0.0988a}$.

To determine

The point at which the axis of wrench intersects the xz-plane.

Answer to Problem 3.140P

The axis of wrench intersects the xz-plane at $\underset{\_}{x=2a,z=-1.99a}$.

Explanation of Solution

Refer fig 1,

Write the equation for the force couple system for the wrench.

${\overrightarrow{M}}_B^R={\overrightarrow{M}}_1+{\overrightarrow{M}}_2$ (VI)

Here, the momentum is ${\overrightarrow{M}}_2$.

Write the expression for the momentum at which the wrench intersects the xz-plane.

${\overrightarrow{M}}_2={\overrightarrow{r}}_{Q/O}\times \overrightarrow{R}$ (VII)

Here, the position vector is ${\overrightarrow{r}}_{Q/O}$.

Write the expression for the position vector is,

${\overrightarrow{r}}_{Q/O}=x\widehat{i}+z\widehat{k}$

Here, the coordinates are $x,z$.

Conclusion:

Substitute, $\left(24a\right)\widehat{j}$ for ${\overrightarrow{r}}_1$, $\left(\frac{-4P}{5}-\frac{3P}{5}\right)\widehat{k}$ for ${\overrightarrow{F}}_1$, $\left(20a\right)\widehat{j}$ for ${\overrightarrow{r}}_2$, $\left(\frac{3P}{5}\widehat{i}-\frac{4P}{5}\widehat{j}\right)$ for ${\overrightarrow{F}}_2$, $\left(20a\right)\widehat{j}$ for ${\overrightarrow{r}}_3$, $\left(-\frac{9P}{25}\widehat{i}-\frac{4P}{5}\widehat{j}+\frac{12P}{25}\widehat{k}\right)$ for ${\overrightarrow{F}}_3$ in equation (IV).

$\begin{array}{c}{\overrightarrow{M}}_o^R=\left[\begin{array}{l}\left\{\left(24a\right)\widehat{j}\times \left(\frac{-4P}{5}-\frac{3P}{5}\right)\widehat{k}\right\}+\left\{\left(20a\right)\widehat{j}\times \left(\frac{3P}{5}\widehat{i}-\frac{4P}{5}\widehat{j}\right)\right\}\\ +\left\{\left(20a\right)\widehat{j}\times \left(-\frac{9P}{25}\widehat{i}-\frac{4P}{5}\widehat{j}+\frac{12P}{25}\widehat{k}\right)\right\}\end{array}\right]\\ =\left[\left(\frac{24Pa}{5}\right)\left(-\widehat{i}-\widehat{k}\right)\right]\end{array}$

Substitute, $\left[\left(\frac{24Pa}{5}\right)\left(-\widehat{i}-\widehat{k}\right)\right]$ for ${\overrightarrow{M}}_o^R$, $\left(\frac{-8Pa}{675}\right)\left(2\widehat{i}-20\widehat{j}-\widehat{k}\right)$ for ${\overrightarrow{M}}_1$ in equation (VI).

$\begin{array}{l}\left[\left(\frac{24Pa}{5}\right)\left(-\widehat{i}-\widehat{k}\right)\right]=\left[\left(\frac{-8Pa}{675}\right)\left(2\widehat{i}-20\widehat{j}-\widehat{k}\right)\right]+{\overrightarrow{M}}_2\\ {\overrightarrow{M}}_2=\left\{\left(\frac{8Pa}{675}\right)\left[-430\widehat{i}-20\widehat{j}-406\widehat{k}\right]\right\}\end{array}$

Substitute, $\left\{\left(\frac{8Pa}{675}\right)\left[-430\widehat{i}-20\widehat{j}-406\widehat{k}\right]\right\}$ for ${\overrightarrow{M}}_2$, $\left\{\left(\frac{3P}{25}\right)\left[2\widehat{i}+20\widehat{j}-\widehat{k}\right]\right\}$ for $\overrightarrow{R}$, $\left(x\widehat{i}+z\widehat{k}\right)$ for ${\overrightarrow{r}}_{Q/O}$ in equation (VII).

$\left\{\left(\frac{8Pa}{675}\right)\left[-430\widehat{i}-20\widehat{j}-406\widehat{k}\right]\right\}=\left(x\widehat{i}+z\widehat{k}\right)\times \left\{\left(\frac{3P}{25}\right)\left[2\widehat{i}+20\widehat{j}-\widehat{k}\right]\right\}$

Comparing the coefficients of the x and z components both sides,

$x=2a,z=-1.99a$

Therefore, he axis of wrench intersects the xz-plane at $\underset{\_}{x=2a,z=-1.99a}$.