Chapter 3.4, Problem 3.140P

(a)

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

(b)

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 $\underset{\_}{-0.0988a}$.

(c)

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 $\underset{\_}{x=2a,z=-1.99a}$.