Chapter 3.4, Problem 3.139P

(a)

To determine

The resultant force.

Answer to Problem 3.139P

The resultant force is $\underset{\_}{\left(2300N\right)\widehat{i}+\left(500\text{N}\right)\widehat{j}+\left(1850\text{N}\right)\widehat{k}}$.

Explanation of Solution

Write the equation of the distance between AC.

$d_{AC}=\sqrt{r_1^2+r_2^2+r_3^2}$ (I)

Here, the distance between AC is $d_{AC}$, the coordinates are $r_1,r_2,r_3$

Write the equation of the distance between BD.

$d_{BD}=\sqrt{r_4^2+r_5^2+r_6^2}$ (II)

Here, the distance between BD is $d_{BD}$, the coordinates are $r_4,r_5,r_6$

Write the equation of the momentum about AC.

${\overrightarrow{T}}_{AC}=\left(\frac{{\overrightarrow{d}}_{AC}}{d_{AC}}\right)F_1$ (III)

Here, the momentum about AC is ${\overrightarrow{T}}_{AC}$, the force is $F_1$.

Write the equation of the momentum about BD.

${\overrightarrow{T}}_{BD}=\left(\frac{{\overrightarrow{d}}_{BD}}{d_{BD}}\right)F_2$ (IV)

Here, the momentum about BD is ${\overrightarrow{T}}_{BD}$, the force is $F_2$.

Write the equation of resultant force.

$\overrightarrow{R}={\overrightarrow{T}}_{AC}+{\overrightarrow{T}}_{BD}$ (V)

Here, the resultant force is $\overrightarrow{R}$.

Conclusion:

Substitute, $6\text{m}$ for $r_1$, $2\text{m}$ for $r_2$, $9\text{}\text{m}$ for $r_3$ in equation (I)

$\begin{array}{c}{d}_{AC}=\sqrt{{\left(6\text{m}\right)}^2+{\left(2\text{m}\right)}^2+{\left(9\text{m}\right)}^2}\\ =11\text{m}\end{array}$

Substitute, $14\text{m}$ for $r_4$, $2\text{m}$ for $r_5$, $5\text{m}$ for $r_5$ in equation (II)

$\begin{array}{c}{d}_{BD}=\sqrt{{\left(14\text{m}\right)}^2+{\left(2\text{m}\right)}^2+{\left(5\text{m}\right)}^2}\\ =15\text{m}\end{array}$

Substitute, $\left(6\text{m}\right)\widehat{i}+\left(2\text{m}\right)\widehat{j}+\left(9\text{m}\right)\widehat{k}$ for ${\overrightarrow{d}}_{AC}$, $\left(11\text{m}\right)$ for $d_{AC}$, $1650\text{N}$ for $F_1$ in equation (III).

$\begin{array}{c}{\overrightarrow{T}}_{AC}=\left(\frac{\left[\left(6\text{m}\right)\widehat{i}+\left(2\text{m}\right)\widehat{j}+\left(9\text{m}\right)\widehat{k}\right]}{\left(11\text{m}\right)}\right)\left(1650\text{N}\right)\\ =\left[\left(900\text{N}\right)\widehat{i}+\left(300\text{N}\right)\widehat{j}+\left(1350\text{N}\right)\widehat{k}\right]\end{array}$

Substitute, $\left(14\text{m}\right)\widehat{i}+\left(2\text{m}\right)\widehat{j}+\left(5\text{m}\right)\widehat{k}$ for ${\overrightarrow{d}}_{BD}$, $\left(15\text{m}\right)$ for $d_{BD}$, $1500\text{N}$ for $F_2$ in equation (IV).

$\begin{array}{c}{\overrightarrow{T}}_{BD}=\left(\frac{\left[\left(14\text{m}\right)\widehat{i}+\left(2\text{m}\right)\widehat{j}+\left(5\text{m}\right)\widehat{k}\right]}{\left(15\text{m}\right)}\right)\left(1500\text{N}\right)\\ =\left[\left(1400\text{N}\right)\widehat{i}+\left(200\text{N}\right)\widehat{j}+\left(500\text{N}\right)\widehat{k}\right]\end{array}$

Substitute, $\left[\left(1400\text{N}\right)\widehat{i}+\left(200\text{N}\right)\widehat{j}+\left(500\text{N}\right)\widehat{k}\right]$ for ${\overrightarrow{T}}_{BD}$, $\left[\left(900\text{N}\right)\widehat{i}+\left(300\text{N}\right)\widehat{j}+\left(1350\text{N}\right)\widehat{k}\right]$ for ${\overrightarrow{T}}_{AC}$ in equation (V).

$\begin{array}{c}\overrightarrow{R}=\left[\left(1400\text{N}\right)\widehat{i}+\left(200\text{N}\right)\widehat{j}+\left(500\text{N}\right)\widehat{k}\right]+\left[\left(900\text{N}\right)\widehat{i}+\left(300\text{N}\right)\widehat{j}+\left(1350\text{N}\right)\widehat{k}\right]\\ =\left[\left(2300\text{N}\right)\widehat{i}+\left(500\text{N}\right)\widehat{j}+\left(1850\text{N}\right)\widehat{k}\right]\end{array}$

The magnitude of the resultant force,

$\begin{array}{c}\left|\overrightarrow{R}\right|=\sqrt{{\left(2300\text{N}\right)}^2+{\left(500\text{N}\right)}^2+{\left(1850\text{N}\right)}^2}\\ =2993.7\text{N}\end{array}$

Thus, the resultant force is $\underset{\_}{\left(2300N\right)\widehat{i}+\left(500\text{N}\right)\widehat{j}+\left(1850\text{N}\right)\widehat{k}}$.

(b)

To determine

The pitch of the wrench.

Answer to Problem 3.139P

The pitch of the wrench is $\underset{\_}{-0.201\text{m}}$.

Explanation of Solution

Write the equation of pitch of the wrench.

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

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,

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

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

Write the expression for the constant is,

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

Write the equation of momentum.

${\overrightarrow{M}}_o^R=\left[\left({\overrightarrow{r}}_A\times {\overrightarrow{T}}_{AC}\right)+\left({\overrightarrow{r}}_B\times {\overrightarrow{T}}_{BD}\right)\right]$ (VIII)

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

Rewrite the expression for the momentum of the wrench is,

${\overrightarrow{M}}_1=\left(\frac{\overrightarrow{R}}{R}\right)\left[\left({\overrightarrow{r}}_A\times {\overrightarrow{T}}_{AC}\right)+\left({\overrightarrow{r}}_B\times {\overrightarrow{T}}_{BD}\right)\right]$ (IX)

Conclusion:

Substitute, $\left[\left(2300N\right)\widehat{i}+\left(500\text{N}\right)\widehat{j}+\left(1850\text{N}\right)\widehat{k}\right]$ for $\overrightarrow{R}$, $2993.7\text{N}$ for $R$, $\left(12\text{m}\right)\widehat{k}$ for ${\overrightarrow{r}}_A$, $\left[\left(900\text{N}\right)\widehat{i}+\left(300\text{N}\right)\widehat{j}+\left(1350\text{N}\right)\widehat{k}\right]$ for ${\overrightarrow{T}}_{AC}$, $\left(\text{9}\text{m}\right)\widehat{i}$ for ${\overrightarrow{r}}_B$, $\left[\left(1400\text{N}\right)\widehat{i}+\left(200\text{N}\right)\widehat{j}+\left(500\text{N}\right)\widehat{k}\right]$ for ${\overrightarrow{T}}_{BD}$ in equation (IX).

$\begin{array}{c}{M}_1=\left(\frac{\left[\left(2300N\right)\widehat{i}+\left(500\text{N}\right)\widehat{j}+\left(1850\text{N}\right)\widehat{k}\right]}{\left(2993.7\text{N}\right)}\right)\left[\begin{array}{l}\left(\left(12\text{m}\right)\widehat{k}\times \left[\left(900\text{N}\right)\widehat{i}+\left(300\text{N}\right)\widehat{j}+\left(1350\text{N}\right)\widehat{k}\right]\right)+\\ \left(\left(\text{9}\text{m}\right)\widehat{i}\times \left[\left(1400\text{N}\right)\widehat{i}+\left(200\text{N}\right)\widehat{j}+\left(500\text{N}\right)\widehat{k}\right]\right)\end{array}\right]\\ =\left[\left(-461.93\text{N-m}\right)\widehat{i}+\left(-100.42\text{N-m}\right)\widehat{j}+\left(-371.56\text{N-m}\right)\widehat{k}\right]\\ =-601.26\text{N-m}\end{array}$

Substitute, $-601.26\text{N-m}$ for $M_1$, $\left(\text{2993}\text{.7}\text{N}\right)$ for $R$ in equation (VI)

$\begin{array}{c}p=\frac{\left(-601.26\text{N-m}\right)}{\left(\text{2993}\text{.7}\text{N}\right)}\\ =-0.201\text{m}\end{array}$

Thus, the pitch of the wrench is $\underset{\_}{-0.201\text{m}}$.

(c)

To determine

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

Answer to Problem 3.139P

The axis of wrench intersects the yz-plane at $\underset{\_}{y=-0.944m,z=2.78\text{m}}$.

Explanation of Solution

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

Refer fig 1,

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

${\overrightarrow{M}}_o^R={\overrightarrow{M}}_1+{\overrightarrow{M}}_2$ (X)

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

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

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

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

Write the expression for the position vector is,

${\overrightarrow{r}}_{Q/O}=\left(y\widehat{j}+z\widehat{k}\right)$

Here, the coordinates are $y,z$.

Conclusion:

Substitute, $\left(12\text{m}\right)\widehat{k}$ for ${\overrightarrow{r}}_A$, $\left[\left(900\text{N}\right)\widehat{i}+\left(300\text{N}\right)\widehat{j}+\left(1350\text{N}\right)\widehat{k}\right]$ for ${\overrightarrow{T}}_{AC}$, $\left(\text{9}\text{m}\right)\widehat{i}$ for ${\overrightarrow{r}}_B$, $\left[\left(1400\text{N}\right)\widehat{i}+\left(200\text{N}\right)\widehat{j}+\left(500\text{N}\right)\widehat{k}\right]$ for ${\overrightarrow{T}}_{BD}$ in equation (VIII).

$\begin{array}{c}{M}_o^R=\left[\begin{array}{l}\left(\left(12\text{m}\right)\widehat{k}\times \left[\left(900\text{N}\right)\widehat{i}+\left(300\text{N}\right)\widehat{j}+\left(1350\text{N}\right)\widehat{k}\right]\right)+\\ \left(\left(\text{9}\text{m}\right)\widehat{i}\times \left[\left(1400\text{N}\right)\widehat{i}+\left(200\text{N}\right)\widehat{j}+\left(500\text{N}\right)\widehat{k}\right]\right)\end{array}\right]\\ =-\left[\left(3600\text{N-m}\right)\widehat{i}+\left(6300\text{N-m}\right)\widehat{j}+\left(1800\text{N-m}\right)\widehat{k}\right]\end{array}$

Substitute, $-\left[\left(3600\text{N-m}\right)\widehat{i}+\left(6300\text{N-m}\right)\widehat{j}+\left(1800\text{N-m}\right)\widehat{k}\right]$ for ${\overrightarrow{M}}_o^R$, $\left[\left(-461.93\text{N-m}\right)\widehat{i}+\left(-100.42\text{N-m}\right)\widehat{j}+\left(-371.56\text{N-m}\right)\widehat{k}\right]$ for ${\overrightarrow{M}}_1$ in equation (X).

$\begin{array}{l}-\left[\left(3600\text{N-m}\right)\widehat{i}+\left(6300\text{N-m}\right)\widehat{j}+\left(1800\text{N-m}\right)\widehat{k}\right]=\left[\begin{array}{l}\left(-461.93\text{N-m}\right)\widehat{i}+\left(-100.42\text{N-m}\right)\widehat{j}\\ +\left(-371.56\text{N-m}\right)\widehat{k}\end{array}\right]+{\overrightarrow{M}}_2\\ {\overrightarrow{M}}_2=\left[\left(-3138.1\text{N-m}\right)\widehat{i}+\left(6400.4\text{N-m}\right)\widehat{j}+\left(2171.6\text{N-m}\right)\widehat{k}\right]\end{array}$

Substitute, $\left[\left(-3138.1\text{N-m}\right)\widehat{i}+\left(6400.4\text{N-m}\right)\widehat{j}+\left(2171.6\text{N-m}\right)\widehat{k}\right]$ for ${\overrightarrow{M}}_2$, $\left[\left(\text{2300N}\right)\widehat{i}+\left(500\text{N}\right)\widehat{j}+\left(1850N\right)\widehat{k}\right]$ for $\overrightarrow{R}$, $\left(x\widehat{i}+z\widehat{k}\right)$ for ${\overrightarrow{r}}_{Q/O}$ in equation (XI).

$\left[\left(-3138.1\text{N-m}\right)\widehat{i}+\left(6400.4\text{N-m}\right)\widehat{j}+\left(2171.6\text{N-m}\right)\widehat{k}\right]=\left(y\widehat{j}+z\widehat{k}\right)\times \left[\left(\text{2300N}\right)\widehat{i}+\left(500\text{N}\right)\widehat{j}+\left(1850N\right)\widehat{k}\right]$

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

$y=-0.944m,z=2.78\text{m}$

Therefore, the axis of wrench intersects the yz-plane at $\underset{\_}{y=-0.944m,z=2.78\text{m}}$.