Chapter 3.3, Problem 3.76P

If P = 0 in the figure, replace the two remaining couples with a single equivalent couple, specifying its magnitude and the direction of its axis.

To determine

The net equivalent couple vector shown in the triangular co-ordinates and its direction.

Answer to Problem 3.76P

The net equivalent couple vector is $M=604\text{lb}\cdot \text{in}$ and the direction of the couple formed is ${\theta }_x=72.8°,{\theta }_y=27.3°,{\theta }_z=110.5°$.

Explanation of Solution

Write the equation of moment of first couple about an arbitrary axis.

${\overrightarrow{M}}_1={\overrightarrow{r}}_c\times {\overrightarrow{F}}_1$ (I)

Here, the first couple vector formed due to the force $F_1$ is ${\overrightarrow{M}}_1$, perpendicular distance of the line of action of force is $r_c$, and the vector force acting on the side $C$ is ${\overrightarrow{F}}_1$.

Substitute $\left(30\text{in}\right)\widehat{i}$ for $r_c$ and $\left(-16\text{lb}\right)\widehat{j}$ for ${\overrightarrow{F}}_1$ in equation (I).

$\begin{array}{c}{\overrightarrow{M}}_1=\left(30\text{in}\right)\widehat{i}\times \left(-16\text{lb}\right)\widehat{j}\\ =-\left(480\text{lb}\cdot \text{in}\right)\widehat{k}\end{array}$ (II)

Write the equation of moment of second couple about an arbitrary axis.

${\overrightarrow{M}}_2={\overrightarrow{r}}_{E/B}\times {\overrightarrow{F}}_2$ (III)

Here, the second couple vector formed due to the force $F_2$ is ${\overrightarrow{M}}_2$, perpendicular distance of the vector line drawn from $B$ to $E$ is $r_{E/B}$, and the vector force acting on the side $EB$ is ${\overrightarrow{F}}_2$.

The vector line drawn from $B$ to $E$ is,

${\overrightarrow{r}}_{E/B}=r_2\widehat{i}-r_1\widehat{j}$

Here, the distance between the points $DC$ is $r_1$ and distance between the points $EC$ is $r_2$.

Substitute $\left(5\text{in}\right)$ for $r_1$ and $\left(15\text{in}\right)$ for $r_2$.

${\overrightarrow{r}}_{E/B}=\left(15\text{in}\right)\widehat{i}-\left(5\text{in}\right)\widehat{j}$

The distance of the line drawn from $DE$ is,

$\begin{array}{c}{d}_{DE}=\sqrt{{\left(0\right)}^2+{\left(5\right)}^2+{\left(10\right)}^2}\\ =5\sqrt{5}\text{in}\end{array}$

The resultant magnitude of the each force is,

$F_2=\frac{M_{DE}}{d_{DE}}$ (IV)

Here, moment of couple formed by the force at the points $DE$ is $M_{DE}$ and perpendicular distance of the line drawn from $DE$ is $d_{DE}$.

Substitute $40\text{lb}$ for $M_{DE}$ and $5\sqrt{5}\text{in}$ for $d_{DE}$ in equation (IV).

$\begin{array}{c}{F}_2=\frac{40\text{lb}}{5\sqrt{5}\text{in}}\left(5\widehat{j}-10\widehat{k}\right)\\ =8\sqrt{5}\left[\left(1\text{lb}\right)\widehat{j}-\left(2\text{lb}\right)\widehat{k}\right]\end{array}$

Write the vector form for the moment of second couple $M_2$.

$\begin{array}{c}{\overrightarrow{M}}_2=8\sqrt{5}\left|\begin{array}{l}\widehat{i}\widehat{j}\widehat{k}\\ 15-5\text{ 0}\\ \text{0 1 }-\text{2}\end{array}\right|\\ =8\sqrt{5}\left[\left(10\text{lb}\cdot \text{in}\right)\widehat{i}+\left(30\text{lb}\cdot \text{in}\right)\widehat{j}+\left(15\text{lb}\cdot \text{in}\right)\widehat{k}\right]\end{array}$ (V)

Write the equation of net moment of couple about an arbitrary axis.

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

Substitute equation (II) and (V) in equation (VI).

$\begin{array}{c}\overrightarrow{M}=-\left(480\text{lb}\cdot \text{in}\right)\widehat{k}+8\sqrt{5}\left[\left(10\text{lb}\cdot \text{in}\right)\widehat{i}+\left(30\text{lb}\cdot \text{in}\right)\widehat{j}+\left(15\text{lb}\cdot \text{in}\right)\widehat{k}\right]\\ =\left(178.885\text{lb}\cdot \text{in}\right)\widehat{i}+\left(536.66\text{lb}\cdot \text{in}\right)\widehat{j}-\left(211.67\text{lb}\cdot \text{in}\right)\widehat{k}\end{array}$

The magnitude of the net moment of couple is,

$\begin{array}{c}\left|\overrightarrow{M}\right|=\sqrt{{\left(178.885\text{lb}\cdot \text{in}\right)}^2+{\left(536.66\text{lb}\cdot \text{in}\right)}^2+{\left(-211.67\text{lb}\cdot \text{in}\right)}^2}\\ M=603.99\text{lb}\cdot \text{in}\end{array}$

Conclusion:

Write the equation for the unit vector of the couple.

${\lambda }_{\text{axis}}=\frac{\overrightarrow{M}}{M}$ (VII)

Here, the unit vector is represented as ${\lambda }_{\text{axis}}$.

Substitute $\left(178.885\text{lb}\cdot \text{in}\right)\widehat{i}+\left(536.66\text{lb}\cdot \text{in}\right)\widehat{j}-\left(211.67\text{lb}\cdot \text{in}\right)\widehat{k}$ for $\overrightarrow{M}$ and $603.99\text{lb}\cdot \text{in}$ for $M$ in equation (VII).

$\begin{array}{c}{\lambda }_{\text{axis}}=\frac{\left(178.885\text{lb}\cdot \text{in}\right)\widehat{i}+\left(536.66\text{lb}\cdot \text{in}\right)\widehat{j}-\left(211.67\text{lb}\cdot \text{in}\right)\widehat{k}}{603.99\text{lb}\cdot \text{in}}\\ =0.29617\widehat{i}+0.88852\widehat{j}-0.35045\widehat{k}\end{array}$

The resultant couple formed along the x direction.

$\begin{array}{c}\cos {\theta }_x=0.29617\\ {\theta }_x={\cos }^{-1}\left(0.29617\right)\\ =72.8°\end{array}$

The resultant couple formed along the y direction.

$\begin{array}{c}\cos {\theta }_y=0.88852\\ {\theta }_y={\cos }^{-1}\left(0.88852\right)\\ =27.3°\end{array}$

The resultant couple formed along the z direction.

$\begin{array}{c}\cos {\theta }_z=-0.35045\\ {\theta }_z={\cos }^{-1}\left(-0.35045\right)\\ =110.5°\end{array}$

Therefore, the net equivalent couple vector is $M=604\text{lb}\cdot \text{in}$ and the direction of the couple formed is ${\theta }_x=72.8°,{\theta }_y=27.3°,{\theta }_z=110.5°$.