Chapter 3.2, Problem 3.61P

To determine

The moment of force on the point $\overrightarrow{P}$ about the edge $OA$.

Answer to Problem 3.61P

The moment of force on the point $\overrightarrow{P}$ about the edge $OA$ is $M_{OA}=aP/\sqrt{2}$.

Explanation of Solution

The tetrahedron has six sides of length as $a$.

Formula to calculate the moment along the edge $OA$.

${\overrightarrow{M}}_{OA}={\overrightarrow{\lambda }}_{OA}\left({\overrightarrow{r}}_{C/O}\times \overrightarrow{P}\right)$ (I)

Here, moment of force along the edge $OA$ is ${\overrightarrow{M}}_{OA}$, unit vector along $OA$ is ${\overrightarrow{\lambda }}_{OA}$, the position vector drawn from the point $O$ to $C$ is ${\overrightarrow{r}}_{C/O}$, and the force acting along the axis is $\overrightarrow{P}$.

Sketch the tetrahedron as shown in Figure 1.

Refer From fig 1, the length of the $OA$ along $x$ axis.

${\left(OA\right)}_x=\frac{a}{2}$ (II)

Here, the length of the edge is $a$.

From the triangle $OBC$, refer to Figure 1.

$\begin{array}{c}\tan \theta =\frac{opp}{adj}\\ \tan \theta =\frac{BC}{OC}\\ =\frac{{\left(OA\right)}_z}{{\left(OA\right)}_x}\end{array}$

Substitute $30°$ for $\theta$ and $a/2$ for ${\left(OA\right)}_x$.

$\begin{array}{c}\tan 30°=\frac{{\left(OA\right)}_z}{{\left(OA\right)}_x}\\ {\left(OA\right)}_z={\left(OA\right)}_x\tan 30°\\ =\frac{a}{2}\left(\frac{1}{\sqrt{3}}\right)\\ =\frac{a}{2\sqrt{3}}\end{array}$ (III)

Since apply the Pythagoras theorem for the point $OA$ along $x$, y and z axis.

${\left(OA\right)}^2={\left(OA\right)}_x^2+{\left(OA\right)}_y^2+{\left(OA\right)}_z^2$

Substitute $a$ for $\left(OA\right)$, $a/2$ for ${\left(OA\right)}_x$, and $a/2\sqrt{3}$ for ${\left(OA\right)}_z$.

$\begin{array}{c}{\left(a\right)}^2={\left(\frac{a}{2}\right)}^2+{\left(OA\right)}_y^2+{\left(\frac{a}{2\sqrt{3}}\right)}^2\\ {\left(OA\right)}_y^2=a^2-{\left(\frac{a}{2}\right)}^2-{\left(\frac{a}{2\sqrt{3}}\right)}^2\end{array}$

Rewrite the above relation for ${\left(OA\right)}_y$.

$\begin{array}{c}{\left(OA\right)}_y=\sqrt{a^2-\left(\frac{a^2}{4}\right)-\left(\frac{a^2}{12}\right)}\\ =a\sqrt{\frac{2}{3}}\end{array}$

The position vector drawn from the point $O$ to $A$ is,

${\overrightarrow{r}}_{A/O}=\frac{a}{2}\widehat{i}+a\sqrt{\frac{2}{3}}\widehat{j}+\frac{a}{2\sqrt{3}}\widehat{k}$

Sketch the tetrahedron along the position vector drawn from the point $O$ to $C$ as shown in Figure 2.

Refer From fig 2, the unit vector along $OA$.

${\lambda }_{OA}=\frac{1}{2}\widehat{i}+\sqrt{\frac{2}{3}}\widehat{j}+\frac{1}{2\sqrt{3}}\widehat{k}$

The force acting along the point $BC$.

$\overrightarrow{P}={\overrightarrow{\lambda }}_{BC}P$

Here, the unit vector along $BC$ is ${\overrightarrow{\lambda }}_{BC}$.

Replace $\left(a_x\widehat{i}-a_z\widehat{k}\right)/a$ for ${\overrightarrow{\lambda }}_{BC}$, $a\sin \theta$ for $a_x$, and $a\cos \theta$ for $a_z$.

$\begin{array}{c}\overrightarrow{P}=\frac{\left(a_x\widehat{i}-a_z\widehat{k}\right)}{a}P\\ =\frac{\left(a\sin 30°\widehat{i}-a\cos 30°\widehat{k}\right)}{a}P\\ =\frac{P}{2}\left(\widehat{i}-\sqrt{3}\widehat{k}\right)\end{array}$

Refer From fig 2, the position vector drawn from the point $O$ to $C$.

$r_{C/O}=a\widehat{i}$

Conclusion:

From the equation (I) the moment along the edge $OA$ is,

${\overrightarrow{M}}_{OA}={\overrightarrow{\lambda }}_{OA}\left({\overrightarrow{r}}_{C/O}\times \overrightarrow{P}\right)$

The moment along the edge $OA$ in determinant form is,

$\begin{array}{c}{\overrightarrow{M}}_{OA}=\left|\begin{array}{l}\frac{1}{2}\frac{\sqrt{2}}{3}\frac{1}{2\sqrt{3}}\\ 1\text{ 0 0}\\ 1\text{ 0 -}\sqrt{3}\end{array}\right|\left(a\right)\left(\frac{P}{2}\right)\\ =\frac{aP}{2}\left(-\sqrt{\frac{2}{3}}\right)\left(1\right)\left(-\sqrt{3}\right)\\ =\frac{aP}{\sqrt{2}}\end{array}$

Therefore, the moment along the edge $OA$ is $M_{OA}=aP/\sqrt{2}$.