Chapter 3.1, Problem 3.9P

Rod AB is held in place by the cord AC. Knowing that the tension in the cord is 1350 N and that c = 360 mm, determine the moment about B of the force exerted by the cord at point A by resolving that force into horizontal and vertical components applied (a) at point A, (b) at point C.

To determine

The moment about B of the force exerted by the cord at point A.

Answer to Problem 3.9P

The moment about B of the force exerted by the cord at point A is $292\text{N}\cdot \text{m}$ in the clockwise direction.

Explanation of Solution

Refer Figure 1.

Write an expression to calculate the length of cord.

$AC=\sqrt{{\left(OA\right)}^2+{\left(OB+BC\right)}^2}$ (I)

Here, AC is the length of the cord, OA is the horizontal distance between A and C, OB is the distance between the points O and B and BC is the distance between B and C.

Write an expression to calculate angle $\alpha$.

$\alpha ={\tan }^{-1}\left(\frac{OC}{OA}\right)$ (II)

Here, $\alpha$ is the angle $\angle CAO$ and OC is the distance between O and C.

Write an expression to calculate the tension acting along the cord.

$F=F\cos \alpha i+F\sin \alpha j$ (III)

Here, $F$ is the force vector acting along the cord and $F$ is the magnitude of the force vector $F$.

Write an expression to calculate the position vector between point A and B.

$r_{A/B}=-OAi-OBj$ (IV)

Here, $r_{A/B}$ is the position vector between point A and B.

Write an expression to calculate the moment at point B.

$M_B=r_{A/B}\times F$ (V)

Here, $M_B$ is the moment at point B.

Conclusion:

Substitute $450\text{mm}$ for $OA$, $240\text{mm}$ for $OB$, and $360\text{mm}$ for $BC$ in equation (I) to find $AC$.

$\begin{array}{c}AC=\sqrt{{\left(\left(450\text{mm}\right)\left(\frac{1\text{m}}{1000\text{mm}}\right)\right)}^2+{\left(\left(240\text{mm}\right)\left(\frac{1\text{m}}{1000\text{mm}}\right)+\left(360\text{mm}\right)\left(\frac{1\text{m}}{1000\text{mm}}\right)\right)}^2}\\ =0.75\text{m}\end{array}$

Substitute $450\text{mm}$ for $OA$, and $\text{600}\text{mm}$ for $OC$ in equation (II) to find $\alpha$.

$\begin{array}{c}\alpha ={\tan }^{-1}\left(\frac{600\text{mm}}{450\text{mm}}\right)\\ =53.1°\end{array}$

Substitute $1350\text{N}$ for $F$, and $53.1°$ for $\alpha$ in equation (III) to find $F$.

$\begin{array}{c}F=\left(1350\text{N}\right)\cos \left(53.1°\right)i+\left(1350\text{N}\right)\sin \left(53.1°\right)j\\ =\left(810\text{N}\right)i+\left(1080\text{N}\right)j\end{array}$

Substitute $450\text{mm}$ for $OA$, and $240\text{mm}$ for $OB$ in equation (IV) to find $r_{A/B}$.

$\begin{array}{c}{r}_{A/B}=-\left(\left(450\text{mm}\right)\left(\frac{1\text{m}}{1000\text{mm}}\right)\right)i-\left(\left(240\text{mm}\right)\left(\frac{1\text{m}}{1000\text{mm}}\right)\right)j\\ =-\left(0.45\text{m}\right)i-\left(0.24\text{m}\right)j\end{array}$

Substitute $\left(810\text{N}\right)i+\left(1080\text{N}\right)j$ for $F$, and $-\left(0.45\text{m}\right)i-\left(0.24\text{m}\right)j$ for $r_{A/B}$ in equation (V) to find $M_B$.

$\begin{array}{c}{M}_B=\left(-\left(0.45\text{m}\right)i-\left(0.24\text{m}\right)j\right)\times \left(\left(810\text{N}\right)i+\left(1080\text{N}\right)j\right)\\ =-\left(486\text{N}\cdot \text{m}\right)k+\left(194\text{N}\cdot \text{m}\right)k\\ =-\left(292\text{N}\cdot \text{m}\right)k\end{array}$

Therefore, the moment about B of the force exerted by the cord at point A is $292\text{N}\cdot \text{m}$ in the clockwise direction.

To determine

The moment about B of the force exerted by the cord at point C.

Answer to Problem 3.9P

The moment about B of the force exerted by the cord at point C is $292\text{N}\cdot \text{m}$ in the clockwise direction.

Explanation of Solution

Refer Figure 2.

Write an expression to calculate the position vector between point A and B.

$r_{C/B}=BCj$ (VI)

Here, $r_{C/B}$ is the position vector between point C and B.

Write an expression to calculate the moment at point B.

$M_B=r_{C/B}\times F$ (VII)

Conclusion:

Substitute $360\text{mm}$ for $BC$ in equation (VI) to find $r_{C/B}$.

$\begin{array}{c}{r}_{C/B}=\left(\left(360\text{mm}\right)\left(\frac{1\text{m}}{1000\text{mm}}\right)\right)j\\ =\left(0.36\text{m}\right)j\end{array}$

Substitute $\left(810\text{N}\right)i+\left(1080\text{N}\right)j$ for $F$, and $\left(0.36\text{m}\right)j$ for $r_{C/B}$ in equation (VII) to find $M_B$.

$\begin{array}{c}{M}_B=\left(\left(0.36\text{m}\right)j\right)\times \left(\left(810\text{N}\right)i+\left(1080\text{N}\right)j\right)\\ =-\left(292\text{N}\cdot \text{m}\right)k\end{array}$

Therefore, the moment about B of the force exerted by the cord at point C is $292\text{N}\cdot \text{m}$ in the clockwise direction.