Chapter 2.9, Problem 25FP

Determine the angle θ between the force and the line AO.

Prob. F2-25

To determine

The angle $\theta$ between the force and the line AO.

Answer to Problem 25FP

The angle $\theta$ between the force and the line AO is $\underset{\_}{57.72°}$.

Explanation of Solution

Given:

The force in the form of Cartesian vector, $F=\left\{-6i+9j+3k\right\}$.

Explanation:

Draw the free body diagram of forces F and line AO as in Figure (1).

Let us establish a position vector for points A and O. The coordinate points $A\left(1,-2,2\right)\text{m}$ and point O $\left(0,0,0\right)$ m.

Express the position vector r in the direction of AO using Figure (1).

$r_{AO}=\left(x_O-x_A\right)i+\left(y_O-y_A\right)j+\left(z_O-z_A\right)k$ (I)

Here, the coordinates of A and O are $x_A,x_O,y_A,y_O,z_A,\text{and}{z}_O$ respectively.

Express the magnitude of the direction of AO.

$r_{AO}=\sqrt{{\left(x_O-x_A\right)}^2+{\left(y_O-y_A\right)}^2+{\left(z_O-z_A\right)}^2}$ (II)

Express the force in Cartesian vector form.

$F=\left\{-6i+9j+3k\right\}$

Express the magnitude of the direction of F.

$\begin{array}{c}F=\sqrt{{\left(-6\right)}^2+{\left(9\right)}^2+{\left(3\right)}^2}\\ =\sqrt{36+81+9}\\ =11.22\text{kN}\end{array}$

Determine the unit vector of AO.

$u_{AO}=\left(\frac{r_{AO}}{r_{AO}}\right)$ (III)

Apply the dot product to find the angle between F and line AO.

$\begin{array}{l}{u}_{AO}\cdot F=r_{AO}\cdot F\cos \theta \\ \theta ={\cos }^{-1}\frac{\left(u_{AO}\cdot F\right)}{F}\end{array}$ (IV)

Conclusion:

Substitute 1 m for $x_A$, ­2 m for $y_A$, 2 m for $z_A$, 0 for $x_O$, 0 for $y_O$, and 0 m for $z_O$ in Equation (I).

$\begin{array}{c}{r}_{AO}=\left(0-1\right)i+\left(0-\left(-2\right)\right)j+\left(0-2\right)k\\ =-i+2j-2k\end{array}$

Substitute 1 m for $x_A$, ­2 m for $y_A$, 2 m for $z_A$, 0 for $x_O$, 0 for $y_O$, and 0 m for $z_O$ in Equation (II).

$\begin{array}{c}{r}_{AO}=\sqrt{{\left(0-1\right)}^2+{\left(0-\left(-2\right)\right)}^2+{\left(0-2\right)}^2}\\ =\sqrt{1+4+4}\\ =3\text{m}\end{array}$

Substitute $-i+2j-2k$ for $r_{AO}$ and 3 m for $r_{AO}$ in Equation (III).

$\begin{array}{c}{u}_{AO}=\left(\frac{-i+2j-2k}{3\text{m}}\right)\\ =-\frac{1}{3}i+\frac{2}{3}j-\frac{2}{3}k\end{array}$

Substitute $-\frac{1}{3}i+\frac{2}{3}j-\frac{2}{3}k$ for $u_{AO}$, $\left\{-6i+9j+3k\right\}$ for $F$, and 11.22 kN for F in Equation (IV).

$\begin{array}{c}\theta ={\cos }^{-1}\frac{\left(\left\{-\frac{1}{3}i+\frac{2}{3}j-\frac{2}{3}k\right\}\text{m}\cdot \left\{-6i+9j+3k\right\}\text{kN}\right)}{\left(11.22\text{kN}\right)}\\ ={\cos }^{-1}\frac{\left(2+6-2\right)}{11.22}\\ ={\cos }^{-1}0.534\\ =57.72°\end{array}$

Thus, the angle $\theta$ between the force F and the line AO is $\underset{\_}{57.72°}$.