Question

2-113. if f = {16i + 10j - 14k} n, determine the magnitude of the projection of f along the axis of the pole and perpendicular to it. problem 2-113

Explanation:

Step1: Find the unit - vector along the pole

First, find the position vector of point $A$ relative to $O$. $\vec{r}_{OA}=(4\mathbf{i}+2\mathbf{j}+0\mathbf{k})$ m. The magnitude of $\vec{r}_{OA}$ is $r_{OA}=\sqrt{4^{2}+2^{2}+0^{2}}=\sqrt{16 + 4}=\sqrt{20}=2\sqrt{5}$ m. The unit - vector along the pole $\vec{u}_{OA}=\frac{\vec{r}_{OA}}{r_{OA}}=\frac{4\mathbf{i}+2\mathbf{j}+0\mathbf{k}}{2\sqrt{5}}=\frac{2}{\sqrt{5}}\mathbf{i}+\frac{1}{\sqrt{5}}\mathbf{j}+0\mathbf{k}$.

Step2: Calculate the projection of $\vec{F}$ along the pole

The projection of $\vec{F}=\{16\mathbf{i}+10\mathbf{j}-14\mathbf{k}\}$ N along the pole is $F_{OA}=\vec{F}\cdot\vec{u}_{OA}=(16\mathbf{i}+10\mathbf{j}-14\mathbf{k})\cdot(\frac{2}{\sqrt{5}}\mathbf{i}+\frac{1}{\sqrt{5}}\mathbf{j}+0\mathbf{k})=\frac{16\times2}{\sqrt{5}}+\frac{10\times1}{\sqrt{5}}+0=\frac{32 + 10}{\sqrt{5}}=\frac{42}{\sqrt{5}}\approx18.8$ N.

Step3: Calculate the projection of $\vec{F}$ perpendicular to the pole

The magnitude of $\vec{F}$ is $|\vec{F}|=\sqrt{16^{2}+10^{2}+(-14)^{2}}=\sqrt{256+100 + 196}=\sqrt{552}=2\sqrt{138}$ N. Using the Pythagorean theorem, the projection of $\vec{F}$ perpendicular to the pole is $F_{\perp}=\sqrt{|\vec{F}|^{2}-F_{OA}^{2}}$. First, $|\vec{F}|^{2}=552$ and $F_{OA}^{2}=\frac{42^{2}}{5}=\frac{1764}{5}=352.8$. Then $F_{\perp}=\sqrt{552-352.8}=\sqrt{199.2}\approx14.1$ N.

Answer:

The magnitude of the projection of $\vec{F}$ along the axis of the pole is $\frac{42}{\sqrt{5}}\approx18.8$ N, and the magnitude of the projection of $\vec{F}$ perpendicular to the axis of the pole is $\sqrt{199.2}\approx14.1$ N.