determine the magnitude of the projection of the force f acting along line oa. f = 420 n. express your…

Explanation:

Step1: Find the unit - vector along OA

First, find the position vector of point A with respect to O. The coordinates of point A in the x - y - z system (assuming O as the origin) are: $x = 300\sin30^{\circ}\text{ mm}$, $y=- 300 - 300\cos30^{\circ}\text{ mm}$, $z = 300\sin30^{\circ}\text{ mm}$. The position vector $\vec{r}{OA}=300\sin30^{\circ}\hat{i}+(-300 - 300\cos30^{\circ})\hat{j}+300\sin30^{\circ}\hat{k}$ (in mm). The magnitude of $\vec{r}{OA}$ is $r_{OA}=\sqrt{(300\sin30^{\circ})^2+(-300 - 300\cos30^{\circ})^2+(300\sin30^{\circ})^2}$. $r_{OA}=\sqrt{(150)^2+(-300 - 259.81)^2+(150)^2}=\sqrt{22500+( - 559.81)^2+22500}=\sqrt{22500 + 313407.24+22500}=\sqrt{358407.24}\approx598.67\text{ mm}$. The unit - vector $\hat{u}{OA}=\frac{\vec{r}{OA}}{r_{OA}}=\frac{300\sin30^{\circ}\hat{i}+(-300 - 300\cos30^{\circ})\hat{j}+300\sin30^{\circ}\hat{k}}{598.67}$.

Step2: Express the force vector

The force vector $\vec{F}=420(\sin30^{\circ}\hat{i}-\cos30^{\circ}\hat{j})\text{ N}$ (assuming the direction of the force in the x - y plane as shown).

Step3: Calculate the projection of the force along OA

The projection of $\vec{F}$ along $\vec{r}{OA}$ is given by $F{OA}=\vec{F}\cdot\hat{u}{OA}$. $\vec{F}\cdot\hat{u}{OA}=\frac{420}{598.67}[(300\sin30^{\circ}\sin30^{\circ})+(-300 - 300\cos30^{\circ})(-\cos30^{\circ})+(300\sin30^{\circ})\times0]$. First term: $300\sin30^{\circ}\sin30^{\circ}=300\times\frac{1}{2}\times\frac{1}{2}=75$. Second term: $(-300 - 300\cos30^{\circ})(-\cos30^{\circ})=(300 + 259.81)\times0.866=(559.81)\times0.866 = 484.8$. $F_{OA}=\frac{420}{598.67}(75 + 484.8)=\frac{420}{598.67}\times559.8=\frac{235116}{598.67}\approx393\text{ N}$.

Answer:

$393\text{ N}$