problem 2.4 - find the angle θ between force vectors f₁ and f₂. f₁ = 800 n 70° θ f₂ = 300 n (7 m, 3 m, 6 m)…

Explanation:

Step1: Express vectors in component - form

First, find the components of $\vec{F}_1$. The direction of $\vec{F}_1$: Let's assume it lies in a plane. If we consider the given angles, we can write $\vec{F}_1 = 800\left(-\sin25^{\circ}\vec{i}-\cos25^{\circ}\vec{j}\right)\text{ N}$. $\vec{F}_1=800\left(- 0.4226\vec{i}-0.9063\vec{j}\right)=(-338.08\vec{i} - 725.04\vec{j})\text{ N}$.

The position vector of the end - point of $\vec{F}_2$ is $\vec{r}=(7\vec{i}+3\vec{j}+6\vec{k})\text{ m}$. Assuming $\vec{F}_2$ starts from the origin, $\vec{F}_2$ has magnitude $300\text{ N}$ and direction along the position vector $\vec{r}$. The unit vector along $\vec{r}$ is $\hat{r}=\frac{\vec{r}}{\vert\vec{r}\vert}$, where $\vert\vec{r}\vert=\sqrt{7^{2}+3^{2}+6^{2}}=\sqrt{49 + 9+36}=\sqrt{94}\approx9.7$. $\hat{r}=\frac{7\vec{i}+3\vec{j}+6\vec{k}}{9.7}$, so $\vec{F}_2 = 300\times\frac{7\vec{i}+3\vec{j}+6\vec{k}}{9.7}=(216.49\vec{i}+92.78\vec{j}+185.57\vec{k})\text{ N}$.

Step2: Use the dot - product formula

The dot - product formula is $\vec{F}_1\cdot\vec{F}_2=\vert\vec{F}_1\vert\vert\vec{F}_2\vert\cos\theta$. $\vec{F}_1\cdot\vec{F}_2=(-338.08)\times(216.49)+(-725.04)\times(92.78)+0\times(185.57)$ $=-73122.9 +(- 67279.9)=-140402.8$. $\vert\vec{F}_1\vert = 800\text{ N}$ and $\vert\vec{F}_2\vert = 300\text{ N}$. $\cos\theta=\frac{\vec{F}_1\cdot\vec{F}_2}{\vert\vec{F}_1\vert\vert\vec{F}_2\vert}=\frac{- 140402.8}{800\times300}=\frac{-140402.8}{240000}\approx - 0.585$. $\theta=\cos^{-1}(-0.585)\approx125.8^{\circ}$.

Answer:

$\theta\approx125.8^{\circ}$