Question20. A telephone pole has three cables pulling as shown from above, with $\vec{F}_{1}=(300.0 \hat{i}+500.0 \hat{j}), \vec{F}_{2}=-200.0 \hat{i}$, and $\vec{F}_{3}=-800.0 \hat{j}$. (a) Find the net force on the telephone pole in component form. (b) Find the magnitude and direction of this net force.

Studdy Solution
Calculate the direction of the net force using the arctangent function.
The direction $\theta$ of the net force is given by: $$\theta = \tan^{-1}\left(\frac{F_{\text{net}, \hat{j}}}{F_{\text{net}, \hat{i}}}\right)$$ $$\theta = \tan^{-1}\left(\frac{-300.0}{100.0}\right)$$ $$\theta = \tan^{-1}(-3.0)$$ $$\theta \approx -71.57^\circ$$
The negative angle indicates the direction is below the positive $\hat{i}$-axis.
The net force on the telephone pole in component form is $\vec{F}_{\text{net}} = (100.0 \hat{i} - 300.0 \hat{j})$, with a magnitude of $316.23$ and a direction of approximately $-71.57^\circ$.