Question

Determine the magnitude and direction of the resultant force of the following free body diagram.

Answer 1

Answer:

The magnitude and direction of the resultant force are approximately 599.923 newtons and 36.405°.

Explanation:

First, we must calculate the resultant force ($\vec F$), in newtons, by vectorial sum:

$\vec F = [(-200\,N)\cdot \cos 60^{\circ}+(400\,N)\cdot \cos 45^{\circ}+300\,N]\,\hat{i} + [(200\,N)\cdot \sin 60^{\circ} + (400\,N)\cdot \sin 45^{\circ}-100\,N]\,\hat{j}$ (1)

$\vec F = 182.843\,\hat{i} + 356.048\,\hat{j}$

Second, we calculate the magnitude of the resultant force by Pythagorean Theorem:

$\|\vec F\| = \sqrt{(482.843\,N)^{2}+(356.048\,N)^{2}}$

$\|\vec F\| \approx 599.923\,N$

Let suppose that direction of the resultant force is an standard angle. According to (1), the resultant force is set in the first quadrant:

$\theta = \tan^{-1}\left(\frac{356.048\,N}{482.843\,N} \right)$

Where $\theta$ is the direction of the resultant force, in sexagesimal degrees.

$\theta \approx 36.405^{\circ}$

The magnitude and direction of the resultant force are approximately 599.923 newtons and 36.405°.