# Three forces act on a point: 3 N at 0°, 4 N at 90°, and 5 N at 217°. What is the net force?

Mar 30, 2018

The resultant force is $\text{1.41 N}$ at ${315}^{\circ}$.

#### Explanation:

The net force $\left({F}_{\text{net}}\right)$ is the resultant force $\left({F}_{\text{R}}\right)$. Each force can be resolved into an $x$-component and a $y$-component.

Find the $x$-component of each force by multiplying the force by the cosine of the angle. Add them to get the resultant $x$-component.

Sigma(F_"x")=("3 N"*cos0^@) + ("4 N"*cos90^@) + ("5 N"*cos217^@)"="-1 "N"

Find the $y$-component of each force by multiplying each force by the sine of the angle. Add them to get the resultant $x$-component.

$\Sigma \left({F}_{y}\right)$$=$("3 N"*sin0^@) + ("4 N"*sin90^@) + ("5 N"*sin217^@)"="+1 "N"

Use the Pythagorean to get the magnitude of the resultant force.

$\Sigma \left({F}_{R}\right)$$=$$\sqrt{{\left({F}_{x}\right)}^{2} + {\left({F}_{y}\right)}^{2}}$

$\Sigma \left({F}_{R}\right)$$=$$\sqrt{{\left(- 1 \text{N")^2+(1 "N}\right)}^{2}}$

$\Sigma \left({F}_{R}\right)$$=$$\sqrt{{\text{1 N"^2 + "1 N}}^{2}}$

$\Sigma \left({F}_{R}\right)$$=$$\sqrt{{\text{2 N}}^{2}}$

$\Sigma \left({F}_{R}\right)$$=$$\text{1.41 N}$

To find the direction of the resultant force, use the tangent:

$\tan \theta = \frac{{F}_{y}}{{F}_{x}} = \left(\text{1 N")/(-"1 N}\right)$

${\tan}^{- 1} \left(\frac{1}{- 1}\right) = - {45}^{\circ}$

Subtract ${45}^{\circ}$ from ${360}^{\circ}$ to get ${315}^{\circ}$.

The resultant force is $\text{1.41 N}$ at ${315}^{\circ}$.