# An object with a mass of 2 kg is acted on by two forces. The first is F_1= < -6 N , 1 N> and the second is F_2 = < 4 N, 9 N>. What is the object's rate and direction of acceleration?

Jul 21, 2017

$a = 5.10$ ${\text{m/s}}^{2}$

$\theta = {101}^{\text{o}}$

#### Explanation:

We're asked to find the magnitude and direction of an object's acceleration, given two forces that act on it.

To do this, let's first split up the forces into their components:

${F}_{1 x} = - 6$ $\text{N}$

${F}_{1 y} = 1$ $\text{N}$

${F}_{2 x} = 4$ $\text{N}$

${F}_{2 y} = 9$ $\text{N}$

We'll now find the components of the net force that acts on the body, by adding components:

$\sum {F}_{x} = - 6$ $\text{N}$ $+ 4$ $\text{N}$ $= - 2$ $\text{N}$

$\sum {F}_{y} = 1$ $\text{N}$ $+ 9$ $\text{N}$ $= 10$ $\text{N}$

Now, we can use Newton's second law to find the components of the object's acceleration:

$\sum {F}_{x} = m {a}_{x}$

${a}_{x} = \frac{\sum {F}_{x}}{m} = \left(- 2 \textcolor{w h i t e}{l} \text{N")/(2color(white)(l)"kg}\right) = - 1$ ${\text{m/s}}^{2}$

${a}_{y} = \frac{\sum {F}_{y}}{m} = \left(10 \textcolor{w h i t e}{l} \text{N")/(2color(white)(l)"kg}\right) = 5$ ${\text{m/s}}^{2}$

The magnitude of the acceleration is

$a = \sqrt{{\left({a}_{x}\right)}^{2} + {\left({a}_{y}\right)}^{2}} = \sqrt{{\left(- 1 \textcolor{w h i t e}{l} {\text{m/s"^2)^2 + (5color(white)(l)"m/s}}^{2}\right)}^{2}}$

= color(red)(5.10 color(red)("m/s"^2

And the direction is

theta = arctan((a_y)/(a_x)) = arctan((5cancel("m/s"^2))/(-1cancel("m/s"^2))) = -78.7^"o" + 180^"o"

= color(blue)(101^"o"

Always be sure to check your direction calculation, as it could be ${180}^{\text{o}}$ off! (by noting what direction the acceleration is relative to the object).