# An object with a mass of 4 kg is acted on by two forces. The first is F_1= < -5 N , 2 N> and the second is F_2 = < 7 N, 6 N>. What is the object's rate and direction of acceleration?

Jul 21, 2017

$a = 2.06$ ${\text{m/s}}^{2}$

$\theta = {76.0}^{\text{o}}$

#### 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:

${F}_{1 x} = - 5$ $\text{N}$

${F}_{1 y} = 2$ $\text{N}$

${F}_{2 x} = 7$ $\text{N}$

${F}_{2 y} = 6$ $\text{N}$

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

$\sum {F}_{x} = - 5$ $\text{N}$ $+ 7$ $\text{N}$ $= 2$ $\text{N}$

$\sum {F}_{y} = 2$ $\text{N}$ $+ 6$ $\text{N}$ $= 8$ $\text{N}$

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

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

${a}_{x} = \frac{\sum {F}_{x}}{m} = \left(2 \textcolor{w h i t e}{l} \text{N")/(4color(white)(l)"kg}\right) = 0.5$ ${\text{m/s}}^{2}$

${a}_{y} = \frac{\sum {F}_{y}}{m} = \left(8 \textcolor{w h i t e}{l} \text{N")/(4color(white)(l)"kg}\right) = 2$ ${\text{m/s}}^{2}$

The magnitude of the acceleration is

$a = \sqrt{{\left({a}_{x}\right)}^{2} + {\left({a}_{y}\right)}^{2}} = \sqrt{{\left(0.5 \textcolor{w h i t e}{l} {\text{m/s"^2)^2 + (2color(white)(l)"m/s}}^{2}\right)}^{2}}$

= color(red)(2.06 color(red)("m/s"^2

And the direction is

theta = arctan((a_y)/(a_x)) = arctan((2cancel("m/s"^2))/(0.5cancel("m/s"^2))) = color(blue)(76.0^"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).