# An object with a mass of 9 kg is acted on by two forces. The first is F_1= < -7 N , 1 N> and the second is F_2 = < 4 N, -5 N>. What is the objects rate and direction of acceleration?

Jul 12, 2017

$a = \frac{5}{9}$ ${\text{m/s}}^{2}$

$\theta = {233}^{\text{o}}$

#### Explanation:

We're asked to find the acceleration (magnitude and direction) of an object with a known mass and two forces that act on it.

We can use the famous equation

$\sum \vec{F} = m \vec{a}$

to help solve this problem. In component form, this becomes

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

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

We therefore need to find the components of the net force acting on the object.

The net force components are the vector sums of the components of the individual forces:

$\sum {F}_{x} = - 7$ $\text{N}$ $+ 4$ $\text{N}$ $= - 3$ $\text{N}$

$\sum {F}_{y} = 1$ $\text{N}$ $- 5$ $\text{N}$ $= - 4$ $\text{N}$

The components of the acceleration are

${a}_{x} = \frac{\sum {F}_{x}}{m} = \left(- 3 \textcolor{w h i t e}{l} \text{N")/(9color(white)(l)"kg}\right) = - \frac{1}{3}$ ${\text{m/s}}^{2}$

${a}_{y} = \frac{\sum {F}_{y}}{m} = \left(- 4 \textcolor{w h i t e}{l} \text{N")/(9color(white)(l)"kg}\right) = - \frac{4}{9}$ ${\text{m/s}}^{2}$

The magnitude of the acceleration is thus

$a = \sqrt{{\left({a}_{x}\right)}^{2} + {\left({a}_{y}\right)}^{2}} = \sqrt{{\left(- \frac{1}{3} \textcolor{w h i t e}{l} {\text{m/s"^2)^2 + (-4/9color(white)(l)"m/s}}^{2}\right)}^{2}}$

= color(red)(5/9 color(red)("m/s"^2

The direction of the acceleration is

theta = arctan((a_y)/(a_x)) = arctan((-4/9cancel("m/s"^2))/(-1/3cancel("m/s"^2))) = 53.1^"o" + 180^"o"

= color(blue)(233^"o"

(The ${180}^{\text{o}}$ was added because the acceleration is directed toward the third quadrant relative to the object.)