# An object with a mass of 14 kg is acted on by two forces. The first is F_1= < 5 N , 3 N> and the second is F_2 = < 2 N, 9 N>. What is the object's rate and direction of acceleration?

Jul 18, 2017

$a = 0.993$ ${\text{m/s}}^{2}$

$\theta = {59.7}^{\text{o}}$

#### Explanation:

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

To do this, we'll split the problem into $x$- and $y$-components, so we have

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

${F}_{1 y} = 3$ $\text{N}$

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

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

We can add respective components to find the net force components:

$\sum {F}_{x} = {F}_{1 x} + {F}_{2 x} = 5$ $\text{N}$ $+ 2$ $\text{N}$ = color(red)(7color(white)(l)"N"

$\sum {F}_{y} = {F}_{1 y} + {F}_{2 y} = 3$ $\text{N}$ $+ 9$ $\text{N}$ = color(green)(12color(white)(l)"N"

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

a_x = (sumF_x)/m = color(red)(7color(white)(l)"N")/(14color(white)(l)"kg") = color(purple)(0.5color(white)(l)"m/s"^2

a_y = (sumF_y)/m = color(green)(12color(white)(l)"N")/(14color(white)(l)"kg") = color(orange)(6/7color(white)(l)"m/s"^2

The magnitude of the acceleration is thus

a = sqrt((a_x)^2 + (a_y)^2) = sqrt((color(purple)(0.5color(white)(l)"m/s"^2))^2 + (color(orange)(6/7color(white)(l)"m/s"^2))^2

= color(blue)(0.993 color(blue)("m/s"^2

and the direction is

theta = arctan((a_y)/(a_x)) = arctan((color(orange)(6/7color(white)(l)"m/s"^2))/(color(purple)(0.5color(white)(l)"m/s"^2))) = color(blue)(59.7^"o"