# 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 = < 3 N, 7 N>. What is the object's rate and direction of acceleration?

Aug 5, 2017

$a = 0.91$ $m / {s}^{2}$ at ${51.3}^{o}$ above the horizontal.

#### Explanation:

We can make use of Newton's second law to solve this problem:

${\vec{F}}_{n e t} = m \vec{a}$

where $m$ is the object's mass and $a$ is the object's acceleration

We are given ${\vec{F}}_{1}$ and ${\vec{F}}_{2}$ in vector notation, where each force is broken up into its parallel (x) and perpendicular (y) components. In order to calculate the acceleration of the object, we will need the net force acting on the object, which is the resultant force caused by ${\vec{F}}_{1}$ and ${\vec{F}}_{2}$.

${\vec{F}}_{n e t} = {\vec{F}}_{1} + {\vec{F}}_{2}$

Since our vectors are already broken up into components, we can do simple vector addition to find ${\vec{F}}_{n e t}$

${\vec{F}}_{n e t} = < 5 N , 3 N > + < 3 N , 7 N >$

$= < 8 N , 10 N >$

This is the net force expressed as a vector showing both its parallel and perpendicular components. We will combine them to get a single value (find the magnitude).

${F}_{n e t} = \sqrt{{\left({F}_{x}\right)}^{2} + {\left({F}_{y}\right)}^{2}}$

$= \sqrt{{\left(8 N\right)}^{2} + {\left(10 N\right)}^{2}}$

$= \sqrt{164}$ $N$

$= 2 \sqrt{41}$ $N$

Now we can use the second law to calculate the acceleration.

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

$= \frac{2 \sqrt{41} N}{14 k g}$

$= 0.91 \frac{m}{s} ^ 2$

We can find the direction (angle) of the net force using basic trigonometry.

$\tan \left(\theta\right) = \frac{{F}_{y}}{{F}_{x}}$

$\implies \theta = \arctan \left(\frac{{F}_{y}}{{F}_{x}}\right)$

$= \arctan \left(\frac{10}{8}\right)$

$= {51.3}^{o}$ (above the horizontal/positive x-axis)

The acceleration occurs in the direction of the net force.

Note that you could also express the acceleration as a vector if this is desired.

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

$= \frac{1}{14} \cdot < 8 , 10 >$

$= < \frac{4}{7} , \frac{5}{7} >$ $m / {s}^{2}$

You could then find the vector's magnitude:

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

$= \sqrt{{\left(\frac{4}{7}\right)}^{2} + {\left(\frac{5}{7}\right)}^{2}}$

$= 0.91$ $m / {s}^{2}$