# An object with a mass of 4 kg is acted on by two forces. The first is F_1= < -2 N , -6 N> and the second is F_2 = < 1 N, 3 N>. What is the object's rate and direction of acceleration?

Oct 10, 2017

$\vec{a} = 0.79 \frac{m}{s} ^ 2 , {251.6}^{o}$

(If sig figs were to be accounted for, you can imagine the answer you'd get)

#### Explanation:

It always best to draw a diagram with the info you know:

You can already visualize that the final vector will most likely point down and to the left relative to the object.

Since you're already given the force vectors in component form, we can use the formula:

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

Where the mass is given (and already in the proper units): $4 k g$

But, we will first find the sum of each vector to find the acceleration in the x- and y- direction, then use basic trigonometry to find the final magnitude and direction.

$\sum {F}_{x} = \left(- 2 N\right) + \left(1 N\right) = - 1 N$

$\sum {F}_{y} = \left(- 6 N\right) + \left(3 N\right) = - 3 N$

We can then use these to find acceleration in each direction:

${a}_{x} = \frac{- 1 N}{4 k g} = - 0.25 \frac{m}{s} ^ 2$

${a}_{y} = \frac{- 3 N}{4 k g} = - 0.75 \frac{m}{s} ^ 2$

$a = \sqrt{{\left(- 0.25 \frac{m}{s} ^ 2\right)}^{2} + {\left(- 0.75 \frac{m}{s} ^ 2\right)}^{2}} = 0.79 \frac{m}{s} ^ 2$

Then to find the vector $\vec{a}$ we need the direction, though using $\arctan$, we have to be careful because we will get a "relative" angle, thus:

$\arctan \left(\frac{- 0.75}{- 0.25}\right) = {71.6}^{o}$

The angle we have here represents $\theta$ in the photo below:

So, we will add ${180}^{o}$ to have our angle in standard form where
+x-axis $= {0}^{o}$

${71.6}^{o} + {180}^{o} = {251.6}^{o}$