# An object with a mass of 9 kg is acted on by two forces. The first is F_1= < -2 N , -1 N> and the second is F_2 = < 8 N, -5 N>. What is the object's rate and direction of acceleration?

Jan 23, 2017

The object is accelerating at a rate of $1.602 \frac{m}{s} ^ 2$
at an angle of $- {33.69}^{\circ}$

#### Explanation:

The total force acting upon an object is the sum of all the forces acting on that object.

${F}_{\text{tot}} = \sum {F}_{i}$

In this case

${F}_{\text{tot}} = {F}_{1} + {F}_{2}$

${F}_{\text{tot}} = < - 2 N , 1 N > + < 8 N , - 5 N >$

$= < 6 N , - 4 N >$

and by Newton's 2nd Law

$F = m a$

Then we divide by $m = 9 k g$

${F}_{\text{tot}} = 9 k g \cdot < \frac{2}{3} \frac{m}{s} ^ 2 , - \frac{4}{9} \frac{m}{s} ^ 2 >$

Then

$a = < \frac{2}{3} \frac{m}{s} ^ 2 , - \frac{4}{9} \frac{m}{s} ^ 2 >$

The rate of acceleration is the magnitude of a.

$| a | = \sqrt{{\left(\frac{2}{3}\right)}^{2} + {\left(- \frac{4}{9}\right)}^{2}} = \sqrt{\frac{4}{9} + \frac{16}{81}} = \sqrt{\frac{36}{81} + \frac{16}{81}} = \sqrt{\frac{52}{81}} = \frac{4 \sqrt{13}}{9} \approx 1.602 \frac{m}{s} ^ 2$

The direction is down and to the right and the angle $\theta$

$\theta = \arctan \left(- \frac{2}{3}\right) \approx . - 0.588 r a \mathrm{da} p p r \otimes - {33.69}^{\circ}$