# An object with a mass of 4 kg is acted on by two forces. The first is F_1= < -4 N , 5 N> and the second is F_2 = < 1 N, -8 N>. What is the object's rate and direction of acceleration?

Sep 27, 2016

$\vec{a} = \frac{3 \sqrt{2}}{4} \text{ } \frac{m}{s} ^ 2$

$\alpha = 180 + 45 = {225}^{o} \text{ with positive x direction}$

#### Explanation:

${F}_{1} = < - 4 N , 5 N >$
${F}_{1 x} = - 4 N$
${F}_{1 y} = 5 N$

${F}_{2} = < 1 N , - 8 N >$
${F}_{2 x} = 1 N$
${F}_{2 y} = - 8 N$

$\text{The vectorial sum of the horizontal components :}$

$\Sigma {\vec{F}}_{x} = {\vec{F}}_{1 x} + {\vec{F}}_{2 x}$

$\Sigma {\vec{F}}_{x} = - 4 + 1 = - 3 N$

$\text{The vectorial sum of the vertical components :}$

$\Sigma {\vec{F}}_{y} = {\vec{F}}_{1 y} + {\vec{F}}_{2 y}$

$\Sigma {\vec{F}}_{y} = 5 - 8 = - 3 N$

$\text{The magnitude of the resultant Force :}$

${F}_{R} = \sqrt{\Sigma {\vec{F}}_{x} + \Sigma {\vec{F}}_{y}}$

${F}_{R} = \sqrt{{\left(- 3\right)}^{2} + {\left(- 3\right)}^{2}}$

${F}_{R} = \sqrt{9 + 9}$

${F}_{R} = 3 \sqrt{2} \text{ N}$

$\text{You can use the Newton's second law to calculate acceleration}$

$\vec{a} : \text{acceleration}$
$m : \text{mass}$
${\vec{F}}_{R} : \text{Force}$

$\vec{a} = {\vec{F}}_{R} / m$

$\vec{a} = \frac{3 \sqrt{2}}{4} \text{ } \frac{m}{s} ^ 2$

$\text{Direction :}$

$\tan \alpha = \frac{- 3}{- 3} = 1$

$\alpha = 180 + 45 = {225}^{o} \text{ with positive x direction}$