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

Apr 9, 2016

The acceleration, $\vec{a} = \frac{1}{4} \left[\begin{matrix}3 \\ 1\end{matrix}\right] \frac{m}{s} ^ 2$
and the magnitude is:
$| \vec{a} | = \frac{1}{4} \sqrt{{\left(3\right)}^{2} + {\left(1\right)}^{2}} N = \frac{1}{4} \sqrt{10} \frac{m}{s} ^ 2$
You can write the acceleration, $\vec{a} = \frac{\sqrt{10}}{4} \left[\begin{matrix}\frac{3}{\sqrt{10}} \\ \frac{1}{\sqrt{10}}\end{matrix}\right] \frac{m}{s} ^ 2$
Where$\left[\begin{matrix}\frac{3}{\sqrt{10}} \\ \frac{1}{\sqrt{10}}\end{matrix}\right] =$ is the unit vector of $\vec{a}$ #### Explanation:

Given: Forces on an object of mass, $m = 4 k g$
$\vec{{F}_{1}} = \left[\begin{matrix}6 \\ - 2\end{matrix}\right] N$ and $\vec{{F}_{2}} = \left[\begin{matrix}- 3 \\ 3\end{matrix}\right] N$

Required: Object acceleration, $\vec{a}$

Solution Strategy:
A) Draw a "Free Body Diagram" and add the two force vector
B) Use Newton's law sum_l(F_xhati + F_yhatj)=m(a_xhati+a_yjhat)
That is:
1) $\text{ "sum_l F_x = ma_x" }$ $\mathmr{and} \text{ }$ 2) $\text{ } {\sum}_{l} {F}_{y} = m {a}_{y}$

A) See Image Below The FBD yield:
${F}_{x} = {F}_{{1}_{x}} + {F}_{{2}_{x}} = 6 - 3 = 3 N$
${F}_{y} = {F}_{{1}_{y}} + {F}_{{2}_{y}} = - 2 + 3 = 1 N$
Thus the net force, $\vec{{F}_{\text{net}}} = \left[\begin{matrix}3 \\ 1\end{matrix}\right] N$

B) Newton Law:
From equation 1) we have:
1) F_(x)=ma_x; a_x=3/4 m/s^2
F_(y)=ma_y; a_y=1/4 m/s^2

Thus the net acceleration, $\vec{a} = \frac{1}{4} \left[\begin{matrix}3 \\ 1\end{matrix}\right] \frac{m}{s} ^ 2$
and the magnitude is:
$| \vec{a} | = \frac{1}{4} \sqrt{{\left(3\right)}^{2} + {\left(1\right)}^{2}} N = \frac{1}{4} \sqrt{10} \frac{m}{s} ^ 2$