# An object with a mass of 14 kg is acted on by two forces. The first is F_1= < 4 N , 6 N> and the second is F_2 = < 2 N, 9 N>. What is the object's rate and direction of acceleration?

Jan 8, 2017

$\vec{a} = \left(\begin{matrix}\frac{6}{14} \\ \frac{15}{14}\end{matrix}\right) m {s}^{- 2} = \frac{1}{14} \left(6 \hat{\vec{i}} + 15 \hat{\vec{j}}\right) m {s}^{- 2}$

direction to $x -$axis: ${\tan}^{- 1} \left(\frac{5}{2}\right)$

#### Explanation:

Newton's second law states

Net force =mass $\times$acceleration

In this case we have

${\vec{F}}_{1} = \left(\begin{matrix}4 \\ 6\end{matrix}\right) N$

${\vec{F}}_{2} = \left(\begin{matrix}2 \\ 9\end{matrix}\right) N$

net force

$= \Sigma \vec{F} = {\vec{F}}_{1} + {\vec{F}}_{2} = \left(\begin{matrix}4 \\ 6\end{matrix}\right) + \left(\begin{matrix}2 \\ 9\end{matrix}\right) N$

$\Sigma F = \left(\begin{matrix}6 \\ 15\end{matrix}\right) N$

$N 2 L$

$\left(\begin{matrix}6 \\ 15\end{matrix}\right) = 14 \vec{a}$

$\vec{a} = \left(\begin{matrix}\frac{6}{14} \\ \frac{15}{14}\end{matrix}\right) m {s}^{- 2}$

direction to $x -$axis

${\tan}^{- 1} \left(\frac{\frac{15}{14}}{\frac{6}{14}}\right) = {\tan}^{- 1} \left(\frac{15}{6}\right) = {\tan}^{- 1} \left(\frac{5}{2}\right)$