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

Jul 30, 2016

rate:$| \vec{a} | = \sqrt{2}$

direction; $\theta = \frac{3 \pi}{4}$

Explanation:

Newton's 2nd law

$\Sigma \vec{F} = m \vec{a}$

$\Sigma \vec{F} = \left(\begin{matrix}- 2 \\ - 3\end{matrix}\right) + \left(\begin{matrix}- 1 \\ 6\end{matrix}\right) = \left(\begin{matrix}- 3 \\ 3\end{matrix}\right)$

$= m \vec{a} = 3 \vec{a} \setminus q \quad \left[= \left(\begin{matrix}- 3 \\ 3\end{matrix}\right)\right]$

So

$\vec{a} = \left(\begin{matrix}- 1 \\ 1\end{matrix}\right)$

Breaking that vector into a scalar and a direction:

SCALAR:

$| \vec{a} | = \sqrt{{\left(- 1\right)}^{2} + {\left(1\right)}^{2}} = \sqrt{2}$

DIRECTION:

in terms of direction, we can dot product it against the unit x vector , $\hat{x} = \left(\begin{matrix}1 \\ 0\end{matrix}\right)$, to get

$\vec{a} \cdot \hat{x}$

$= \left(\begin{matrix}- 1 \\ 1\end{matrix}\right) \cdot \left(\begin{matrix}1 \\ 0\end{matrix}\right) = - 1 q \quad \triangle$

$= | \vec{a} | | \hat{x} | \cos \theta = | \vec{a} | \cos \theta$

$= | \left(- 1\right) , \left(1\right) | \cos \theta$

$= \sqrt{2} \cos \theta q \quad \square$

reconciling $\square$ and $\triangle$ means that

$\cos \theta = - \frac{1}{\sqrt{2}}$

$\theta = \frac{3 \pi}{4}$