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 = <2 N, -8 N>. What is the object's rate and direction of acceleration?

Aug 28, 2016

The object is accelerating at $0.90 {\text{m/s}}^{2}$ in the direction ${56.3}^{\text{o}}$ counterclockwise from negative $x$-axis.

Explanation:

The net force, $\vec{{F}_{\text{net}}}$, is given by

$\vec{{F}_{\text{net}}} = \vec{{F}_{1}} + \vec{{F}_{2}}$

$= < - 4 \text{N", 5 "N" > + < 2 "N", -8 "N} >$

$= < - 2 \text{N", -3 "N} >$

Recall Newton's ${2}^{\text{nd}}$ law

$\vec{{F}_{\text{net}}} = m \vec{a}$

where $m$ is the mass of an object and $\vec{a}$ is its acceleration.

Substituting the values in, the resulting equation is

$< - 2 \text{N", -3 "N" > = (4 "kg") * < a_"x" , a_"y} >$

Solving for ${a}_{\text{x}}$ and ${a}_{\text{y}}$,

${a}_{\text{x" = frac{-2 "N"}{4 "kg"} = -0.5 "m/s}}^{2}$
${a}_{\text{y" = frac{-3 "N"}{4 "kg"} = -0.75 "m/s}}^{2}$

The object's acceleration is found to be

$\vec{a} = < - 0.5 {\text{m/s"^2, -0.75 "m/s}}^{2} >$

Its magnitude is

$| | \vec{a} | | = \sqrt{{\left(- 0.5 {\text{m/s"^2)^2 + (-0.75 "m/s}}^{2}\right)}^{2}}$

$= 0.90 {\text{m/s}}^{2}$

Its direction is

tan^{-1}(abs(frac{-0.75 "m/s"^2}{-0.5 "m/s"^2})) = 56.3^"o"

The angle lies in the third quadrant since both the $x$ and the $y$ components of the vector are negative. The direction of the acceleration is ${56.3}^{\text{o}}$ counterclockwise from negative $x$-axis.