# Question #e4d20

Sep 2, 2017

I would begin by drawing a force diagram.

We are given that:

• $\mapsto {\vec{F}}_{1} = 30 \text{N}$

• $\mapsto {\vec{F}}_{2} = 50 \text{N}$

The net force is given by:

$\textcolor{b l u e}{\left\mid {\vec{F}}_{\text{net}} \right\mid = \sqrt{{\left({F}_{x}\right)}^{2} + {\left({F}_{y}\right)}^{2}}}$

We will first need to find the parallel (x, horizontal) and perpendicular (y, vertical) components of the given forces.

• For the northward force, there is no parallel component, as the force vector is entirely vertical, i.e. aligned with the y(north)-axis.

• For the northwest force, the force vector lies at an angle, and so this force has both parallel and perpendicular components. We can calculate these using basic trigonometry.

• Because the force is said to act northwest and we are not given a specific angle, we assume it acts at exactly the midpoint between North and West, i.e. at ${45}^{o}$ above the negative x-axis.

$\sin \left(\theta\right) = \text{opposite"/"hypotenuse}$

$\sin \left(\theta\right) = {F}_{2 y} / {F}_{2}$

Solving for ${F}_{2 y}$:

$\implies \textcolor{b l u e}{{F}_{2 y} = {F}_{2} \sin \left(\theta\right)}$

Similarly, using the cosine, we find that $\textcolor{b l u e}{{F}_{2 x} = {F}_{2} \cos \left(\theta\right)}$. Note that because the parallel component is to the left of the origin, we will say that this is negative.

Therefore, we have:

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

${F}_{\text{x net}} = \sum {F}_{x} = - {F}_{2 x}$

So, we know that:

• ${F}_{y} = 30 + 50 \sin \left({45}^{o}\right)$

• ${F}_{x} = - 50 \cos \left({45}^{o}\right)$

And so, the net force is:

${F}_{\text{net}} = \sqrt{{\left(- 50 \cos \left({45}^{o}\right)\right)}^{2} + {\left(30 + 50 \sin \left({45}^{o}\right)\right)}^{2}}$

$\implies \textcolor{b l u e}{{F}_{\text{net"~~74"N}}}$

Which is closest to answer choice c. The slight difference in values is due to rounding (I used exact values).

Hope that helps!