Determine the resultant for the vector sum: 10N at 045° and then 8 N at 068°?

Question

So if you draw it out on a quadrant the angle between 10N and 8N should be

68 - 45 = 23° right?

So can i do cosine law to solve the resultant?

c^2 = 10^2 + 8^2 - 2(10)(8)cos 23 c = 4.09

is this wrong?

how do i find the direction of the resultant?

Answer 1

You are using the wrong angle. We have $AB=10, BC=8 \angle CAB=\angle DBE23^\circ$ To get the resultant you want $AD$, so the angle to use in the cosine law is $\angle ABD$

To the resultant, I would put $A$ at the origin and calculate the vectors $AB, AC$, then add them.

Added: One way is to calculate the vectors. Assuming three place accuracy is acceptable, $10$N at $45^\circ$ is $(7.07,7.07)$N and $8$N at $68^\circ$ is $(8\cos 68^\circ, 8\sin 68^\circ)N=(3.00,7.42)$N, so the resultant is $(10.07,14.49)N$ with an angle of $\arctan(\frac {14.49}{10.07})=55.20^\circ$ The other way to get the resultant is the cosine law: $AD^2=AB^2+BD^2-2AB\cdot BD \cos \theta$. As $\angle DBE=23^\circ, \angle ABD=157^\circ$, so $AC^2=164-160\cdot (-0.921), AC=17.643$ There are rounding errors.