Two dogs pull horizontally on ropes attached to a post; the angle between the ropes is 60.7.
Dog A exerts a force of 280 and dog B exerts a force of 310.
Using this given information I was able to calculate the magnitude of the resultant force to be 509 N by finding Rx= A+Bcos(60.7), Rx=431.71N, and Ry= Bsin(60.7), Ry=270.34N, then using the pythagorean theorem to calculate c.
However, I am now having difficulties finding the angle the resultant force makes with dog A’s rope. Could someone please point me in the right direction? I would really appreciate it!
Thanks! You guys helped immensely. I had the right idea, but instead of finding the arctan(Fj/Fi) I was finding the arctan(IFI/Fi). Now, I was able to find the correct answer. I’ll remember to think of planes in i and j, and I will definitely be studying the sine and cosine rules even more extensively.
3 Answers

Do your own home work!
Good luck buddy.

You should get used to using i and j for directions…
easier way to think of it is ‘i’ is in the x direction and ‘j’ is in the y direction
FA = 280 i + 0 j N
FB = 310cos(60.7) i + 310sin(60.7) j N
Sum of the forces in the i direction = Rx = 280 N + 310 N * cos(60.7)
Sum of the forces in the j direction = Ry = 0 N + 310 N*sin(60.7)
F = (280 N + 310 N * cos(60.7)) i + (0 N + 310 N*sin(60.7)) j
F = sqrt(Fi^2 + Fj^2)
Direction = atan (Fj / Fi)

If you just want a pointer, then I suggest you review the Sine rule and Cosine rule for any triangle.
a/SinA = b / SinB = c / SinC
a^2 = b^2 + c^2 – 2bcCosA
where
side a is opposite angle A
side b is opposite angle B &
side c is opposite angle C
Nearly all of these sorts of problems reduce to using one or both of these rules.