Resultant Magnitude & Direction: 158.98 N, -86.04° E of S

[cowgiljl]

It wants me to find the magnitude and direction of the resultant
F1 = 75.0 Newtons, 25 degrees east of north
F2 = 100 Newtons 25 degrees east of south

i drew a picture again
I took the for F1 sin 25 * 75 and got 31.69
F2 cos 25 * 75 and got 67.97

Sin 25 * 100 = 42.67
cos 25 * 100 = 90.63

So then i added the 31.69+(-42.67) = -10.98
and added 67.97+90.63 = 158.60

used the C^2 = A^2 + b^2 C^2 =158.60^2+10.98^2
C = 158.98 N
Tan -1 (158.60/-10.98)
and theata was = to a -86.04 degrees east of south

is that one right?

 

[jamesrc]

I think you should double-check how you added the components. The x-component of the force (if we label east as x and north as y) should by (F1+F2)sin(25) and the y-component of the force should be (F1-F2)*cos(25). From your diagram, you should ree that the resultant vector points roughly ESE

 

[NateTG]

Originally posted by cowgiljl
It wants me to find the magnitude and direction of the resultant
F1 = 75.0 Newtons, 25 degrees east of north
F2 = 100 Newtons 25 degrees east of south

i drew a picture again

One of the ways to add verctors is to draw them head to tail with each other. You might not get an exact answer, but it often makes it easy to check your answer.

I suspect that you aren't keeping track of which component is which well enough.

You might do better if you drew them in and labled them x₁ y₁ and x₂ y₂ for F1 and F2 respectively. It would also make it easier for others to follow your work.

From looking at the exercise, I would guess that the resultant force should have a magnitude of about 50N and be pointing roughly southeast

 

[HallsofIvy]

So then i added the 31.69+(-42.67) = -10.98
and added 67.97+90.63 = 158.60

You might find that it helps YOU (not to mention us) if you labeled these better. The 31.69 and 42.67 you got using sin(25): is that "east-west" or "north-south". Once you are sure of that, think about which ways the forces are in the "same direction" and which opposite.