The angle between two horizontal forces?

1. Oct 17, 2006

zcm5000

I need help with this problem (my work is at the bottom):

Here is what I tried:

$$a_x$$ from graph = 3 m/s^2

$$Fnetx$$ = m* $$a_x$$

$$F_n_e_t_x$$ = 3.9 N

$$F_n_e_t_x$$ = -3 + $$F_2*\cos(\theta)$$

6.9 = 9.0*$$\cos(\theta)$$

$$\cos(\theta)$$ = 0.7667
$$\theta$$ = 39.9 degrees from x-axis
degrees between forces = 180+39.9 = 219.9 degrees

This is wrong and I've been trying to figure out what to do for a long time so I'm hoping someone can help me out please? Thanks

Last edited: Oct 17, 2006
2. Oct 17, 2006

BishopUser

Typically you use the smaller angle between two things. Have you tried that?

3. Oct 17, 2006

If the force F1 points in the negative x direction, and F2 has an angle of 39.9 with the positive x direction, then the angle between them is 90 + (90 - 39.9) = 180 - 39.9 = 140.1 degree, unless I missed something.

4. Oct 17, 2006

zcm5000

Yes, I believe that is right. However, I tried inputting 140 and it is incorrect. I am unsure if $$F_1$$ is actually going at 180 degrees. It only says it is going in the negative x direction? Also, I am unsure if I am setting up the equation correctly, and also if I am supposed to use $$F_2\cos(\theta)$$.

Can anyone confirm or deny this?

Last edited: Oct 17, 2006
5. Oct 17, 2006

Using $$F_{2} \cos(\theta)$$ is correct, since $$\vec{F}_{2}$$ makes some angle $$\theta$$ with the x axis. Further on, $$\vec{F}_{1}$$ is not 'going' at 180 degrees. It is a simple calculation of the angle between the two force vectors, and 180 arises in it.

6. Oct 17, 2006

zcm5000

Well I am not sure what I did wrong because neither 140 or 220 degrees is correct. Is the equation supposed to be $$F_n_e_t_x$$ = $$-F_1$$ + $$F_2\cos(\theta)$$? Hopefully someone can show me where I went wrong

Last edited: Oct 17, 2006
7. Oct 17, 2006

civil_dude

I also get that the angle between them is 140.06 degrees.

8. Oct 17, 2006

zcm5000

Thanks guys, the answer WAS 140 degrees :rofl: . I thought I had tried that answer before I tried 220, but I only tried 220 :grumpy: Thanks for making my physics experience a little better

At least I understand why I got that answer now