# The angle between two horizontal forces? (1 Viewer)

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#### zcm5000

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

http://img96.imageshack.us/img96/9823/physicsquestiondd9.jpg [Broken]

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

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#### BishopUser

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

Homework Helper
zcm5000 said:
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
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.

#### zcm5000

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.
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?

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Homework Helper
zcm5000 said:
Yes, I believe that is right. However, 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?
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.

#### zcm5000

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.
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

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#### civil_dude

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

#### 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

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