The angle between two horizontal forces?

[zcm5000]

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

http://img96.imageshack.us/img96/9823/physicsquestiondd9.jpg

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

 

[BishopUser]

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

 

[radou]

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?

 

[radou]

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

 

[civil_dude]

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

 

[zcm5000]

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

At least I understand why I got that answer now