The angle between two horizontal forces? (1 Viewer)

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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:

[tex]a_x[/tex] from graph = 3 m/s^2

[tex]Fnetx[/tex] = m* [tex]a_x[/tex]

[tex]F_n_e_t_x[/tex] = 3.9 N

[tex]F_n_e_t_x[/tex] = -3 + [tex]F_2*\cos(\theta)[/tex]

6.9 = 9.0*[tex]\cos(\theta)[/tex]

[tex]\cos(\theta)[/tex] = 0.7667
[tex]\theta[/tex] = 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? :confused: Thanks
 
Last edited by a moderator:
Typically you use the smaller angle between two things. Have you tried that?
 

radou

Homework Helper
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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? :confused: 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.
 
radou said:
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 [tex]F_1[/tex] 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 [tex]F_2\cos(\theta)[/tex].

Can anyone confirm or deny this?
 
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radou

Homework Helper
3,085
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zcm5000 said:
Yes, I believe that is right. However, I am unsure if [tex]F_1[/tex] 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 [tex]F_2\cos(\theta)[/tex].

Can anyone confirm or deny this?
Using [tex]F_{2} \cos(\theta)[/tex] is correct, since [tex]\vec{F}_{2}[/tex] makes some angle [tex]\theta[/tex] with the x axis. Further on, [tex]\vec{F}_{1}[/tex] is not 'going' at 180 degrees. It is a simple calculation of the angle between the two force vectors, and 180 arises in it.
 
radou said:
Using [tex]F_{2} \cos(\theta)[/tex] is correct, since [tex]\vec{F}_{2}[/tex] makes some angle [tex]\theta[/tex] with the x axis. Further on, [tex]\vec{F}_{1}[/tex] 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 [tex]F_n_e_t_x[/tex] = [tex]-F_1[/tex] + [tex]F_2\cos(\theta)[/tex]? Hopefully someone can show me where I went wrong
 
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I also get that the angle between them is 140.06 degrees.
 
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 :rolleyes:

At least I understand why I got that answer now
 

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