# Resolving Forces

1. Mar 3, 2009

### skaboy607

Hi,

Attached is a link to an image:

http://i423.photobucket.com/albums/pp315/skaboy607/Image.jpg [Broken]

Probably a very easy question but I cant work out how they get the vertical and horizontal reaction forces to be F/cos(fi) and Ftan(fi). Any help would be most appreciated.

Thanks

Last edited by a moderator: May 4, 2017
2. Mar 4, 2009

### tiny-tim

Hi skaboy607!

(have a phi: φ )

I assume there's no friction at the top, so taking vertical components (assuming there's no acceleration) should give you an almost-vertical force of F/cosφ

3. Mar 4, 2009

### skaboy607

Hi,

Thanks for the phi φ!

When I take vertical forces I get Fcosφ. Not F/cosφ . And I have no idea where the Ftanφ comes from.

4. Mar 4, 2009

### tiny-tim

No … if the unknown force is G, vertical components give you F = Gcosφ

5. Mar 4, 2009

### skaboy607

oooooooooohhhhhhhhhhhh. Thanks! Just confused me because they used the same F. Any ideas on the Ftanφ.

Thanks

6. Mar 4, 2009

### tiny-tim

erm … I was going to ask you that!

(probably got something to do with resolving horizontally )

(btw, is the bottom fixed?)

7. Mar 4, 2009

### skaboy607

oh I don't know, it is the horizontal force but how they got to that i dont know. i'm thinking along the lines sin/cos=tan?

Yea point Q doesnt move.

8. Mar 4, 2009

### tiny-tim

Hint: what is the horizontal component of the F/cosφ force?

9. Mar 4, 2009

### skaboy607

Sorted-think ive got it. Horizontal component of F/cosφ force (F/cosφ)(sinφ) which is equal to Ftanφ!

10. Mar 4, 2009

Woohoo!