Resolving Forces: F/cos(fi) & Ftan(fi) Explained

[skaboy607]

Hi,

Attached is a link to an image:

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

Probably a very easy question but I can't 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

 

[tiny-tim]

I can't work out how they get the vertical and horizontal reaction forces to be F/cos(fi) and Ftan(fi).

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

 

[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.

Thanks for your help.

 

[tiny-tim]

When I take vertical forces I get Fcosφ. Not F/cosφ

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

 

[skaboy607]

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

Thanks

 

[tiny-tim]

Any ideas on the Ftanφ.

erm … I was going to ask you that!

(probably got something to do with resolving horizontally )

(btw, is the bottom fixed?)

 

[skaboy607]

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

Yea point Q doesn't move.

 

[tiny-tim]

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

 

[skaboy607]

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

 

[tiny-tim]

Woohoo! ​