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

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Discussion Overview

The discussion revolves around the resolution of forces in a physics problem, specifically how to derive the vertical and horizontal reaction forces expressed as F/cos(φ) and Ftan(φ). The scope includes conceptual clarification and mathematical reasoning related to force components.

Discussion Character

  • Conceptual clarification
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant questions how the vertical and horizontal reaction forces are derived as F/cos(φ) and Ftan(φ).
  • Another participant suggests that if there is no friction at the top, vertical components should yield an almost-vertical force of F/cos(φ).
  • A different participant asserts that when taking vertical forces, they arrive at Fcos(φ) instead of F/cos(φ), expressing confusion about the origin of Ftan(φ).
  • One participant proposes that the unknown force G relates to the equation F = Gcos(φ) when considering vertical components.
  • Another participant hints that the horizontal component of the F/cos(φ) force may be relevant to understanding Ftan(φ).
  • One participant connects the horizontal force to the tangent function, suggesting a relationship between sin(φ) and cos(φ) leading to tan(φ).
  • A later reply indicates that the horizontal component of the F/cos(φ) force can be expressed as (F/cos(φ))(sin(φ)), which equals Ftan(φ).

Areas of Agreement / Disagreement

Participants express differing views on the derivation of the forces, with no consensus reached on the correct approach or interpretation of the equations involved.

Contextual Notes

Some assumptions regarding friction and the fixed nature of point Q are mentioned, but these aspects remain unresolved in the discussion.

skaboy607
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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
 
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skaboy607 said:
I can't work out how they get the vertical and horizontal reaction forces to be F/cos(fi) and Ftan(fi).

Hi skaboy607! :smile:

(have a phi: φ :wink:)

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 :wink:
 
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.
 
skaboy607 said:
When I take vertical forces I get Fcosφ. Not F/cosφ

No … if the unknown force is G, vertical components give you F = Gcosφ :wink:
 
oooooooooohhhhhhhhhhhh. Thanks! Just confused me because they used the same F. Any ideas on the Ftanφ.

Thanks
 
skaboy607 said:
Any ideas on the Ftanφ.

erm … I was going to ask you that! :wink:

(probably got something to do with resolving horizontally :rolleyes:)

(btw, is the bottom fixed?)
 
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.
 
Hint: what is the horizontal component of the F/cosφ force?
 
Sorted-think I've got it. Horizontal component of F/cosφ force (F/cosφ)(sinφ) which is equal to Ftanφ!
 
  • #10
:biggrin: Woohoo! :biggrin:
 

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