# Force on the top screw in a hanging basket scenario

1. Nov 23, 2009

1. The problem statement, all variables and given/known data

Hi and Thanks to anyone that takes time to help!

This is not a homework thing but a 'discussion' with friends.

The measurments are not important only that it is the same frame either with strut or reversed for tie.

The difference in opinion is simply that one says there is more force pulling the top screw from the wall in the tie senarios of each design, while the other says there is no difference in force on the top screw either way round.

(We are both in our late 30's so it was some time since working out any actual equations- let alone remembering which ones are relevant)

2. Relevant equations

Errr...vectors, moments..??

3. The attempt at a solution

It's been too long but only exact maths is gonna conclude this little discussion!

Please, please can you prove the error in one of our heads and make the other safe in the knowledge they are not a complete idiot!

EDITave I put this in teh wrong section? It just seemed the best place fer an exact answer.

Last edited: Nov 23, 2009
2. Nov 23, 2009

### A.T.

If we assume the lower screw as the pivot, then the moment around it and the pull on the upper screw are the same in each case.

3. Nov 24, 2009

Thanks A.T.,

Would the pivot not be the bottom end of the bar against the wall and does that make any difference?

What equations need to be used to work out exact numbers for the force acting on the top screw in both senarios for each design?

4. Nov 24, 2009

### A.T.

That might be more sensible. Then both screws are pulled.
Not between the designs, if the distance of the screws to the bottom end of the bar is the same for all of them
http://en.wikipedia.org/wiki/Torque
You compute the torque created by the given force (M = F1 * r1) and then you use this torque to compute the pull force on the screw (F2 = M / r2). Where r is the shortest distance between the pivot and the line of action of F.

5. Nov 24, 2009