What is the force on the exerted edge of the cap

[pauln]

Homework Statement

To remove a bottle cap, a student exerts a force of 40 N on the opener. What is the force on the exerted edge of the cap by the opener?

http://img225.imageshack.us/my.php?image=22965751hd6.png

Homework Equations

sigma(Tccw) = sigma(Tcw)
t=r*d

The Attempt at a Solution

t= 40*.35
t= 40*.30
+
= 26 n*m

i know its supposed to be in Newtons but I am not sure how to calculate force when torque is involved

 

[Shooting Star]

Since the torque is the same, force*dist is the same in both cases.

 

[pauln]

im still confused
i got torque for the whole bottle opener (t=40*.3) and i got torque for the opener coming in contact with the edge of the lid (t=40*.25)
the answer is 240 N but i come no where close to it

 

[Shooting Star]

Torque of 40 N is acting at a dist of 30 cm from the point = torque of F acting at a dist of 5 cm from the pint =>
F*0.05 = 40*0.30 =>
F = 240. (in N)

 

[pauln]

argh i got it, I was thinking the start of the handle was a pivot so i was backwards in thinking for terms of distance
thanks for clearing it up :)