Force applied by someones hand on a lever

  • #1
I am redesigning a bicycle wheel quick release lever for a design project. I am trying to put together a force diagram as part of the project and I need to know the average force someones hand could apply to the tip of the lever to calculate the force exerted on the locking mechanism. Could anyone point me in the direction of a rough estimate of the force someones hand could exert, sources would be great but I haven't found anything after extensive googling.

Thanks
 

Answers and Replies

  • #2
stewartcs
Science Advisor
2,177
3
I am redesigning a bicycle wheel quick release lever for a design project. I am trying to put together a force diagram as part of the project and I need to know the average force someones hand could apply to the tip of the lever to calculate the force exerted on the locking mechanism. Could anyone point me in the direction of a rough estimate of the force someones hand could exert, sources would be great but I haven't found anything after extensive googling.

Thanks

The maximum force one can apply statically is equal to one's weight.

CS
 
  • #3
Thanks stewartcs, that will work for the calculations.

On a slightly different question. Can I make these assumptions in my force diagram?
(R is the resistance force caused by friction)

F1x(45x10^-3)=F2

F2-R=N

Or do I have to take torque into account, I'm trying to keep it as simple as possible and the project doesn't require me to go into to much detail, its more a question of if I know how to do a project then anything else.
 

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  • #4
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9
The lever -quick release devices I am familiar with operate on a hand grip principle, the lever being pushed with the thumb against the pull of the other fingers on the rest of the wheel.

The strength of hand grip is well documented eg

http://www.brianmac.co.uk/grip.htm
 
  • #5
stewartcs
Science Advisor
2,177
3
On a slightly different question. Can I make these assumptions in my force diagram?
(R is the resistance force caused by friction)

F1x(45x10^-3)=F2

F2-R=N

Or do I have to take torque into account, I'm trying to keep it as simple as possible and the project doesn't require me to go into to much detail, its more a question of if I know how to do a project then anything else.

Generally speaking, you'll need to sum all of the forces and moments acting on they body in order to find the missing force. So yes you'll need to consider the torque (i.e. moments).

F1 x (0.045) = Torque, not force. Now the torque at the fulcrum will be constant since you are applying a constant force from F1. So if you have some type of lever arrangement then you can calculated the force at F2 based on that torque (i.e. F1 x 0.045 = F2 x d). Then you'll have the horizontal force. From that you can sum the remaining horizontal forces to find the net horizontal force (magnitude and direction).

I'm not sure what parts are moving or how they are connected in your diagram. Can you label it or explain it a bit better?

CS
 
  • #6
Sorry I didn't really explain how it works. Hopefully these pictures explain it better then words.

The first one is when the mech is open and the second when it's closed. The third the detail of the lever.
 

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  • #7
....and the exploded views, could only upload 3 files at a time.
 

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