Find the Cutting Force Generated At A

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SUMMARY

The discussion centers on calculating the cutting force generated at point A in a pruner using a compound leverage mechanism. The initial calculation yielded a cutting force of 65 N, which was incorrect due to a misunderstanding of the blade's attachment. After clarifying the free body diagram (FBD) and considering the forces acting on the blade, the correct cutting force at point A was determined to be 169 N. This solution involved analyzing moments at points B and C and correctly accounting for the forces acting on the blade.

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  • Understanding of free body diagrams (FBD)
  • Knowledge of moment calculations in mechanics
  • Familiarity with compound leverage mechanisms
  • Basic trigonometry for resolving forces
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Homework Statement



The pruner multiplies blade-cutting power with the compound leverage mechanism.
If F = 26 is applied to the handles, determine the cutting force generated at A . Assume that the contact surface at is smooth.
Hibbler.ch6.p85.jpg

Homework Equations


M = F x d

The Attempt at a Solution



I drew the free body diagram of the upper handle. I took the moment at point B.

M_b = -26N (150 mm) + (Cutting Force) (60 mm) = 0 -> Cutting force = 65 N.
The answer is incorrect. I don't understand how that could be incorrect. There's no other force acting on the upper handle.

I tried looking at the lower handle, but there's point C and D which I can't figure out the FBD for. Please help! I really don't understand what is going on. I've been looking at that picture for half and hour and nothing. Dumb it down for me. I don't need an answer. I just need to know what is missing or going on. Please!
 
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gocats741 said:
I drew the free body diagram of the upper handle. I took the moment at point B.

M_b = -26N (150 mm) + (Cutting Force) (60 mm) = 0 -> Cutting force = 65 N.
The answer is incorrect. I don't understand how that could be incorrect.

Hi gocats741, http://img96.imageshack.us/img96/5725/red5e5etimes5e5e45e5e25.gif

The grey blade is not attached to the upper handle, apart from being free to pivot at point B. Apart from this pivot, the grey blade is acted on by the small grey arm exerting force at point E.
 
Last edited by a moderator:
Thank you Oxygen. That was what cleared things up for me. I thought it was attached so it threw off all the FBD's. This is what happens when you get assigned problems with tiny pictures.

Here is the work I did to correct the situation:

FBD of the lower handle only including points C, D and G.

Take moments at C

M_c = (26N)(150mm) - Dsin45(25mm) = 0 --> D = 220.617

FBD of only the Gray Blade including only points A, B, and E

Force at E is the same as D but acting in the opposite direction for this FBD

Take moments at pin B

M_b = F_a(60mm) + Dsin45(55mm) +Dcos45(10mm) = 0

Plug in values for D and solve for F_a

F_a = 169 N.

Correct! Thanks!
 

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