Lever system - brake (counting of forces)

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SUMMARY

The discussion revolves around the application of the lever principle and hydraulic mechanics, specifically addressing the equation F1D1=F2D2. A participant incorrectly divided the force of 60N by 0.06m instead of using the correct lever arm of 0.2m, leading to confusion regarding the calculation of forces on a piston. The correct approach involves understanding that the question pertains to forces rather than pressures, and emphasizes the use of torque equilibrium for solving such problems.

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Bublifuk
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Hello.
I tried to solve this question:
3586m3n.jpg

So F1D1=F2D2. Thus:
60N/ 0,06m x 0,2 m= f/0,02m x 0,04m
=100 Newton applied to the piston, but this doesn't fit to the answers offered.
What did I wrong?:(
 
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Why are you dividing the 60 N force by 6 cm?
 
Because it applies some force on the area, isn't it?

It's so easy, that I just need to divide 60N by 0,2M?
 
Bublifuk said:
Because it applies some force on the area, isn't it?

It's so easy, that I just need to divide 60N by 0,2M?

I think you are mixing the lever rule with the hydraulic press (Pascal principle).
The question is about forces. You don't need pressures. And dividing by 6 cm will not give pressure anyway (it's piston's diameter not its area).
Are there more questions related to the same diagram?
 
No, only this one.
Have you managed to solve it?
 
Use the equation for the lever or just equilibrium of torques around the pivot.
 

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