Lever system - brake (counting of forces)

AI Thread Summary
The discussion centers on solving a lever system problem involving forces and distances. The initial equation used, F1D1=F2D2, was questioned due to confusion over the application of force and area. It was clarified that the problem pertains to forces rather than pressures, and dividing by the piston diameter does not yield pressure. Participants emphasized the importance of using the lever equation or torque equilibrium for accurate calculations. The conversation concludes with a focus on correctly applying the principles of levers and hydraulic systems to solve the problem.
Bublifuk
Messages
5
Reaction score
0
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?:(
 
Physics news on Phys.org
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.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top