Recent content by powdernose

Okay. I assume the mechanical advantage problem has been adequately solved, thanks for the guidance

I'm not sure I follow. I think it would be easier for me to understand taking the cylinder pressure in isolation. F1 applied to the piston (3.8cm) on top of a column of water, applies pressure P1, For equilibrium at the bottom (4.4cm) must have a pressure P2 P1=P2 in case1 and case3 only...

Yes, I'm sorry about the initial mistake with the angles, I tried to edit the post when I sorted the angles out, but I couldn't edit the post by then. Yes that is correct. F.l.sin(Θ) + F4.sin(Θ).l/2=F1.r.sin(B)/cos(C) + (F1-F-F4).r.sin(B2)/cos(C2) -> F1=(F(l.sin(Θ)+r.sin(B2)/cos(C2)) +...

Yes, I believe I said you were right in post #5, While my estimation of 55 deg was close, because I was looking at the effort arm, when looking at the bottom position I initially, absent minded, looked at the shorter side of the V bar. I did use a protractor later to get better estimations, for...

Okay, so the crank disk needs to apply an equal and opposite upward force on the other end of the lever, and that force pairs with a downward force on the crank disk? And in the case that the force applied on the lever is always vertical, then the force will =F I guess I kept thinking the torque...

Okay, I think I see what you mean. If I don't assume optimal operation with the force applied perpendicular to the lever, then I will have a linear force along the lever towards the crank, and this force will have a vertical component, which should be F3=cos^2(Θ).F And so it would follow that...

Thanks for the link. Actually, the sleeve is never in contact with the crank, my first diagram is inaccurate in that aspect, but I'd said the crank has a diameter of 3.3cm, while the cylinder is 3.8cm in diameter. How does that change things? So to continue: τ=τ1+τ2 F.l.sin(Θ)=Fp.r.sin(B) +...

By the disk do you mean the crank? Or the piston disk? I suppose it doesn't matter as they move downwards in unison. Assuming equilibrium, F=m. a Acceleration =0 so net Force =0 The Lever rod l only creates torque, So vertical component F1 must equal vertical component F2, for the crank to be...

Probably, or at least closer to the perpendicular, but I want the vertical force because I can put a scale under the device and measure the vertical force. I have no idea. At this point I'm really looking more for answers than questions. Is F1=F2? Because of F2, δx displacement along x2...

I plotted the x displacement versus crank angle A for both rod h and rod h2 but I lost the file as the laptop crashed, in any case it was theoretical validation of the piston travel, allowing me to see if there were a full 180 deg range of motion the full travel would be 4cm, and that with...

Thanks, the nickname is a play on perfume terms, powdery scent being one category. The application is a lever espresso machine. I'll attach some pics Yes, that is probably right, as you can see the Lever is actually in a v type shape, my estimations of angle were perhaps not the best...

I am not a student btw, this is seeking extra knowledge regarding a hobby of mine, and it has been too long since I've tried to solve a physics problem. 1. Homework Statement I have the above system in which Lever L (200mm) applies force to a crank D (diameter 33mm), which has a reverse...

Just saying hello, hoping to get some help now and then with practical physics