New answers tagged

1

Progressing cavity pumps. The rotor executes a rolling motion in the stator, because of the different auger like shapes of rotor and stator the cavity between moves along the axis. This video shows the movement.


1

Our body has several examples, our throat when swallowing, our intestines would be another example. Arteries in our legs pump the blood by our leg muscles squeezing them.


2

Rolling diphragms have been around for many years now. See... https://www.diacom.com/rolling-diaphragm-theory With or without the spring. (With enough vacuum in the top you can get the membrane to slip upwards, but it is easily reset by cycling.) Fabric reinforcement helps prevent stretch, but generally everything is elastic to some extent. Additionally, ...


0

Q1 - Should a normal and friction force be applied at the base of the weightless bar? A: Yes. Assume the triangle is a non-yielding/rigid body, the reactions on the bar are graphically shown as below. Note: Although the bar is barely resting on the contact surface, however, as long as the ground can provide the required friction force, the bar will be ...


3

In general this is impossible, because for a given value of $F_e$, the friction damper dissipates a fixed amount of energy per cycle of vibration independent of the vibration frequency. This is (nonlinear) hysteretic damping, not (nonlinear) viscous damping. The only way to approximate this with a viscous damper would be to make $C$ a function of both the $...


1

I think a hydraulic device could approximate it. General concept: For each direction: a spring-loaded check valve (ie with a cracking pressure) in series with a small-diameter-tube would realize a roughly constant pressure difference, as long as there is some very small flow (i.e. cylinder movement). This could then operate a pilot valve, opening the larger ...


2

The problem when trying to replace coulomb damping (friction) with viscous damping is that it produces a relatively constant force ($- \mu \cdot N$). (the sign is opposite to the sign of the velocity). Viscous damping on the other hand is proportional to the velocity. $c\cdot \dot{x}$. Therefore, its not always possible to substitute the one for the other. ...


0

This is not an answer but to express how would I attack this problem. At any time $t$, there is a resistant force at the toe of the moving body, and the motion will continue if it is surpassed. Assume the sketch below is at $t = 0$, and $d = 0$, your job is then to find the energy accumulated between $t - t_f$ and $d - L$, in which $L$ is the length of the ...


Top 50 recent answers are included