System Dynamics and Control: Module 4b - Modeling Mechanical Systems Examples

Three examples of modeling mechanical systems are presented employing a Newton's second law type approach (sum of forces, sum of moments). Both translational and rotational examples are presented for mass-spring-damper type systems, in particular.

[@leemontgomery7914]

Well explained. You’re the type of professor that can produce top-notch engineers.

[@Ovar9Thousand]

Thank you so much for this series. Incredibly well detailed and explained, the modelling of systems was a part of my course that was rushed through quickly so this is a huge help. Keep up the great work.

[@jeremywall575]

Thank's for going over everything in gruesome detail, I was super confuzed by this stuff (especially the directions and positives and negatives).

[@zainmughal3841]

Positive and Negative signs used here with respect to the movement of mass and arrow given in data. If the arrow head of mass is in the direction of force than it will be positive otherwise it will be negative.

[@HellOnGames]

Thanks for the video, I was very confused with the direction of the forces but got it thanks to you.

[@prasadj2696]

If u really want to get something out of this, then u shouldn't have half knowledge about this, Great expalanation

[@AJ-et3vf]

Awesome lecture!!!

[@qtip3998]

Good video. Simple. But good.

[@naumanfayyaz9413]

Great explanation!!

[@altuber99_athlete]

5:12 Something doesn't look clear to me: A system is said to be in *static equilibrium* if it is either at rest or moves a at constant velocity (in both cases the acceleration is zero, which is the criteria for a system to be in static equilibrium, which by Newton's second law and its analogous law it occurs if the sum of the forces and the sum of the torques are both zero.) Also, gravity is always actuating over the masses m_1 and m_2, and so they have weight. In this example, I assume that when you said _static equlibrium_ you referred that the system is at _rest._ Thus, in static equilibrium of the system, each weight pulls down (gravity) which compresses the two springs. In other words, *when the **_system_** is in static equilibrium, the two springs are compressed, i.e. the springs **_aren't_** in their equilibrium position,* which means they're exerting force over the masses because of Hooke's law, so as to try to put themselves in their equilibrium position. (Notice the difference between "static equilibrium of the system" and "equilibrium position of each spring".) Now, you've considered that x=0 and y=0 when the _whole system is_ in static equilibrium and so the _springs aren't_ in their equilibrium positions. Thus, when x=0 and y=0, the springs _must_ be exerting forces because they're deformed. But according to the equation you put in min 5:12, which is F_s = k_2 × (y-x), when x=0 and y=0 (system is in static equilibrium and springs aren't in their equilibrium positions), the spring k_2 isn't exerting a force since F_s2 = k_2 × (0-0) = k_2 × 0 = 0. But that is false, because the spring _must_ be exerting force on mass m_2 since the spring is compressed in the static equilibrium of the system. What did I get wrong? I hope you can explain this. Thanks.

[@baconcheese4381]

thank you , thank you , thank you for this lecture!

[@Austin-nt5pu]

Amazing explanation! Thank you!!

[@wontmk]

I really enjoyed this course at the U of M. Thanks for the vids dude! Pleases keep it up

[@surla420]

Good video..Thank you sir

[@6glob]

U state that F_k will be positive (which it will, since y-x is positive), but in the diagram you depict it against the positive direction you defined (it's going downwards, while positive is upwards). This makes me think that F_k should be -k(y-x). Am I missing something?

[@simplyshernade4165]

What is a mathematical model that can be use to predict the distance that a car will go when the engine is shut off after 5m?

[@Aguvika]

So force due to gravity can be neglected because of the opposing normal force from the ground? Yes??

[@hillrickc]

The force due to gravity (the weight) can be neglected because of the opposing force generated due to the static deflection of the springs. You can imagine that if the system is in static equilibrium the springs deflect under the weight of the masses such that all forces balance (weight forces = spring forces). Therefore, if you choose the reference for the deflection of the springs to be from this point of static equilibrium, then the weight forces can be thought of as being cancelled by the static spring forces.

[@slenderman432]

@@hillrickc Thank you!

[@anmolzaidi8041]

then how we will make matrix from two equations when we have 3 variables x,y and u??

[@vadimonezou7920]

I do have issues to know at which moment this (y-x) or (x-y) Any tip to help deal with that

[@hillrickc]

You can express a spring force as k2(y-x) or k2(x-y), the only difference is the sign. The key is that when you draw your free-body diagram, you want the direction of the force to match when your expression is positive. When y > x, the spring is stretched and the force matches the free-body diagram. When the mass oscillates and y < x, then the force is negative and has a direction opposite that shown in the free-body diagram.

[@lumina_india]

How is this different from module 2?

[@davidsonfellipe5286]

very good your video and its explanation, although it is in English and I do not understand rsrsrs I am student of engineering of automation and I am having a lot of difficulties in modeling of dynamic systems quarter car system, you could make an example of this with values ​​of the masses and constants of k, and c. Thank you very much in advance

[@TjipzPK]

At around 5:35 you say that the way that you have modelled it, the force from the spring is positive, but you still drew it in the negative direction, how come?

[@hillrickc]

I know that is a little confusing. What I said is that the force is positive down and by that I mean that when the expression k2(y-x) is positive (when y is larger than x), then the force will be downward. When the expression is negative (when x is larger than y), then the force will be in the opposite direction. I hope that helps.

[@TjipzPK]

Okay, thanks. So what it comes down to is, if I apply a force on mass m1 then the spring force on mass m1 has to negative and positive on mass m2, if the positive direction is in the direction of the force applied? So x1-> is positive direction then (m1)f-> then (-k1)x1, and k2(x2), if m2 is hold still and m1 is moved that is. Is this correct? Thanks again Rick, wonderful videos.

[@anasadaileh3670]

you are wonderful

[@malshwwaf]

thank you for this wonderful lecture what is the book that i can use??

[@hillrickc]

Mohammed AL-Shawaf You can try System Dynamics by Ogata.

[@jandlouhy6914]

Bit confusing due to coordinate system overlap .

[@austinwalker5176]

According to my professor, the brake pedal example is wrong.

[@auds_e7]

What about mg?

[@hillrickc]

In the suspension problem, we have defined the coordinates so that they equal zero from the point the system is in static equilibrium. This means that the deflections of the springs in this state generate forces that exactly offset the weights of the masses. Since the weights are canceled, we leave them out of our equations.

[@egw-su6km]

you have the most generic american name i have ever heard in my 23 years of living in this planet in this universe

[@syedafatimakazmi6431]

Plz don't put big subtitles they are annoying

[@hillrickc]

Those are generated by KZbin. You toggle them on/off with the closed captioning (cc) option in the lower righthand corner of the screen.