Mechanics naman ta bi. Let us discuss first the condition called equilibrium. Equilibrium is a state when three truths are established.

1. Summation of forces in the vertical axis is zero
2. Summation of forces in the horizontal axis is zero
3. Summation of moments/torque applied to the body is zero

If these conditions are established, the body is said to be in equilibrium. If the body is now in equilibrium, there are only two possibilities that it can be.

1. It is at rest, not moving at all
2. Or it can be moving at a constant speed along a straight path


So, logically we say:

If an object is at rest or moving in constant velocity, then it is in equilibrium. If it is in equilibrium, then summation of forces and torques result to zero.

balik ta sa energy conservation formula....






energy in = energy out




:thumb:

what's about energy loss?

friction and other losses are included in the equation..... depends on the energy balance equation but ussually its included in the "energy out" portion

Back to the lesson proper...


The summation of forces is called "resultant force". Therefore:

- if the body is in equilibrium, the resultant force must be zero

- if the resultant force is not zero, the body will start to move in the direction of the force. If this force is continuously applied what will happen to the body under consideration?

guess.....

if the body is not in equilibrium, then it will move... then we will jump into dynamics.... :bounce

Tama! There will be movement and one of our best mentors discovered the law governing this. His name was Sir Isaac Newton. He was able to establish the relationship between forces and motion and the result of his investigations:

F = ma

and we all know that

F = Force
m = mass
a = acceleration in the direction of the force


From here on, he derived the equations of energy due to movement (which we now know as kinetic energy) and impulse as well as momentum.


Will anybody present the derivation of kinetic energy (KE) from the basic force equation above?

ari ... sadya ni nga thread.... brain teaser... try ko ah...


using the Work theorem..... W = delta KE = F * R

KE = kinetic energy
F = Force = mass (acceleration)
r = distance

delta KE = delta W = m*delta a*delta r = m * dR *dV/dt ; a = dV/dT

dR = differential of distance
dV - differential of velocity
dT = differential of time

rearranging since dR/dT = V

dKE = m* VdV

integrating....

dKE = m* v^2/s with limits 0 to final V; and since @ initial V=0

KE = m*V^2/2


insakto man? hehehehehe.....

Perfect! Amo na ya ang TUPVian!

So as done by Toto Rostan, Sir Isaac needed to develop calculus in order to derive the resulting equations. That was 300 years ago!

Tuloy mo na ang impulse and momentum equation To Rostan. :)

tsk tsk.. mariit ba.. :worthy: :worthy: :worthy: :worthy:


di na ko ya kadumdum sang calculus ko ya!

karun naman kay ari na ko sa meeting hehehehehe :thumb:


pero it starts with:

momentum = force * velocity

impluse = change in momentum or force applied over a period of time

pirti kasagad ah!!! matuon lang kami diri ah!!hehehe!!

Huo, matuon ta tanan. Very good ning ginbuhat sang aton mga mods nga may section nga amo sini.

agree para makalakakas ang tulktuk sa mga bagul ta hehehehe


plug ko lang gali....

if anyone in need of a facility design (CSA, MEP, as such) and project management services, please feel free to PM me..... gaan ko kamu discount as a fellow TUPVian...

nagasaka na ang mga presyo so dapat i exploit ta ang natun-an.... hehehehehe

:thumb:

Momentum ---- is scalar.... means, with directions
----- applicable at constant acceleration
----- considers what will happen AT AN INSTANT

on some collosion incidents, this can be used to approximate the speed by trigonometry...

using pythagorean theorem (if collides on right angle) or trigo functions, and given the mass (or weight) of each object....

:smoke

Ok. So impulse and momentum...

F = ma

F = mv/t

Ft = mv

and Ft is the impulse, mv is momentum

And since the relationship between forces and motion has been established by Sir Newton, we now have awareness about moving things around us. We now know that the momentum of a truck is bigger than that of a car, even though they are both moving at 5km/hr. In the industry where we now work, can we see applications of these concepts? Can we give examples of how we can take advantage of this knowledge, apply it in our jobs and get a promotion in the end? :hmmm:

i think got one previously...

not literally MV or FT but more on like this...

handling small but fast projects is just the same as managing large projects but takes longer time to complete.... say:

PhP 3M project in 2 weeks and PhP450M project in 9 months....

projects can all be completed, but its the class on how and when it was completed differs...

and talking of MV and FT....

there were a lot of applications on my scope (at least for me).. like in piping systems, ducting systems, and as such... so i am very thankful to have known and learned the derivations and not memorizing formulas.....

(does it make sense? wahehehehehe :brr )

practice problems?.... :shot: :cheers: :smoke

Practice problem kuno:

I am a process engineer in the diebond area of semiconductor assembly. I am currently encountering die cracks found after the die attach process itself. What am I supposed to look at and correct? :( :(

bond force?

ejector pin?

contamination?

:naughty:

Why would the die crack? Very seldom/almost nil nagka-crack sa bond site. During the bond process, the die will be under compression loading. As the die is a very brittle material, it will not practically be prone to compression stress failure. It will be very susceptible however to tensile stress failure. So sa diin may tensile loading during die bonding as a whole? It should be in the pick up process. If we do a free body diagram on the die during the pick process, there will be a uniformly distributed load by the tape adhesion, there will be a single acting force due to the ejector, and the downward side forces on the edge of the die due to the pick up collet. Easily, we would see that the die is under bending or flexural stress. So now that we know the forces, what can we do to reduce these forces and thus reduce the risk of die crack? (Actual stress < Su of the silicon die)

what's the die size?

i am not a process engr nor an eqpt engr... i just happeed to like some lady process and eqpt engrs then.... hehehehehe pero back to the question, since the load is concentrated, the very low hanging fruit to slve the problem is to distribute the concentration of the load.... :smoke


ambot kung insakto ah :cheers:

@ iceman: what's the die size?

No particular number... :)

Correct ka gid toto Rostan!

To distribute the load, we need to use multiple needle ejection system if the die size allows us. Mechanics in action!

Dugangan ta pa bi!

:thumb: :thumb: :thumb:


in facility utility standpoint, multiple injection needles means multiple posible point of leakage on the compressed dry air system wthin the machine.... :D :shot: :cheers:

daw makayuhum man ko ba...sa R&D ko sang development sang Die Bonding machines pero wala ko kaityindi sang discussion parti sa bond force..hik hik hik..

daw iuli ko man sweldo ko ba..he he he

ti kay sa automation ka man ya cigurado na assign.. hehehehehehe..... :thumb: :thumb: :thumb:

Gin labugay bi ni toto Rostan mo. Injection needles ya ang iya gin sugpon. Ejection needles ang gin-istoryahan pro. Ti kundi nadula and hwisyo ni bolitski! hehehe :lol: :lol: :lol:

wala ko lang mabsahan sang matul-id.... pasensya ... :lol: :lol: :lol:


next topic?.... :smoke :cheers:

Die crack correction due to kinematics alone:

1. pick force must be reduced
2. ejection height must be reduced
3. multiple ejector needles to distribute the load
4. ejector needle acceleration reduced (remember, F=ma)
5. ejector needle tip diameter increased
6. more factors pa sa ejector cap, collet, collet body bearing load, vacuum, etc


;)

indi ko na ya kabalo... ti pareho dyapun ang cfm kag pressure needed sang CDA? :D

kumlpleto gid ba... kanami na sang science involve dira sa ejection and pick-up pa lang... hehehehe... and you must consider man die size and die thickness...

Another application:

After a certain process, processed leadframes are being ejected out into magazines. The final pusher is driven by a servo motor allowing programmable motion. The engineers notice that the leadframes always go over the other side of the magazine (meaning outside) even though the pusher distance is just right. Using motion laws, how will the engineers correct this issue?

what do you mean of the pusher distance, is it the follow through?

Reduce the initial velocity, make the pusher movement as smooth as possible, no jerking...


:naughty:

Yup. but that is on the material side, not on the kinematic side. We can include the die material for ultimate stress values, the tape material for adhesion properties, etc... :thumb:

yup corect, but it should be your basis in computing your ejector pin size, on how many ejector pins you will use in case the die size is too big for a single ejector pin, the ejection height, ejection motion profile, speed, ejection to pick-up...etc...

Die dimensions is one of the basis of calculations. It "should" not be the only one. Again, one of the major considerations is die cracking. Hypothetically, if the die is big but does not crack at all, I can just use a single pin ejector even if I can fit two or four pins. Examples of this kind are "spacer" dies having no circuitry at all. As for speed, our basis for calculation is die crack and die placement (no considerations on epoxy yet). If there is no crack and placement issues, I would just put my machine to maximum speeds and get the highest output possible. We will not even need to bother with die size and thickness. If your current calculations are based on die sizes, probability is that you are following a certain rule of the thumb in your process. There is a chance that you can further optimize the process considering the real mechanics/kinematics while still meeting the required quality criteria. Baka ma-promote ka pagkatapos. :)

i dont know pero maybe pusher velocity (or basically power) is quite high.... dont know how to reduce it leterally... another thing is the "friction" AFTER the pusher stroke....

sensya, kay facilities design ang specialty ko wahehehehehe :lol:

hmmmn... i am imagining a really large and thin die with an injector pin and a really nasty back tape pulling on the edges of the die against the pushing ejector pin on the die center... :D

Hehe, you are imagining a crack! Siling ko gid daw! :D