I would like to know what the dynamic force load is....

In summary: I don't understand how to find the value of tetha from the given info.In summary, the blue boxes weigh 50 kg, the system is rotating about point A, and the resultant force at point A is given by the equation: A= √(Ay2 + Ax2).
  • #36
Divya Shyam Singh said:
wouldn't it be a safe assumption to ignore the mass of the piston
I was referring to the lack of mention of the weight of the load in this equation:
omarmorocci said:
I can use Sum of moment = Inertia * angular accelration right ?

By- Ay = I * a
But maybe omarmorocci was taking moments about the mass centre.
Also, it seems to me that Ay and By are forces, not moments.
And I cannot find what omar is using for I.
 
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  • #37
haruspex said:
I was referring to the lack of mention of the weight of the load in this equation:

But maybe omarmorocci was taking moments about the mass centre.
Also, it seems to me that Ay and By are forces, not moments.
And I cannot find what omar is using for I.

I believe that you are right.
Btw please just look at my latest answer and ignore my previous workings.
For mass moment of inertia I took the moments about point A and used I = mr^2 ( hence I believe I might have made a mistake there)
But i'
 
  • #38
omarmorocci said:
I believe that you are right.
Btw please just look at my latest answer and ignore my previous workings.
For mass moment of inertia I took the moments about point A and used I = mr^2 ( hence I believe I might have made a mistake there)
But i'
omarmorocci said:
3267*0.086 - 640cos 20 * 0.44 = mr^2*a
754 * 0.520 - 3267 *0.086 = 64 *0.44^2 * a
As I have pointed out a couple of times, you are using the wrong expression for moment of inertia about A.
Do you know the formula for moment of inertia of a uniform rod (or rectangular plate) about one end? Or, if you don't know that, the formula for the moment of inertia about the centre of the rod, and the parallel axis theorem?
 
  • #39
haruspex said:
As I have pointed out a couple of times, you are using the wrong expression for moment of inertia about A.
Do you know the formula for moment of inertia of a uniform rod (or rectangular plate) about one end? Or, if you don't know that, the formula for the moment of inertia about the centre of the rod, and the parallel axis theorem?

Using the parallel axis therom
To find the mass moment of inertia about point A , I assume the total weight acts as a point mass at 440mm from A ( for simplicity reasons)

Hence I = mass (64 kg) * distance from mass to axis which I am taking ( 0.44^2)

Is there something that I am not seeing?
 
  • #40
omarmorocci said:
assume the total weight acts as a point mass at 440mm from A
But that is not correct. That gives you mL2/4 for a rod of length L about one end. It should be mL2/3. This will get you closer to your measured result.
 
  • #41
haruspex said:
But that is not correct. That gives you mL2/4 for a rod of length L about one end. It should be mL2/3. This will get you closer to your measured result.

But is actually a rectangle , is possible to just take consider it a rod ?

And yes that does give me a value closer to the actual one
 
  • #42
omarmorocci said:
But is actually a rectangle , is possible to just take consider it a rod
Yes, it's the same formula. Think of it as a lot of rods side by side.
 
  • #43
The mass of 64kg is that of the rotating rod? or we have a rotating rod (or plating) that supports a mass of 64kg at distance 440mm from the point of rotation? Cause some of the figures at the first posts indicate the latter...

Anyway, if the mass of 64kg has dimensions comparable to the moment arm of 440mm then you just can not approximate sufficiently enough its moment of inertia by just 64*(0.440)^2, that's a very raw approximation..
 
  • #44
haruspex said:
But that is not correct. That gives you mL2/4 for a rod of length L about one end. It should be mL2/3. This will get you closer to your measured result.
Actually, he is assuming the whole system as a point mass denoted by a square mass of 64kg. Hence its moment of inertia will be Mr2.
Although you are right in saying that the wooden plate has to be treated correctly by using correct formulae and parallel axis theorem for correct moment of inertia for more precise value. However he is assuming the system to be a point mass.
 
  • #45
haruspex said:
Yes, it's the same formula. Think of it as a lot of rods side by side.

Alright then, I am surprised I got such a close value even though I estimated and assummed many things for simplicity ( for example that they are all a comined mass )

Delta² said:
The mass of 64kg is that of the rotating rod? or we have a rotating rod (or plating) that supports a mass of 64kg at distance 440mm from the point of rotation? Cause some of the figures at the first posts indicate the latter...

Anyway, if the mass of 64kg has dimensions comparable to the moment arm of 440mm then you just can not approximate sufficiently enough its moment of inertia by just 64*(0.440)^2, that's a very raw approximation..

The 64 Kgs is the weight of the plate + the sheet metal + the weight on the rotating plate

Divya Shyam Singh said:
Actually, he is assuming the whole system as a point mass denoted by a square mass of 64kg. Hence its moment of inertia will be Mr2.
Although you are right in saying that the wooden plate has to be treated correctly by using correct formulae and parallel axis theorem for correct moment of inertia for more precise value. However he is assuming the system to be a point mass.
Yes, you are right , that was the assumption that I have made, but treating it like a rod does infact give a closer number to the actual value
 
  • #46
Can you tell us each weight separately? What is the weight of the plate, what is the weight of the sheet metal and what is the weight of the weight on the rotating plate?
 
  • #47
Delta² said:
Can you tell us each weight separately? What is the weight of the plate, what is the weight of the sheet metal and what is the weight of the weight on the rotating plate?
1.png

Ws = weight of sheet metal plate = 6 kg
Ww= weight of wood = 8 kg
Wp= weight on the wood is 50 kg
 
  • #48
So does the "main weight" Wp of 50Kg extends over the wood like the figure shows, or is it concentrated at a small region.

From the figure (I suppose Wp is the blue body) it looks like its moment of inertia is that of a rod with length 86+115+230+444
And ofcourse different moment of inertia for the wood (slightly bigger than the blue body) and another for the metal plate, cause each has different length as I see from this figure.

(Sorry if I made you repost the figure)..

The best approximation would be to take the moment of inertia as 64*(0.875^2)/3 I believe that would take you much closer to the result.
 
Last edited:
  • #49
Delta² said:
So does the "main weight" Wp of 50Kg extends over the wood like the figure shows, or is it concentrated at a small region.

From the figure (I suppose Wp is the blue body) it looks like its moment of inertia is that of a rod with length 86+115+230+444
And ofcourse different moment of inertia for the wood (slightly bigger than the blue body) and another for the metal plate, cause each has different length as I see from this figure.

(Sorry if I made you repost the figure)..

The weight is assumed to be equally distributed over the whole plate.
So you are suggesting to treat the 3 bodies as 3 different rods with different lengths using the formula mL2/3

(No need to appologize , thank you guys for everything so far)
 
  • #50
omarmorocci said:
The weight is assumed to be equally distributed over the whole plate.
So you are suggesting to treat the 3 bodies as 3 different rods with different lengths using the formula mL2/3

(No need to appologize , thank you guys for everything so far)

yes but I edited my post, I believe the best approximation would be to take the moment of inertia as 64*(0.875^2)/3 I believe that would take you much closer to the result.
 
  • #51
omarmorocci said:
The weight is assumed to be equally distributed over the whole plate.
So you are suggesting to treat the 3 bodies as 3 different rods with different lengths using the formula mL2/3
you could do that, but I don't think it will make enough difference to matter. There are probably other aproximations being made that are more significant, like the way the torque from the load decreases as the angle increases. (How did you measure the angular acceleration?)
 
  • #52
haruspex said:
you could do that, but I don't think it will make enough difference to matter. There are probably other aproximations being made that are more significant, like the way the torque from the load decreases as the angle increases. (How did you measure the angular acceleration?)
For that I will make an excel document and formula that connects the angle to the torque of the load hence giving me different acceleration.
But for now it's just the initial position,
 
  • #53
What I am actually also confused about is how to calculate the dynamic load on point A ( pivoting point )
now that I know the acceleration should I just use the following formula
By - Ay - Wt cos tetha = mass * acceleration ?

given eveything is know , I should be able to calculate Ay , which is the dynamic load right ?
 
  • #54
omarmorocci said:
What I am actually also confused about is how to calculate the dynamic load on point A ( pivoting point )
now that I know the acceleration should I just use the following formula
By - Ay - Wt cos tetha = mass * acceleration ?

given eveything is know , I should be able to calculate Ay , which is the dynamic load right ?
The problem with that formula is that not all the forces and accelerations are parallel. So it works as a vector equation (if you drop the cos theta) but not as a scalar one.
Choose two perpendicular directions to resolve the forces in (either vertical and horizontal, or parallel to the plate and perpendicular to the plate) and write one equation for each.
 
  • #55
haruspex said:
The problem with that formula is that not all the forces and accelerations are parallel. So it works as a vector equation (if you drop the cos theta) but not as a scalar one.
Choose two perpendicular directions to resolve the forces in (either vertical and horizontal, or parallel to the plate and perpendicular to the plate) and write one equation for each.
ssssss.png

I am talking about the tangential acceleration tangential acceleration = angular acceleration * radius
and I am resolving the forces in the direction of the plate and perpendicular to it ( The same as the Ax and Ay)
 
  • #56
omarmorocci said:
What I am actually also confused about is how to calculate the dynamic load on point A ( pivoting point )
now that I know the acceleration should I just use the following formula
By - Ay - Wt cos tetha = mass * acceleration ?

given eveything is know , I should be able to calculate Ay , which is the dynamic load right ?
What you calculated is the angular acceleration ##a##. The acceleration in your equation above should be ##a_y## which is equal ##a_y=0.440a##

But there is also the ##A_x## component of the load which relates to the centripetal force and centripetal acceleration ##a_x## which doesn't relate to the angular acceleration, but it relates to the angular velocity ##\omega## it is ##a_x=\omega^2(0.440)## and it should be
##A_x+W_tsin(\theta)=M_ta_x## , where ##M_t## total mass of 64kg.

The total load will be ##A=\sqrt{A_x^2+A_y^2}##
 
  • #57
omarmorocci said:
View attachment 103561
I am talking about the tangential acceleration tangential acceleration = angular acceleration * radius
and I am resolving the forces in the direction of the plate and perpendicular to it ( The same as the Ax and Ay)
Sorry, yes, I see what you are doing. As Delta2 says, you need to consider Ax as well. However, I don't think you need to worry about centripetal acceleration. It will be insignificant.
 
  • #58
Excellent work guys. I will tidy up all the calculations and make less assumptions since now I know how to work it out.

One last question I have is how to I get the compressive pressure ?
http://www.igus.com/wpck/3726/iglidur_clipslager?C=US
Im looking to use these bearings, and I need to know the compressive pressure.
the static load on each bearing is approx 500 N
and dynamic load is approx 1100 N
the thickness of the sheetmetal is 2 mm

Thanks in advance
 
  • #59
Come in contact with someone from the company and talk to them. They usually have a software where you 'd have to enter your value. Trust me, they are better at selecting the right bearings for you. I tried doing the same calculations, got some values, fixated on some model of bearings, talked to their technical guy and it turns out that whatever i had done was all in vain.
You could also ask them to suggest you the right bearning model for your use.
Please do tell them the correct conditions in which the bearing is to be used. (like environmental conditions, loads, Life of the component that you want)
One important aspect is serviceability which basically means how often can you get the bearing serviced.

Try to get as much information regarding bearing as possible from the person you talk to. He will be more than happy to help.

Hope this helps.
 
  • #60
Before doing anything else work out whether your calculated acceleration figure gives you total motion time for the platen consistent with your practical observations .
 
  • #61
Nidum said:
Before doing anything else work out whether your calculated acceleration figure gives you total motion time for the platen consistent with your practical observations .
Capture1.PNG

I used excel to vary the angle and hence get different angular accelerations,
In real life, the anglar speed seems to be contant but I am not really sure how to confirm this since the whole motion is only about 20 degrees and in 1.5 seconds
 
  • #62
.
 

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  • #63
First of all:
GOOD work with the calculations.
In real life situations, there are resistive frictional forces at every hinge point whose resistive moments have not been considered here.
Also, the actual force output of the actuator is less than the theoretical force given in the actuator manual etc.

One thing i wanted to bring into your concern is that since the angular acceleration is increasing constantly over the range of motion, there will be a "collision" at the end between Hinge at B and the upper most portion of the slot. Over the time, due to constant impact loads on hinge B might be deleterious.
 
  • #64
Divya Shyam Singh said:
First of all:
GOOD work with the calculations.
In real life situations, there are resistive frictional forces at every hinge point whose resistive moments have not been considered here.
Also, the actual force output of the actuator is less than the theoretical force given in the actuator manual etc.

One thing i wanted to bring into your concern is that since the angular acceleration is increasing constantly over the range of motion, there will be a "collision" at the end between Hinge at B and the upper most portion of the slot. Over the time, due to constant impact loads on hinge B might be deleterious.

Thank you for your words,

Fortunetly, the collission does not infact happen as the pneumatic cylinder reaches it's maximum length before the shaft hit the end of the slot, so there is no damage there.
 
  • #65
May i ask where this component is currently used? what is it related to?
 
  • #66
Divya Shyam Singh said:
May i ask where this component is currently used? what is it related to?
It is used in sorting packages for companies like DHL Fedex etc
 
  • #67
Divya Shyam Singh said:
Come in contact with someone from the company and talk to them. They usually have a software where you 'd have to enter your value. Trust me, they are better at selecting the right bearings for you. I tried doing the same calculations, got some values, fixated on some model of bearings, talked to their technical guy and it turns out that whatever i had done was all in vain.
You could also ask them to suggest you the right bearning model for your use.
Please do tell them the correct conditions in which the bearing is to be used. (like environmental conditions, loads, Life of the component that you want)
One important aspect is serviceability which basically means how often can you get the bearing serviced.

Try to get as much information regarding bearing as possible from the person you talk to. He will be more than happy to help.

Hope this helps.
I believe there was some simple calculation to find the pressure ,
It is force/area but I am not sure on what area to take ( is it diameter * thickness)?
 
  • #68
No, actually i was talking about a project i worked on a few months back. Its always better to talk to them ask what all parameters do they require for them to be able to decide for the best bearing.
But yes, generally its Force/Area for deciding the pressure required from the actuator
I don't know about the bearings though. In my opinion it actually depends upon the bearing. In general maximum contact pressure is derived by using hertz law but there are a lot of other factors in play so its quite complicated when designing bearings for practical purposes.
 
  • #69
http://www.astbearings.com/technical-information.html
 

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