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

[omarmorocci]

Homework Statement

I am currently doing an internship and as part of my assignemnt I have to calculate the dynamic load on the bearing of a certain part that is rotating about a fixed point.
https://www.physicsforums.com/attachments/chute-png.103400/
This is the part that moves ( the brown part rotates about the back shaft)https://www.physicsforums.com/attachments/paint-chute-png.103401/ The Ww is 8 kg,
The Ws = 6 kg
the weight of the blue boxes (Wp) on it is total 50 kg and it is centerized
Tetha is 20-60 degrees

I would like to what the dynamic force load is

w = 0.5 rad/s

Homework Equations

Sum of moments = Moment of inertia* angular acceleration
F= M*a

The Attempt at a Solution

I assumed all the masses are combined to become 64 Kg acting at the center of the wood
but I got the force to be tiny.
I am lost on how to start [/B]

 

[Divya Shyam Singh]

Please explain :
The system is rotating about which point, namely A or B?

 

[Divya Shyam Singh]

Assuming the system is rotating about point A(as far as i understood) you can approach the problem in following way:
Calculate the spatial position of the centre of mass.
There will be several forces acting on this point.
The forces acting here will be: 1. Gravitational force due to the combined weight of the system acting downwards
2. Centrifugal force due to rotation acting along the line joining the point A and centre of mass of the system which will be mrw2
Resolve the forces along the direction of Ax and Ay.
There you have it.
Hope it helped

 

[omarmorocci]

Assuming the system is rotating about point A(as far as i understood) you can approach the problem in following way:
Calculate the spatial position of the centre of mass.
There will be several forces acting on this point.
The forces acting here will be: 1. Gravitational force due to the combined weight of the system acting downwards
2. Centrifugal force due to rotation acting along the line joining the point A and centre of mass of the system which will be mrw2
Resolve the forces along the direction of Ax and Ay.
There you have it.
Hope it helped

Hello Divya,
Thank you for your reply,

Yes your assumption is correct,the rotation is about point A but the force in order to rotate is provided by shaft B

In order to simplify the question , I have made a new drawing.

About your point 1 : Don't you account for the reaction forces?
About your point 2 it's not clear which r I should take and do I then add the gravitational to the centrifugal force ?

How about the reaction forces ?

 

[Divya Shyam Singh]

I am writing the eqbm equations:
Ay= Fb(Force at B) - Wcosθ
Ax= Mr(w2)2-Wsinθ

The radius would be the distance between the point A and centre of mass of the system.

You will have to calculate w by the equation
w2=w1+αt......(1)
w1 = intial angular velocity

θ= (w1)t + 0.5αt2.....(2)

Replace these two equations in the above equations of Ax and Ay
The net resultant force at the point A will be
A= √(Ay2 + Ax2)
Note you will have to find the maxima of the above equation by differentiating it with respect to t.
You will have to put that value of the t in eq (2) for solve for θ.
If the value of θ lies between 20 to 60 degrees then that's your answer. If it does not lie in that range then just put the value of θ=20 and θ=60 and solve for A(net force). Compare the two value that you obtain and the greater one will be your answer.

Please note that the angular velocity of the system will change at every point because of the acceleration provided by the force at B.

 

[omarmorocci]

I am writing the eqbm equations:
Ay= Fb(Force at B) - Wcosθ
Ax= Mr(w2)2-Wsinθ

The radius would be the distance between the point A and centre of mass of the system.

You will have to calculate w by the equation
w2=w1+αt......(1)
w1 = intial angular velocity

θ= (w1)t + 0.5αt2.....(2)

Replace these two equations in the above equations of Ax and Ay
The net resultant force at the point A will be
A= √(Ay2 + Ax2)
Note you will have to find the maxima of the above equation by differentiating it with respect to t.
You will have to put that value of the t in eq (2) for solve for θ.
If the value of θ lies between 20 to 60 degrees then that's your answer. If it does not lie in that range then just put the value of θ=20 and θ=60 and solve for A(net force). Compare the two value that you obtain and the greater one will be your answer.

Please note that the angular velocity of the system will change at every point because of the acceleration provided by the force at B.

Hello ,
Thank you very much for your effort.
I assume that everything is a contact expect tetha right ?
Here is where I get stuck in integration

 

[omarmorocci]

Hello ,
Thank you very much for your effort.
I assume that everything is a constant expect tetha right ?
Here is where I get stuck in integration

 

[haruspex]

Ay= Fb(Force at B) - Wcosθ
Ax= Mr(w2)2-Wsinθ

Compared with the diagram, I think you have the signs wrong on both Ax and Ay.
For the mrω² term, what is your r? Remember it is a lamina, not a point mass.

θ= (w1)t + 0.5αt²

That is only valid for constant angular acceleration. If the force at B is constant then the acceleration will not be.

Here is where I get stuck in integration

In your first equation you have made the same mistake as DSS with regard to moment of inertia. It is not a point mass at distance 220mm (r₂ ?) from A.
I don't understand how you got the second equation from the first. If I differentiate the second equation wrt t I do not get back to the first equation.

 

[Divya Shyam Singh]

Compared with the diagram, I think you have the signs wrong on both Ax and Ay.
For the mrω² term, what is your r? Remember it is a lamina, not a point mass.

That is only valid for constant angular acceleration. If the force at B is constant then the acceleration will not be.

In your first equation you have made the same mistake as DSS with regard to moment of inertia. It is not a point mass at distance 220mm (r₂ ?) from A.
I don't understand how you got the second equation from the first. If I differentiate the second equation wrt t I do not get back to the first equation.

Yes my last solution is wrong. The correct way would be solving the following eq:
d2θ/dt2= (Fbr1 - Wcosθ.r2)/I
where r1 and r2 are the distances of the force Fb and weight from the point A respectively.

After double integration with respect to time we will have two unknown variables whose value we can calculate by:
at t=0 , θ = 20
at θ=20, w=0.5rad/s

 

[haruspex]

After double integration with respect to time

How are you doing that? There is a theta on the right hand side.

 

[Divya Shyam Singh]

How are you doing that? There is a theta on the right hand side.

How are you doing that? There is a theta on the right hand side.

There's a way to integrate it to some extent which is just enough for us to get us the answer:
0.5(dθ/dt)2 = aθ - bsinθ + c
is the integrated equation
where
a= By.r1/I r1 is the distance between point A and B
b= Wr2/I r2 is the distance between point A and Centre of mass

Now in the above equation dθ/dt is w.
putting the boundary conditions as...
w=0.5 when θ=20 degree will help us calculate the value of c
Once we have the value of c , we can easily calculate the angular velocity at any angle between 20 to 60 degrees from the same equation
Now we write the expression of A= √(Ax2 + Ay2)
In this equation put the following values of θ= 20, 30, 40, 50, 60 and calculate the value of A
I am very sure that the maximum value of A will be at θ=60° since A is most likely to be an increasing function in the domain θ ∈ [20,60]
Please do calculate and verify.

 

[haruspex]

There's a way to integrate it to some extent which is just enough for us to get us the answer:
0.5(dθ/dt)2 = aθ - bsinθ + c

Ok. You previously said to integrate twice with respect to time.

 

[Delta2]

Ok. You previously said to integrate twice with respect to time.

yes instead he integrates with respect to $\theta$ and uses $\frac{d^2\theta}{dt^2}=\frac{d\omega}{dt}=\frac{d\omega}{d\theta}\frac{d\theta}{dt}=\frac{d\omega}{d\theta}\omega$.

But let me ask another thing, in what frame of reference we do the calculations? Isnt the centrifugal force appearing in a rotating frame of reference?

 

[Divya Shyam Singh]

yes instead he integrates with respect to $\theta$ and uses $\frac{d^2\theta}{dt^2}=\frac{d\omega}{dt}=\frac{d\omega}{d\theta}\frac{d\theta}{dt}=\frac{d\omega}{d\theta}\omega$.

But let me ask another thing, in what frame of reference we do the calculations? Isnt the centrifugal force appearing in a rotating frame of reference?

There is a rotating mass whose point of rotation is about point A. Now in order to write the eqbm equation i applied the corresponding inertial centrifugal force mrw2. So the frame of reference i am using is ground(stationary).
Let me ask one more thing: while application of a corresponding inertial force whose direction is faced radially outwards wouldn't there be a force which is perpendicular to this inertial force?

 

[Delta2]

Let me ask one more thing: while application of a corresponding inertial force whose direction is faced radially outwards wouldn't there be a force which is perpendicular to this inertial force?

What do you mean, if we have tangential acceleration along with centripetal acceleration? In this problem yes we have, but it isn't necessary to happen always.

 

[Nidum]

Some of the help that you have received so far is infinitely unhelpful .

You do not have enough information to work out the forces acting on the platen .

To get any further you need information about how the angular position of the platen varies with time .
or
How the magnitude and direction of the driving force varies with angular position of the platen .

Get the information from the project specification or experiment or use experience and common sense to derive a plausible profile .

If you cannot obtain this information then you will have to use a simple static analysis and apply an appropriate g or usage factor multiplier .

 

[omarmorocci]

Some of the help that you have received so far is infinitely unhelpful .

You do not have enough information to work out the forces acting on the platen .

To get any further you need information about how the angular position of the platen varies with time .
or
How the magnitude and direction of the driving force varies with angular position of the platen .

Get the information from the project specification or experiment or use experience and common sense to derive a plausible profile .

If you cannot obtain this information then you will have to use a simple static analysis and apply an appropriate g or usage factor multiplier .

First I would to stay that all the help received from all the members is greatly appreciated .

How would you suggest I find that information by experiment ? ( experiment set up )
The only information I found from testing is that it takes 1.6 seconds to travel 40 degrees,
I am not sure on how to get an equation relating the angle and time due to the short time (1.6s)

Finally I am just an intern and don't have much experience yet so please be patient

 

[Nidum]

Finally I am just an intern and don't have much experience yet so please be patient

No difficulty with that at all .

Let's see if we can solve your problem using sound engineering principles .

We are already much better off than we were before since you have now given us the information that the complete movement takes 1.6 seconds .

What do you think is happening during that 1.6 seconds ?

 

[Nidum]

Think about acceleration , intermediate motion and deceleration .

We need to get at the maximum input force available as well .

How is this device operated - by hand or by a powered mechanism ? Can you estimate the maximum force that can be applied ?

 

[omarmorocci]

No difficulty with that at all .

Let's see if we can solve your problem using sound engineering principles .

We are already much better off than we were before since you have now given us the information that the complete movement takes 1.6 seconds .

What do you think is happening during that 1.6 seconds ?

As you can see here , the arm is operated by a pneumatic cylinder.
The output of the cylinder is 754 N

The acceleration seems to be constant as the angle of cylinder change as the arm rotates about point A

 

[Nidum]

Excellent . We now have much better information available about the whole set up .

 

[omarmorocci]

Excellent . We now have much better information available about the whole set up .

Here is some more information ,
At rest position shown, the cylinder is at and angle of 7 degrees with the horizontal
and the plate is at an angle of 20 Degrees with the horizontal

What are the next steps now ?

 

[Nidum]

You need to work out what effective force that cylinder applies to the platen lifting shaft .

(B_y in diagram in your first posting)

 

[omarmorocci]

I did Sum of moments about A ( turning point) = 0

754 N*680 = By *110

By = 4661 N

?

 

[Nidum]

Taking moments is correct method but one of the effective lever arm lengths that you have used may be wrong .

Show me a simple line diagram of the forces and lengths that you have used .

 

[omarmorocci]

its the black distance , so it's okay it's right

 

[omarmorocci]

Sorry , you are actually right ,
The distance is 529

so By = 3564 N

 

[Nidum]

I haven't checked the actual numbers but that seems like a realistic figure .

You should ideally repeat the calculation with the mechanism in different positions to see how much the force varies . You can do that later when you do the tidied up calculations for your project submission .

Anyway - we now have a known value for B_y . That's the hard work done .

From that you can work out the motion profile for the platen using standard angular motion calculations .

Need any more ?

 

[omarmorocci]

Alright , I will re-check the calculation to make sure it's as close to the actual value as possible.

So assuming that by is 3564 N

I can use Sum of moment = Inertia * angular accelration right ?

By- Ay = I * a

When I do this , i get approx 4 rad/s2 which more than my expected angular acceleration ( 0.8 rad/s2) but I assume that's fine ?

 

[haruspex]

View attachment 103485 Alright , I will re-check the calculation to make sure it's as close to the actual value as possible.

So assuming that by is 3564 N

I can use Sum of moment = Inertia * angular accelration right ?

By- Ay = I * a

When I do this , i get approx 4 rad/s2 which more than my expected angular acceleration ( 0.8 rad/s2) but I assume that's fine ?

You have not shown your working. Have you accounted for the To fact that the piston force has to overcome gravity as well as produce an acceleration?
Also, as the angle increases, the angle of the piston must change. This may reduce its moment.

 

[Divya Shyam Singh]

You have not shown your working. Have you accounted for the To fact that the piston force has to overcome gravity as well as produce an acceleration?
Also, as the angle increases, the angle of the piston must change. This may reduce its moment.

Yes, you are right but in the light of the complexity of the problem wouldn't it be a safe assumption to ignore the mass of the piston with respect to the system?
Similarly, the change of the angle of the force provided by the piston will change but it will be better to assume to be same.

 

[Divya Shyam Singh]

Also, what is the approximate mass of the actuator of that particular model?

 

[Divya Shyam Singh]

I haven't checked the actual numbers but that seems like a realistic figure .

You should ideally repeat the calculation with the mechanism in different positions to see how much the force varies . You can do that later when you do the tidied up calculations for your project submission .

Anyway - we now have a known value for B_y . That's the hard work done .

From that you can work out the motion profile for the platen using standard angular motion calculations .

Need any more ?

What standard angular motion calculations are we exactly talking about here?
Please elaborate

 

[omarmorocci]

Hello guys,

So I took the part out with the forces in order to find the force B
Using moments about A I get
754 * 0.520 - By *0.12
By= 3267 NThen I used the diagram bellow

And I used the formula
Sum of moment = mr2 * angular accelration
Hence I get

3267*0.086 - 640cos 20 * 0.44 = mr^2*a
754 * 0.520 - 3267 *0.086 = 64 *0.44^2 * a

a= 1.27 rad/ sec and my tested value is approximately 0.7 rad/s2 so I think that this is a good estimate ?Since the angle of the pneumatic cylinder changes as the arm goes up then I assume the force supplied by cylinder is constant
But since the angle of the plate changes , the moment due to the weight will degrees hence results in a higher angular acceleration across time

 

[Divya Shyam Singh]

Yes, i think that's correct.
At the same time, the value of angular acceleration will change with θ(increase, since the cos component of the weight will decrease)

 

[haruspex]

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:

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.

 

[omarmorocci]

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'

 

[haruspex]

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'

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?

 

[omarmorocci]

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?

 

[haruspex]

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.

 

[omarmorocci]

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

 

[haruspex]

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.

 

[Delta2]

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..

 

[Divya Shyam Singh]

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.

 

[omarmorocci]

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 )

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

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

 

[Delta2]

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?

 

[omarmorocci]

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?

Ws = weight of sheet metal plate = 6 kg
Ww= weight of wood = 8 kg
Wp= weight on the wood is 50 kg

 

[Delta2]

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.

 

[omarmorocci]

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)

 

[Delta2]

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.