Determine the position and magnitude of the max bending moment

[haruspex]

View attachment 233612
This is what I expected to find! Thank you so much for taking the time to help me again... The world needs more people like yourself!

Glad to have helped you understand how to apply the method.
However, from a purist perspective, finding the maximum moment by checking the moment at various points along the line is not satisfactory.
A neater method is to consider each segment for which the distributed load is constant (maybe zero) and there are no point loads. In the present question that would be just two segments. For each segment you can throw away the negative displacement terms from the torque equation (they will be the same terms for the whole segment) and differentiate to look for a local maximum torque within the segment. If there is no local maximum within the segment then the maximum for the segment occurs at one of its endpoints.
Having found the maximum for each segment you would then compare them to find the maximum for the whole beam.

 

[Al_Pa_Cone]

That sounds like an interesting method which I am unfamiliar with. I am interested to learn a bit more about calculus as I have not used it a great deal throughout my course. I am going to set myself a further challenge to complete which you may be able to assist me with using alternative methods. I am looking to design a automated system to open an close the lid of the greenhouse using a counterweight and a microservo to tip the balance given by the weight.

I have put together this diagram of how it would look and I have made a diagram for the lid and mechanism to show the calculations required. As this is a fictional project I am going to say the net weight of the lid is 1kg or 1000g to simplify the calculations. The micro servo attached to the pulley will transmit the torque in favour of the counterweight to open or to the lid to close. I have produced a basic diagram of how this would look.

I was going to follow my previous method showing the maximum bending moment required to tip the balance over to the counterweight. Again I can say the maximum angular torque available from the servo is 10g as I am not using genuine parts.
Is there anything you can offer me to get started with this problem? Thank you.

 

[Al_Pa_Cone]

Here is a stripped down version with the facts I have set out to find the maximum force required at the big red arrow. I will do some calculations to try and get some results myself.

 

[haruspex]

That sounds like an interesting method which I am unfamiliar with. I am interested to learn a bit more about calculus as I have not used it a great deal throughout my course. I am going to set myself a further challenge to complete which you may be able to assist me with using alternative methods. I am looking to design a automated system to open an close the lid of the greenhouse using a counterweight and a microservo to tip the balance given by the weight.
View attachment 233880
I have put together this diagram of how it would look and I have made a diagram for the lid and mechanism to show the calculations required. As this is a fictional project I am going to say the net weight of the lid is 1kg or 1000g to simplify the calculations. The micro servo attached to the pulley will transmit the torque in favour of the counterweight to open or to the lid to close. I have produced a basic diagram of how this would look.
View attachment 233885
I was going to follow my previous method showing the maximum bending moment required to tip the balance over to the counterweight. Again I can say the maximum angular torque available from the servo is 10g as I am not using genuine parts.
Is there anything you can offer me to get started with this problem? Thank you.

If you want a real workable design, this is not it. The problem is that the torque exerted by the counterweight is roughly constant, whereas the torque load from the lid will reduce as it rises.
To fix that, hang the counterweight from that stub of lid that sticks out to the left. Now it is just a seesaw.

 

[Al_Pa_Cone]

I have redrawn the diagram to show the idea. I have started with working out the net weight of the lid as a UDL and worked out 1000g into Newtons = 9.81N. as a UDL this would be 9.81/0.6m = 16.35N.

 

[haruspex]

View attachment 233888
View attachment 233889
I have redrawn the diagram to show the idea. I have started with working out the net weight of the lid as a UDL and worked out 1000g into Newtons = 9.81N. as a UDL this would be 9.81/0.6m = 16.35N.

For the purpose of calculating the counterweight's mass, no need to figure out the force in Newtons. Mass ratio is inverse of distance ratio.

 

[Al_Pa_Cone]

So the overall weight of the lid would be 1000g but 100mm of it is beyond the left of the tipping point. So would I be correct by starting to divide the weight by 600mm.
1000g/600mm*500mm = 833.3g to the right of the pivot point.166.7g to the left and this ratio would be 1:5. so the weight required of the counterweight would need to be x4 of the left stub.

 

[haruspex]

So the overall weight of the lid would be 1000g but 100mm of it is beyond the left of the tipping point. So would I be correct by starting to divide the weight by 600mm.
1000g/600mm*500mm = 833.3g to the right of the pivot point.166.7g to the left and this ratio would be 1:5. so the weight required of the counterweight would need to be x4 of the left stub.

It is simpler to leave the lid as a single weight but find its mass centre and use the displacement of that for the torque.

 

[Al_Pa_Cone]

So to find the mass centre would I be looking to 50% of 833.3g mass to the right of the pivot (416.65g).
I thought if the setup is effectively a seesaw then to balance each side equally then it would only take a small change in mass at one side to tip the balance and open the lid?

 

[haruspex]

So to find the mass centre would I be looking to 50% of 833.3g mass to the right of the pivot (416.65g).
I thought if the setup is effectively a seesaw then to balance each side equally then it would only take a small change in mass at one side to tip the balance and open the lid?

The whole lid is 1kg and 0.6m wide. Its mass centre is 0.3m from each edge, so 0.2m from the pivot. Its torque about the pivot is (0.2m)(1kg)(g).
If the suspended mass is m, its torque the other way is (0.1m)(mg).
So to balance you will need m=2kg.

 

[Al_Pa_Cone]

So the torque about the pivot is equal to the counterweight mass effectively making the -2000g(clockwise torque being negative) + 2000g(the counterweight anticlockwise torque) = 0 Torque so no movement.

 

[haruspex]

View attachment 234025
So the torque about the pivot is equal to the counterweight mass effectively making the -2000g(clockwise torque being negative) + 2000g(the counterweight anticlockwise torque) = 0 Torque so no movement.

Yes.

 

[Al_Pa_Cone]

I am trying to understand how the mass at twice the distance to the right of the pivot can be equal to the mass balancing it at half the distance to the left of the pivot?

Its torque about the pivot is (0.2m)(1kg)(g).
If the suspended mass is m, its torque the other way is (0.1m)(mg).
So to balance you will need m=2kg.

From your information I worked out the torque about the pivot to be 0.2 x 1000g x 10 which gives 2000g at 0.2m and the counterweight mass to be 2000g at 0.1 to the left? I would have expected it to be double to compesate for the distance?

 

[haruspex]

0.2 x 1000g x 10 which gives 2000g at 0.2m

No, it gives 200g gram-meters.
(To avoid confusion, let's write out gram in full and reserve 'g' gravitational acceleration. Alternatively, we can switch to kg.)
That's a torque, so it is not "at" anywhere. You have already used the 0.2m info.
It balances the 2000 gram x 0.1m x g = 200g gram-meter (or 2 Newton-meter) torque from the suspended mass.
I note that in your diagram in post #43 you have written that the mass of the lid is now 2000 grams. I assume you meant it has a torque of 2 Newton-meters.

 

[Al_Pa_Cone]

Yes, I was trying to visually represent the Torque as it helps me to try and understand the logic of the situation. I think my fault in understanding your method is because I have not used the units gram-meters anywhere in my course so far and I am unfamiliar with it. I have just googled a gram-meter and know I understand its value to a Newton meter.

Thank you for all your help. I owe you a pint or two!

 

[haruspex]

Yes, I was trying to visually represent the Torque as it helps me to try and understand the logic of the situation. I think my fault in understanding your method is because I have not used the units gram-meters anywhere in my course so far and I am unfamiliar with it. I have just googled a gram-meter and know I understand its value to a Newton meter.

Thank you for all your help. I owe you a pint or two!

It would be better to stick to standard MKS units. (And it looks like I did not need to use US spelling!)
So that'd be 0.57 to 1.14 litres, right?

 

[Al_Pa_Cone]

Yeah If you drink your pints by the litre I always have a couple, so in that case I owe you 3.51951 Pints. Cheers!

 

[Al_Pa_Cone]

I was checking my initial calculations regarding the weight of the cat impacting the roof from 1 metre and I got 442Nm of force in my method. I have just recalculated and got 44.2Nm. Does this look correct in the method?

 

[haruspex]

View attachment 234080
I was checking my initial calculations regarding the weight of the cat impacting the roof from 1 metre and I got 442Nm of force in my method. I have just recalculated and got 44.2Nm. Does this look correct in the method?

Please create a new thread for this. It is a forum standard that each distinct problem has its own thread. I should have insisted on that for the lid counterweight question.
If you want to make sure I see it, put @haruspex in the text.