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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 said:
Determine the position and magnitude of the max bending moment
In summary, Homework Equations:-The Attempt at a Solution:The user is having difficulty in their homework and is looking for help. They are trying to calculate the position and magnitude of the maximum bending moment for a simply supported beam with a concentrated load at the center. They have tried different methods and are not able to find a correct answer. They are working to correct an error in their original method.
- #37
Al_Pa_Cone
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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.
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.
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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 said:
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- #39
Al_Pa_Cone
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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?
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?
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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).Al_Pa_Cone said:
If the suspended mass is m, its torque the other way is (0.1m)(mg).
So to balance you will need m=2kg.
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Yes.Al_Pa_Cone said:
- #43
Al_Pa_Cone
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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?
haruspex said:
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?
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- #44
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No, it gives 200g gram-meters.Al_Pa_Cone said:
(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.
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Al_Pa_Cone
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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!
Thank you for all your help. I owe you a pint or two!
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It would be better to stick to standard MKS units. (And it looks like I did not need to use US spelling!)Al_Pa_Cone said:
So that'd be 0.57 to 1.14 litres, right?
- #47
Al_Pa_Cone
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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!
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