Centre of Mass/Tipping Point Question

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The discussion centers around a potential transcription error in a problem involving the center of mass and tipping point of a double-decker bus. The original question was misinterpreted as 280 degrees instead of the correct 28 degrees from the vertical. This misunderstanding led to incorrect calculations, as the angle affects the relevant formula used. After recalculating with the correct angle, the answer aligns with the book's solution of 2.3m. The conversation highlights the importance of accurate transcription in mathematical problems.
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Homework Statement
The top deck of a bus is loaded with sandbags. The bus is tested to ensure all 4 wheels are in contact with the ground up, to at least 280 degrees to the vertical. The height of the bus is 4.4m and width is 2.4m. Calculate the height of the centre of mass of the loaded bus by assuming it would topple over at larger angles.
Relevant Equations
h = 1.2tan(10)
1653996533504.png

I drew out a small diagram to illustrate my attempt. I interpreted 280 degrees from the vertical as 10 degrees from the horizontal. Using trig I solved for the h, which I get an obviously incorrect answer. The actual answer in the book is 2.3m,
 
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I suggest it means the bus base is tilted 10° to the horizontal.
 
I believe that you have misread the question or that it has been poorly transcribed. It is not 280 degrees. It is 28 degrees (28°)

One can easily imagine a transcription error where a degree symbol is rendered as the digit zero.
https://www.quora.com/Why-do-double-decker-buses-not-tip-over said:
In the UK, British transportation regulations required that double decker buses be tested on a tilting ramp, to an angle of 28-degrees.
Note that this is 28 degrees from the vertical, not 10 degrees from the horizontal. So the relevant formula will change slightly from the one you have used.

With those tweaks in mind, I get the book answer to two significant digits.

Edit to add graphical image
https://www.ltmuseum.co.uk/collections/collections-online/photographs/item/2002-18969 said:
1654008667895.png
 
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jbriggs444 said:
I believe that you have misread the question or that it has been poorly transcribed. It is not 280 degrees. It is 28 degrees (28°)

One can easily imagine a transcription error where a degree symbol is rendered as the digit zero.

Note that this is 28 degrees from the vertical, not 10 degrees from the horizontal. So the relevant formula will change slightly from the one you have used.

With those tweaks in mind, I get the book answer to two significant digits.
Thank you, after re-calculating I get 2.256..., which is consistent with the answer to 2.s.f. It unfortunately appears to be a printing error.
 
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Thread 'Chain falling out of a horizontal tube onto a table'
My attempt: Initial total M.E = PE of hanging part + PE of part of chain in the tube. I've considered the table as to be at zero of PE. PE of hanging part = ##\frac{1}{2} \frac{m}{l}gh^{2}##. PE of part in the tube = ##\frac{m}{l}(l - h)gh##. Final ME = ##\frac{1}{2}\frac{m}{l}gh^{2}## + ##\frac{1}{2}\frac{m}{l}hv^{2}##. Since Initial ME = Final ME. Therefore, ##\frac{1}{2}\frac{m}{l}hv^{2}## = ##\frac{m}{l}(l-h)gh##. Solving this gives: ## v = \sqrt{2g(l-h)}##. But the answer in the book...

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