Recent content by haruspex

Afterthought… I was unconsciously assuming the shelf would be relatively stiff. With a relatively flexible shelf, the principal load could be on the tips of the pins. Messier than I thought.

That would be true if the shelf were correspondingly shorter, so that the length of contact were the same. Even then, it’s not a question of stability in the usual sense but one of not breaking the pins. If the shelf always reaches the wall at both ends there will be minimum pin length...

It is not clear to me. What step breaks if it is not constant wrt time?

I am not sure how to read that. Are you saying that when accelerating as fast as you could you could not prevent the front wheel rising, and that limited how fast you could go, for fear of flipping backwards? Or that you deliberately raised the front wheel a little because that reduced the...

Professional cyclists achieve astonishing acceleration but don’t flip over backwards. It would seem perverse to try to do the lift entirely by pedal power when it would be so much easier to get a start using the body’s inertia. No doubt pedal power comes into play thereafter.

I don’t get how you can do that without first pushing down on them in order to give the body the necessary posture and or angular momentum.

Does that mean they push down on the handlebars while the front wheel is still grounded, throwing the body up and back, so that they can then pull up on the handlebars when the front wheel leaves the ground? Yes, interesting. I'll have a go at that.

Taking the wheelbase as 2L, mass centre at midpoint, and assuming it can only rotate through a small angle before the rear wheel becomes airborne, I get the final ω as ##\frac{2g}v\frac{mL^2}{I+mL^2}##. If v is small then the angle change becomes significant. ##\omega^2\leq \frac{2mgL}{I+mL^2}##

My guess is that is not the ideal. Seems better to spread the vertical impulse by landing the wheels one at a time.

No, that's a Megajoule. A Gigajoule has nine zeroes. But the 275kWh is correct (well, 278).

Ok, but the force at point O due to the suspended weight is W, as you had it in post #98. The error there is that, since the horizontal bar is rigid, there is also a torque ##Wz##. I.e. the force of the weight acting at the end of the bar is equivalent to W acting at O plus a clockwise torque...

That gives the right moments equation, but it is not right as a force diagram. The weight W, applied where it is, is not equivalent to a greater force applied at O. Why not consider the horizontal bar and the diagonal bar as the system and just write ##(y+z)W=xF##?

@gleem is taking moments about P, not O. I see no conflict.

How come you have relocated W to act at point O? The horizontal distance from P to the hanging mass is what matters.

Is it possible to move between sections? If so, maybe some not intending to use phones moved to where phone usage was low.