Recent content by Microcephalus

The energy of the bullet would be E = ½·m·v² So for the same m and v the energy would be equal, regardless of how the bullet was fired.

Thanks for answering! Well, the roof surface should withstand 3 kg without penetration. The hut should withstand 20 kg without the equipment inside taking damage, though it is allowed for the roof to get smashed up, I suppose. And then there is the deposits of snow and ice and wind loads as...

Sorry if I write something redundant here, I'm in a small hurry. Rolling friction is not the same as sliding friction. You need a horizontal force of about F = k·mg where k ≈ 0.01 for car tires on asphalt, give or take ±0.005 or so. So a 100 kg vehicle needs about 10 N to move. Multiply with...

Hello I'm tasked to evaluate the roof of a small power station situated beneath a rather high tower in an alpine landscape, by means of FEA. One requirement is that the roof shall endure chunks of ice falling on it. Problem is, I am stumped as how to find the loads. The Eurocode norms I am...

Ouch indeed. What is that recoiling component called in english, really? I google for cannons and howitzers but the closest I found was "bolt". Breech? English is not my primary language and I never did any military service so the nomenclature and taxonomy of guns is not quite my forte.

Thanks, I think the picture is clearing in my head now. The 50 kN does kick on the bolt, briefly. The 1000 N is then required to catch the motion, over extended time. We're not breaking any Newtonian laws here. Also, it is fully acceptable that one mass gets less kinetic energy than the other.

Ciao people I'm having trouble wrapping my mind around the physics of the follwing: A cannon fires a 1 kg cannonball that leaves the barrel at 1000 m/s The cannon weighs 1000 kg. Now, conservation of momentum suggests that the residual velocity of the cannon would be 1·1000 + 1000·v = 0 → v...