Trying to get hold of the relevant equations/No.s what-have-you, to put on an image of a hammer about to hit a nail.
I'm sure this is quite easy for someone in the know?
Chat Forum
Physics experts . . . Hammer and nail
-
Posted 8 months ago #
-
Newton's second?
Posted 8 months ago # -
kinetic energy of hammer = (friction force of moving nail x distance nail moves) + deformation of wood*.
(*again, force x distance = energy)
Posted 8 months ago # -
Conservation of momentum as a concept should cover it. As mogrim points out above, Newton's second law is about momentum change anyway.
Posted 8 months ago # -
hmm, but the mass of the nail is so small in relation to the mass of the hammer, that the velocity of the hammer isn't really affected by the impact with the nail.
i'd say the velocity of the hammer is affected by the resistance of driving the nail into the wood...
you swing a 1kg hammer at 1m/s at a un-restricted 5g nail. after impact you have a 1.005kg hammer+nail combo travelling at 0.995m/s.
i don't think it's a very helpfull equation/model in this case. but i may be wrong.
Posted 8 months ago # -
Yeah, momentum's pretty irrelevant here. It's friction and deformation - and the former is affected by the latter - TBH, if you want any half-respectable formulae for energy spent in localised material deformation I'd think you're well into FEA rather than simple equations.
Posted 8 months ago # -
Don't forget all the lost sound energy as well
Posted 8 months ago # -
Especially if you hit your thumb.
Posted 8 months ago # -
look back to A-level maths were you have the 2 snooker balls and working out the energy transfer.
You need the weight of the hammer (head will do) and the speed it is being swung at. (you actually want the whole weight applied with an angular velocity to give an angular momentum but if we assume low angle, ie not a huge sledge hammer, we can ignore this).
Going back to the snooker balls there will be a transfer of energy to the nail and a rebound of the hammer but the elasticities of the system will be accounted for using a coefficient of restitution. This will depends on the resistance of the substrate to the nail penetrating, change in momentum of the nail (tiny) and global deformation of the substrate - ie a concrete wall will be solid but a beam of timber will bend.
The best way to work it out would be to have the substrate horizontal. The nail vertical just in the substrate and then pivot the hammer from the end of it's handle so the shaft is horizontal when it strikes the nail. Allow the hammer to fall and strike the nail and then see how high it bounces up (use a camera or something to record it). Catch the hammer before it double strikes (or you have more work to do).
You can then get out all the figures you need, ignoring energy lost to the surroundings, supports and air resistance using the calculations here:
http://www.efm.leeds.ac.uk/CIVE/CIVE1140/section04/mechanics_sec04_full_notes02.html
Posted 8 months ago # -
Don't forget a note reminding people to wear safety glasses when attempting these calculations.
Posted 8 months ago # -
The strongest message to be sent here is about applying various strategies and problem solving approaches. About being versatile and adaptable
"If your only tool is a hammer, every problem looks like a nail"
Posted 8 months ago #
Reply
You must log in to post.
