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A deadblow hammer is a type of mallet with a hollow head filled with shot or sand. When you hit with it, say on an anvil, the mallet does not rebound, but just falls flat and heavy.

deadblow hammer

I don't understand the advantage of this from a physics point of view. Force is force, right? Normally when I compute the force of a hammer I multiply the velocity and the weight (mass). How is the deadblow hammer any different?

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    $\begingroup$ Velocity $\times$ mass is momentum, not force. The force depends on how quickly the hammer stops (i.e., the derivative of momentum w.r.t. time). The more quickly the hammer stops, the more force it delivers. That basically is the magic of hammers: The less your target yields to the hammer blow, the faster it will stop the hammer, and the more force will be delivered. $\endgroup$ – Solomon Slow Jun 25 '15 at 18:10
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A hammer delivers an impulse (i.e., a change in momentum, calculated as the integral of force over time, https://en.wikipedia.org/wiki/Impulse_%28physics%29) to its target. In the case of a hard, rigid hammer striking a hard, rigid surface, the impulse will be a relatively large force delivered for a relatively short duration.

A mallet (including the dead-blow mallet, described in your question) is designed to deliver the same impulse as a lesser force over a longer period of time.

The difference lies in where the energy is going to be absorbed when you strike a piece of wood---say, part of a piece of fine furniture that you are building. The collision is going to be mostly inelastic in either case.

I don't know the physics---I'm not a physicist---but my gut reaction is that the extreme force delivered by the hammer will cause a lot of stress in the wood right at the point of contact, and that the wooden surface is likely to be permanently deformed. Part of the hammer's kinetic energy therefore will be absorbed in the damage site.

I'm guessing that with the mallet, the lesser force and the longer time interval mean that more of the hammer's energy will go into accelerating the whole piece of wood and, ultimately will be absorbed by friction in the joint that you are trying to pound closed.

Sorry, but I don't know how to model the situation with mathematics. I'm thinking that there would be stress and strain tensors---way over my head.

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The hammer can be thought as a mean to deliver enough energy to the nail to deform the underlying material and let it penetrate deeper.

Ideally, all the energy the hammer gets from your arm as kinetic energy is transferred to the material and results in its (hopefully) permanent deformation, but as always, that's not the case in practice.

What should be a totally anelastic collision (e.g. your hammer should stay on the nail after the impact) is instead slightly elastic: part of the kinetic energy is not used for the deformation, thus some of the hammer momentum "comes back" as a recoil.

While in a normal hammer this remaining momentum is transferred back to the entire tool, if you fill the mallet with sand or with something which can move independently from the rest of the hammer, part of this momentum is transferred to this detached mass which thus recoils instead of your hammer, leaving less momentum for the bouncing of the rest of the tool.

The principle is the same used in large cannons: the barrel is detached from the rest of the weapon to let it recoil without displacing the whole cannon.

Another way you can figure it: take a plastic bottle full of sand (but water should work as well) and drop it, you'll see that the bouncing is heavily dampened by the recoiling of its contents, that you should see going around and markedly "jump" notwithstanding the movement of the bottle which contains them.

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    $\begingroup$ Deadblow hammers are not used on nails. $\endgroup$ – Ambrose Swasey Jun 25 '15 at 19:32
  • $\begingroup$ Well my argument holds anyway... $\endgroup$ – mattecapu Jun 25 '15 at 19:40

protected by Qmechanic Sep 3 '15 at 18:10

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