# Hammer vs large mass on nail

Why is a hammer more effective in driving a nail than a large mass resting over the nail ?

I know this has to do with momentum, but cant figure it out.

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Do you mean: Why does striking a nail with a moving hammer (mass=$m$) have more effect than the same mass $m$ at rest on the nail? – Nic Aug 15 '11 at 14:42

The frictional force (F) holding the nail in place is what both the hammer and the large mass must overcome to move the nail. To get the nail to move you need a (Force = mass * acceleration) of the object hitting the nail greater than the (Force) holding the nail in place.

With a large mass just resting on the nail, you are stuck with a constant acceleration gravity, so you will need a larger mass. With a hammer, you can achieve a higher acceleration than gravity, so your mass requirements are not as much.

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Nice and concise, +1. – David Z Aug 15 '11 at 21:04
It is totally possible to drive a nail using mass alone or using the pressure factor (e.g. hydraulic pistons), which should also be in that equation. I know this from experience: If I release the pressure before it hits (i.e. coasting), it doesn't go down as far as if I keep the pressure on it. – Arlen Beiler Aug 8 '12 at 1:05

The key things to remember are:

1.) $F = ma$

2.) $a = \frac{\mathrm{d}v}{\mathrm{d}t}$

For a $100~\text{kg}$ man standing on the nail: $F = 100~\text{kg} \cdot 9.8~\frac{\text{m}}{\text{s}^{2}} = 980~\text{N}$.

For a $\frac{1}{2}~\text{kg}$ hammer head, swung at $10~\frac{\text{m}}{\text{s}}$: $F = 0.5~\text{kg} \cdot a = ?~\text{N}$.

$a$ in this last equation is the deceleration of the hammer head when it hits the nail. Let's say that the hammer drives the nail $x = 2~\text{mm} = 0.002~\text{m}$ with each stroke, and further assume that the deceleration of the hammer head is constant (makes the math easier). Then you get the quadratic:

$t^{2} - \frac{20}{a}t + \frac{4}{1000a} = 0$

Substituting $a = \frac{10~\frac{\text{m}}{\text{s}}}{t}$ into the equation $t = \sqrt{\frac{2x}{a}}$, we get $t = 0.0004~\text{s} = 0.4~\text{ms}$. If we use that $t$ in the quadratic we find that $a = 19060~\frac{\text{m}}{\text{s}^{2}}$.

So $F = 0.5~\text{kg} \cdot 19060~\frac{\text{m}}{\text{s}^{2}} = 9530~\text{N} \implies$ roughly $10$ times the force of standing on the nail.

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I think the last piece to complete this answer is that there has to be enough force to overcome the static friction holding the nail in place. – Peter Shor Aug 16 '11 at 14:34
Of all 10 answers to this question and its duplicate, this is by far the best. – alx9r Feb 9 at 2:57

The equation of just $F=ma$ is lacking the amount of information needed to sufficiently answer this question, so I'll take a shot at this. You'll find most of what you need with a tour around Wikipedia, but I'll try to give some guidance.

First, let me be sure to mention several quantities.

• Energy ($E=\frac{1}{2} m v^2$)
• Impulse ($I=m v$)
• Force ($\frac{dp}{dt}=m \frac{dv}{dt}$)

The hammer head that falls on the nail has all of these quantities. A physics 101 class should teach you how to fluently exercise the algebra to go back and forth between all of these. Impulse is synonymous with momentum, and impulse and energy are the comparatively easy values to find (the low hanging fruit) in the case of a household hammer. The reason is that the velocity of the hammer as it hits the nail isn't particularly difficult and the mass of the hammer head is trivial to assess. As I was saying, the hammer contains some energy and impulse, which result from the mass and velocity - the balance between those two is relevant to the performance of the hammer.

The case of a large mass resting on the nail is a limit case where there is no energy exchanged (unless it pushes the nail) and high impulse

For some simple in-your-head-physics, think of a hammer head that falls without a human pushing it. Energy is $m g h$, where $m$ is the mass, $g$ is the gravity constant, and $h$ is the height it falls from. Impulse is the momentum on contact and could be said to be $m g \Delta t$. In both cases $m g$ is the force of gravity, but energy cares how far it falls and impulse cares how long it falls. In the case of a large mass resting on the nail, gravity continues to impart force on the mass which is continuously resisted by the friction that prevents the nail from going in. This is the friction we wish to overcome. For a more universal picture, think of energy as $F \Delta x$ and impulse as $F \Delta t$, and in our case, $F$ needs to surpass some given threshold. I should add that $\Delta t$ is a direct function of $h$.

The mechanics of the friction can be approximated by the coefficient of friction. The nail is partially in a hole and the wood snugly squeezes on the nail, giving a normal force, so the force the hammer needs to reach is the coefficient of friction times the normal force, $\mu F_{normal}$, which is just some value as far as we are concerned. If I need to move the nail $1 mm$, then a given energy is required because energy is force times distance. However, even if I have enough energy to move it some distance, it might not move because the value of the force never gets high enough.

To get to a force value on a physics 101 level we would use Hooke's Law, because it gives formulas for how the force is distributed over time. If the nail fails to move you may say it is because the nail softens the blow by its inherent spring-like qualities. By the energy we can predict how far an idealized spring will move by $\frac{1}{2} m v^2=\frac{1}{2} k x^2$, and then the maximum force magnitude will be $k x$. This would be fairly valid equations should the nail not move because if it does move we default to the previous equations using coefficient of friction. For the ideal spring, the motion over time will be some constant times $sin(\sqrt{\frac{k}{m}}t)$, from 0 to $\pi \sqrt{\frac{m}{k}}$, which allows to finally apply the impulse concept. The impulse will be equal to the integral of the force over the time it is applied.

I'm not going to solve the full problem, but let's look at the variables that go into it all.

• The mass of the hammer head
• The material stiffness of the nail ($k$)
• The height it falls from

These pretty much sum it up. The combination of $k$ and $m$ determine the time over which the impulse from the hammer is distributed, and should the hammer breakthrough the static friction threshold, the energy will limit how far the hammerhead may push the nail.

Given all this, I can say that we require sufficient stiffness of the spring-like system as well as sufficient impulse from the hammerhead, and we also need sufficient energy if we don't want to be dinging the nail for really small movements all day.

There are plenty of ways you can come up for a way for this to not work. Put silly putting on the head of the hammer and you don't have sufficient stiffness x impulse because of poor stiffness. Also, if you don't "throw" the hammer at the nail, you distribute the time over which the impulse is imparted, so it doesn't work in that case either. In any case, you need a sufficient height or else you won't have sufficient values to move it like you want it.

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To drive a nail into a piece of wood you need to overcome the force of static friction and the force required to push aside the wood (make a hole).

When an object of mass $m$ and velocity $v$ hits a nail, either the nail moves, or the object decelerates very quickly. This sudden change in momentum is what drives the nail. We know that

$$F\Delta t = m\Delta v$$

So if you want to get a greater force you can change any of these parameters:

• increase the mass (heavier hammer)
• move faster (hit harder)
• shorter $\Delta t$

The latter is a function of the elasticity of the hammer and the nail: as the nail is thicker, or less is sticking out of the wood, it will be a stiffer "spring" and deform less during the impact. This means the hammer will exert a greater force. This is one reason why you can keep hammering a nail as it goes deeper into the wood: while there may be more force needed, the shorter nail provides a greater "force amplifier", in the form of shorter $\Delta t$.

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