Given a fixed quantity of work performed, what choice of momentum will maximize how deeply a nail is driven into wood?

Question

Let's say I am trying to drive a nail into a piece of wood by dropping a weight on it.

• I am willing to do some fixed quantity of work to raise up the weight. I can choose the weight and the height, as long as the resulting work is the desired value.

• Concretely, say I'm willing to raise a 1 kg weight up to 1 m, imparting a gravitational potential energy of 1 * 1 * 9.8 = 9.8 J.

• I could also double the height and halve the mass, yielding the same potential energy.

• More generally, as I raise or lower the height, the energy-equivalent mass is 1/h.

• We solve for velocity at impact for a given height h, neglecting air resistance, via: $$ \begin{align} m \cdot v^2 &= m \cdot g \cdot h \\ v^2 &= g \cdot h \\ v &= \sqrt{9.8 \cdot h} \\ &\approx 3.1 \sqrt{h} \end{align} $$

• So that momentum at impact is:

$$ \begin{align} p &= mv \\ &= \frac{1}{h} \cdot 3.1 \cdot \sqrt{h} \\ &= \frac{3.1}{\sqrt{h}} \end{align} $$

So, as height increases and energy stays fixed, momentum at impact decreases.

This seems to suggest that lifting a huge weight a tiny amount is what maximizes momentum under this scenario (except at whatever scale the ideal model becomes too inaccurate).

Is the momentum-maximizing configuration also what would drive the nail the deepest? I've neglected a discussion of the stiffness/spring constant of the materials; does that affect the momentum we'd choose?

Answer 1

Suppose the friction between the nail and the wood is $F(x)$ where $x$ is the length of nail driven into the wood. As a first approximation we will assume $F$ varies linearly with $x$, so $F=\lambda x$. Then the energy required to drive the nail a distance $d$ into the wood is

$\displaystyle E = \int_0^d F(x) \space dx = \int_0^d \lambda x \space dx = \frac 1 2 \lambda d^2$

Note that given $\lambda$ and $d$ then this energy is constant (and this would also be true for any other function $F(x)$). So it appears that the work required to drive a nail a distance $d$ into wood is always the same, no matter how we do that.

However, this calculation does not take into account the fact that each time we hit the nail, some energy is wasted as heat and noise. So the most efficient way to drive a nail into wood is to hit it as few times as possible.

To drive the nail into the wood with a single blow, we would need to raise a weight with mass $m$ by a height $h$ such that

$\displaystyle\text{Work done raising weight} = mgh > E$

Typically, this would require an impossibly heavy weight, or an unfeasibly long lift height, or both. So, in practice, you should lift as heavy as weight as you can by as large a height as you can, in order to drive the nail in with the smallest number of blows and waste the least energy doing so.