# Tag Info

57

There's many different things at work here. First, there's the issue of acceleration. Hammers are very hard and solid, so when you hit the nail head with the hammer, the energy and force of the blow is delivered at almost an instant. Hands, on the other hand, are rather soft, and will spread out the same amount of energy and acceleration over a longer time ...

29

It is just because when you hammer a nail, you apply a force to the nail by changing the momentum of the hammer $F=\dot p$ This force helps in overcoming friction and thus pushing the nail inside. This isn't the case with your hand. Moreover, if you try to develop the same amount of force with your hand, you will get hurt because of the pressure applied to ...

13

Your hands are soft whereas the hammer head is very hard. Therefore your hand will come to rest upon striking a nail in a much greater time as compared to hammer. Now, force is the rate of change of momentum. The hammer is heavier so momentum is more initially even if hand and hammer hit at same velocity. As well as the time taken for the momentum to change ...

8

Looking at the graph you can also see that the displacement is equal to the average velocity $\times$ time.

6

This is in fact possible, as evidenced by the number of videos online of someone hammering a nail in with their bare hands. More to the point, the human hand is plenty hard enough to hammer in a nail into some pretty hard if it is swung fast enough. It's just that the damage to the hand would be rather devastating, and so our pesky brains keep us from ...

4

You can! But first, some precautions: Your hand is squishier than the wood (generally, for wood of a condition to have new nails driven into it and hands prepared to do some hammering). This means that it will take less pressure for the nail to damage your hand than for the nail to enter the wood. The pointy bit at the front of the nail and the broad ...

2

TL;DR: In order for the nail to be driven into the wood, you need to apply sufficient force to split apart the wood fibers ("make a hole") and overcome the force of friction between nail and wood. The hammer can apply a much greater (instantaneous) force than the hand, because it is much harder (has a greater Young's modulus). If you assume a mass $m$ ...

2

Take a simple 1D example. We'll plot gravitational potential energy as a function of horizontal position for the side of a hill: We normally define $F = -\nabla V$ and that means the force points downhill i.e. if we let the particle go it will roll downhill. If instead we defined $F = \nabla V$ the force would point uphill i.e. if we let the particle go ...

2

The inertia matrix for a thin rectangular foil (laying along the xy plane) in body coordinates is $$I_{body} = \begin{vmatrix} \frac{m}{12} b^2 & 0 & 0 \\ 0 & \frac{m}{12} a^2 & 0 \\ 0& 0 & \frac{m}{12}(a^2+b^2) \end{vmatrix}$$ where $a$ and $b$ are the side dimensions. The inertia matrix in world coordinates, while rotated by ...

1

While I have never conducted such an experiment (yet), I conjecture it will be the energy of the projectile what will determine the results. It will also bring momentum, but this would be comparable roughly to three A380 jets at cruise speed, so it probably would not change much on the result. The projectile would start evaporating already when passing ...

1

It depends what you mean by vibrational motion. A point mass indeed does not have internal vibrational degrees of freedom, it can just have translational motion, as you pointed out. However, if you attach this point mass to a spring you create a harmonic oscillator. It has a translational motion, but this oscillating motion can describe vibrations such as ...

1

One way of looking at the change is that you are changing the definition of potential from The potential at a point is the work done on a unit mass in going from an arbitrarily chosen zero of potential to the point to The potential at a point is the work done by a unit mass in going from an arbitrarily chosen zero of potential to the point So ...

1

Your solutions are wrong. As $$\frac{dq}{dt}=\frac{\partial H}{\partial p}\qquad \frac{dp}{dt}=-\frac{\partial H}{\partial q}$$ you get $$\frac{dq}{dt}=10\,p\qquad \frac{dp}{dt}=0$$ i.e. $$q(t)=10\,p_{0}\,t+q_{0}\qquad p=p_{0}$$ The $q$ coordinate flows in time in straight lines, while the $p$ coordinate doesn't change in time. So each phase ...

1

If you're strong enough, you can push the nail in by hand. You only need protection against the "equal and opposite reaction" of the nail against your soft hands. The nail can penetrate the wood because it is sharp, applying high pressure to the surface area it contacts. The nail's head is relatively much wider, so it will not penetrate your hand as easily ...

1

It is a method that is generally used for conservative unidimensional problems (problems with only one degree of freedom, here your angle $\theta$ or cartesian coordinate $x$). You'll notice that it is equivalent to using Newton's second law in this case : let us write the total energy $E = \frac{1}{2} m v^2 + V(x)$, $V$ being potential energy. The problem ...

1

Your system has 2 degrees of freedom, but using $x_1$ and $x_2$ will not be helpful in determining the effective spring rate. To get the spring rate you need the extension $x$ of the connection point with mass $M$ and the tilt angle $\theta$. Do the substitution: \begin{align} x_1 & = x - a \theta \\ x_2 & = x + b \theta \end{align} The ...

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