# Math Help - [SOLVED] Mechanics problem?

1. ## [SOLVED] Mechanics problem?

A nail of mass M is driven into a board against a constant resistive force F by a hammer of mass m, which is allowed to fall freely at each stroke through a height h. The hammer does not rebound after striking the nail. Find the distance the nail is driven in at each blow. Show that the total energy expended in raising the hammer during the operation of driving the nail fully into a depth of d is independent of the value of h and can be decreased by making the hammer more massive.

2. Originally Posted by fardeen_gen
A nail of mass M is driven into a board against a constant resistive force F by a hammer of mass m, which is allowed to fall freely at each stroke through a height h. The hammer does not rebound after striking the nail. Find the distance the nail is driven in at each blow.
The momentum of the hammer when it reaches the nail is
mv = Ft = mg.(2h/g)^1/2 = m(2gh)^1/2
using x = vi*t + 0.5gt^2 (x-displacement, vi-initial velocity=0)
The hammer does not rebound after the collision, i.e. it's final velocity is zero, so using conservation of momentum, the nail's momentum after the collision must be the same as the hammer's just before the collision:
Mv = m(2gh)^1/2
So it's velocity will be: (m/M)(2gh)^1/2
The nail is moving against a constant resistive force F, i.e. this force causes a negative acceleration of F/M on the nail. Using the formula
vf^2 = vi^2 + 2ax (vf-final velocity=0, a-acceleraion=-F/M)
Then re-arrange for displacement x:
x = (vf^2 - vi^2)/(2a) = [0-((m/M)(2gh)^1/2)^2]/(2*(-F/M)) = (m/F)(gh/2)^1/2
Note that this makes sense: the distance the nail is driven in increases if you make m (hammer's mass), g (gravitational acceleration) or h (height hammer falls through) larger, but decreases if the resistive force F is increased.

Show that the total energy expended in raising the hammer during the operation of driving the nail fully into a depth of d is independent of the value of h and can be decreased by making the hammer more massive.
Not sure I understand this part correctly. Are they saying: show that raising the hammer through a vertical distance h requires energy independent of h? That's obviously nonsense.
I can't figure out what this part of the question is asking-- can you give clarification please?

3. ## Energy and momentum

Hello fardeen_gen
Originally Posted by fardeen_gen
A nail of mass M is driven into a board against a constant resistive force F by a hammer of mass m, which is allowed to fall freely at each stroke through a height h. The hammer does not rebound after striking the nail. Find the distance the nail is driven in at each blow. Show that the total energy expended in raising the hammer during the operation of driving the nail fully into a depth of d is independent of the value of h and can be decreased by making the hammer more massive.
The momentum of the hammer just before impact = $m\sqrt{2gh}$. Since the hammer does not rebound, the velocity of the hammer and the nail are equal. If this velocity is $V$, then, using conservation of momentum at impact:

$(m+M)V = m\sqrt{2gh}$

$\Rightarrow V = \frac{m\sqrt{2gh}}{m+M}$

KE of system immediately after impact $= \tfrac12(m+M)V^2 = \frac{m^2gh}{m+M}$

I now make the assumption that the resistive force $F$ is large compared to the combined weights $mg + Mg$. (If not, we shall have to take into account the work done by these weight forces as the nail moves through the wood.)

If the nail is driven in a distance $x$, the work done against resistance $= Fx =$ loss in KE

$\Rightarrow x = \frac{m^2gh}{(m+M)F}$

So if the total distance driven is $d$, the number of blows required $= \frac{d}{x} = \frac{d(m+M)F}{m^2gh}$

Each of these blows requires an energy expenditure of $mgh$ in order to raise the hammer. So the total enery expended is $\frac{d(m+M)F}{m^2gh} \times mgh = \frac{dF(m+M)}{m}$, which is independent of $h$.

$\frac{m+M}{m}= 1 + \frac{M}{m}$, which decreases as $m$ increases. So the total energy expended is decreased by making the hammer more massive.

4. Oops. Looks like I screwed up royally.

5. Hello qspeechc
Originally Posted by qspeechc
Oops. Looks like I screwed up royally.
Not to worry. Keep at it!

6. Originally Posted by Grandad

I now make the assumption that the resistive force $F$ is large compared to the combined weights $mg + Mg$. (If not, we shall have to take into account the work done by these weight forces as the nail moves through the wood.)
Thanks for the help!
The solution given is x(distance the nail is driven in at each blow) = (m^2.g.h)/[(m + M){F - (m + M)g}]. That means we haven't assumed the resistive force to be large as compared to mg + Mg?

7. Hello fardeen_gen
Originally Posted by fardeen_gen
Thanks for the help!
The solution given is (m^2.g.h)/[(m + M){F - (m + M)g}]. That means we haven't assumed the resistive force to be large as compared to mg + Mg?
Yes. Where I have written

If the nail is driven in a distance , the work done against resistance loss in KE

you will have to say $(F - (m+M)g)x$ instead, to take into account the (small) amount of work done when the weight forces $mg$ and $Mg$ move through the distance $x$.