Force produced after landing stiffly

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1. Nov 14, 2014

Jazz

1. The problem statement, all variables and given/known data

Very large forces are produced in joints when a person jumps from some height to the ground. Calculate the magnitude of the force produced if an $\small{80.0\!-\!kg}$ person jumps from a $\small{0.600\!–\!m\!-\!}$high ledge and lands stiffly, compressing joint material $\small{1.50\ cm}$ as a result. (Be certain to include the weight of the person).

Known data:

$m = 80.0\ kg$

$h = 0.600\ m$

$d = 1.50 \times 10^{-2}\ m$

2. Relevant equations

$W_{nc} = \Delta KE + \Delta PE$

3. The attempt at a solution

The work done is negative since it removes mechanical energy from the system:

$- W_{nc} = KE_f - KE_i + PE_f - PE_i$

$- W_{nc} = - PE_i$

$F_{nc}d = mgh$

$F_{nc} = \frac{mgh}{d} = \frac{(80.0\ kg)(9.8\ m/s^2)(0.600\ m)}{1.50 \times 10^{-2}\ m}$

$F_{nc} = 3.13 \times 10^4\ N$

Here is where I thought I was done. The problem asks to include the weight of the person. By doing so, I get $\scriptsize{3.21 \times 10^4\ N}$, which agrees with the answer given to it, but I've not been able to see what is the point of adding their weight.

As fas as I understand the weight is already being included in $F_{nc}$.

If I think about it in terms of compression, when standing their $\scriptsize{(80.0\ kg)(9.8\ m/s^2) = 784\ N}$ are always compressing the joints a little bit.

If I added a barbell weighing $\frac{3.134 \times 10^4\ N - 784\ N}{9.8\ m/s^2} = \scriptsize{3.12 \times 10^3\ kg}$ on top of the man, it would be enough to compress their joints 1.50 cm.

Asking me to add the weight of the man (again, from my view) it's like saying that $F_{nc}$ does not include the man's weight (I can't see why not) and, when standing, the man's weight doesn't compress the joints at all.

Thanks!!

Last edited: Nov 14, 2014
2. Nov 14, 2014

BvU

Hello jazz,

Yes, you included the mass of the man in the Fnc.

I don't know how you go from the 3.13 N to 3.21 by including the man's mass ?

Also, my calculator claims 80 * 9.8 = 784, not 470 !

And now the point: if you decscend 0.6 m and then compress 1.5 cm, how much distance has your center of mass covered ?
So include some of the compression in h. How big a fractgion depends on where the compression takes place. From the answer I'd say they think it's the feet only...

3. Nov 14, 2014

Jazz

Hi BvU,

I had multiplied the man's weight by $\small{0.600\ d}$ too ]: . I've edited that.

The textbook's answer of 3.21 is just adding 80*9.8 to the force obtained as $F_{nc}$.

I still have a hard time understanding this. The man hits the ground with a speed of $\small{\sqrt{2gh}}$ and the collision lasts $\small{\Delta t= \frac{m\sqrt{2gh}}{F_{nc}} = 8.75 \times 10^{-3}\ s}$. During that time the ground is able to distinguish between the weight of the man and the force of the collision, exerting an upward force overcoming these two things. It doesn't make sense to me ):

Can I claim that the textbook's answer is not correct or I need to think about it with another approach?

4. Nov 14, 2014

haruspex

Soapbox: This is a flawed question. The force will not be constant during compression. In the context, the most interesting quantity for many purposes would be the peak force. Often such questions ask for the average force, but to answer that it is not sufficient to know the impact speed and the extent of compression.
Average acceleration is defined as change in velocity divided by change in time: Δv/Δt. Ergo, average force is change in momentum divided by change in time: Δp/Δt. There is insufficient information to calculate that here. You can calculate an "average over distance" from change in energy and displacement, but that's different from the standard meaning of average force.
To answer the question as posed (whether for peak force or for average force) you need to make some assumption about the pattern of deceleration. If it is constant deceleration then the average over distance, peak, and average (over time) all happen to give the same answer, but more likely the deceleration will increase from 0 more-or-less steadily up to a maximum. A reasonable approach would be to treat it as an elastic compression, but with no actual rebound.

Putting all that aside, and accepting ΔE/Δs as an appropriate formula:
You haven't defined Fnc. Is it the force from the floor or the net vertical force on the person? If you choose one, the first equation above is wrong; if you choose the other, the second statement is wrong.

5. Nov 24, 2014

Jazz

I notice I’ve been trying to find the force from the floor. The other case seems similar to the Elevator Problem. Since I actually didn’t see the problem from the latter approach, I should define $F_{nc}$ as $F_{floor}$.

If the duration of the collision were given, then the average force (as it's defined) could be calculated, right?

If I’ve understood this, then the $\Delta KE$ goes into a $\Delta PE_{spring}$ over a $\Delta t$; with the joints of the person serving as the springs:

$-\frac{\Delta KE}{\Delta t} = \frac{\Delta PE_{spring}}{\Delta t} \Rightarrow -\Delta KE = \Delta PE_{spring}$

If that were the case, it would require me to find the spring constant $k$ and then find the force $k\Delta L$, being $\Delta L$ equal to $d$. That yields me twice the force I found above.

I'm somewhat more confused than before about how to use the data I have in order to solve this ]:

6. Nov 24, 2014

PhanthomJay

As has been hinted, your value of h is wrong in your first post. It is not 0.600 m.

7. Nov 24, 2014

haruspex

In that case your equation $F_{nc}d = mgh$ does not comply with this injunction:
Yes, if it's the average that's required. The question as posted does not specify.
Yes.
That would give you the peak force, not the average. To find the average, calculate the time for the compression (assuming SHM).
But I strongly suspect the questioner wants the average over distance, i.e. your first approach.

8. Nov 24, 2014

Jazz

Great. I wasn't sure how average force is exactly defined until now.

Well, now I see what my mistakes were (the wrong answer for a misunderstood question). @BvU pointed out a key part, and I skipped it |:

Thanks guys.

9. Nov 24, 2014

haruspex

There are two ways of looking at it, and they produce the same answer.
Either, the force acts to absorb the total loss in PE: F = mg(h+d)/d,
or, the force acts to absorb the initial loss of PE, plus has to counter gravity while doing so: F = mgh/d + mg
The OP hinted at the second approach, while BvU and PhanthomJay prefer the first.