# What constant force does the floor exert

• HJ^2
In summary: I'm guessing here, but I'd say it's a few Newton.(*) I'm not sure if " (*) " is supposed to be a footnote or something, but it doesn't seem to be doing anything.
HJ^2

## Homework Statement

A 74.4 kg person falls straight down from a 1.96 m height (measured from his feet) and lands with weight distributed equally on both feet. To soften the blow, he bends his knees so that it takes 0.906 s for him to stop once his feet touch the ground. What constant force does the floor exert on each foot while he's stopping? Give your answer in Newtons

## Homework Equations

f=ma
f=(m*v^2) / 2x
F * t = m * Δv
Δp=m(v-vo)
Favg=Δp/Δt
w=mg
Xf=Xi+Vit+(at^2/2)
Vf=Vi+at
Vf^2=Vi^2+2a(Xf-Xi)

## The Attempt at a Solution

The answer is 254.5 and I am trying to work backwards to see what I am doing incorrectly.

Originally, I had lumped all variables into one equation ( f * t = m * Δv ) but now I think it's two parts? First system being A; person falling 1.96m and system B; ground receiving force and pushing back up as the person absorbs shock @ 0.0906s?

I think the equation might be solved by calculating acceleration and solving for force with Newtons 2nd law, but I am unsure and that sounds completely wrong.

What was the speed when he reached the ground ?

BvU said:
What was the speed when he reached the ground ?
well he comes to rest so I would assume 0?

HJ^2 said:
well he comes to rest so I would assume 0?
His speed an instant before he comes to rest.

SteamKing said:
His speed an instant before he comes to rest.
Ah right, so 2.16? (1.96 / 0.906)

HJ^2 said:
Ah right, so 2.16? (1.96 / 0.906)
Why do you say that?

Think. The guy is falling. He falls 1.96 m. You know, gravity. Hint Hint

SteamKing said:
Why do you say that?

Think. The guy is falling. He falls 1.96 m. You know, gravity. Hint Hint

Ok, Vf^2 = Vi^2 + 2ad ?

That would be Vf^2 = 0 + 2(9.8)(1.96) = 38.416, is that right?

HJ^2 said:
Ok, Vf^2 = Vi^2 + 2ad ?

That would be Vf^2 = 0 + 2(9.8)(1.96) = 38.416, is that right?
Yes.

Ok, I am stuck after this. I know initial velocity, final velocity, mass distance and time but i don't know where to go from there.

HJ^2 said:
Ok, I am stuck after this. I know initial velocity, final velocity, mass distance and time but i don't know where to go from there.
You go back to the OP and solve for the unknown force, using the formula listed there.

SteamKing said:
You go back to the OP and solve for the unknown force, using the formula listed there.

f=ma gives me 2858n when the answer should be 254.2?

HJ^2 said:
f=ma gives me 2858n when the answer should be 254.2?
(I get an answer between the two. 254.2 is much too low. The force is greater than that when the person stands still.)

haruspex said:
(I get an answer between the two. 254.2 is much too low. The force is greater than that when the person stands still.)

f=ma as in; f=74.4 * 38.416

Also 254.2 is the answer to the problem as generated via the program I am working this equation through - could this just be incorrect?

HJ^2 said:
f=ma as in; f=74.4 * 38.416
HJ^2 said:
Vf^2 = 0 + 2(9.8)(1.96) = 38.416
That's not an 'a' on the left.
HJ^2 said:
Also 254.2 is the answer to the problem as generated via the program
They've made a classic blunder. See if you can get the right answer, then guess what mistake they made.

haruspex said:

That's not an 'a' on the left.

They've made a classic blunder. See if you can get the right answer, then guess what mistake they made.

Oh whoops, a = (v - v0)/t

a = (38.416 - 0)/0.906 = 42.201

f=ma ---> f=74.4*42.201
so f=3139n / 2 (split evenly between two feet) = 1569n?

HJ^2 said:
Oh whoops, a = (v - v0)/t

a = (38.416 - 0)/0.906 = 42.201

f=ma ---> f=74.4*42.201
so f=3139n / 2 (split evenly between two feet) = 1569n?
You're still guessing here. Go back and read Post #7 carefully. I don't understand why you are having so much trouble understanding your own work.

SteamKing said:
You're still guessing here. Go back and read Post #7 carefully. I don't understand why you are having so much trouble understanding your own work.

I got it! I forgot to square root the final velocity.

Vf = 6.1980

a= 6.1980 / 0.906 = 6.8411

6.8411 * 74.4 = 508.908

508.908 /2 = 254.4

And then after this 0.906 seconds (that's a pretty accurate time measurement they managed there ! (*) ) the force is zero , right ?

Reason I mention this accuracy is: the problem statement clearly says "falls straight down from a 1.96 m height (measured from his feet)" and it also says "bends his knees". Well, I'm too old to practice this, but I guestimate that his center of gravity after these 0.906 s is about 0.5 m lower than when standing up. Somehow that potential energy also has to be compensated. Ignoring that is an inaccuracy of around 25%. Makes those three-digit numbers look funny.
(*) -- and I think it should be considerably shorter... anyone know a suitable Utube clip ? Maybe the last seconds of this one

HJ^2
BvU said:
(*) -- and I think it should be considerably shorter
Indeed. The figures given require the athlete to have legs at least 3m long.
BvU said:
I guestimate that his center of gravity after these 0.906 s is about 0.5 m lower than when standing up... Ignoring that is an inaccuracy of around 25%.
If we take the given data on trust, the error is much greater.
HJ^2, you can either use BvU's method of taking into account the extra PE lost in bending the legs (but be warned you'll get some crazy values), or you can think in terms of the forces acting on the athlete and the net acceleration. The two methods should produce the same result.

Dear haru, we are criticizing the composer of the exercise. HJ is content with the 255 N -- even when standing still is already 365 N per foot...

BvU said:
Dear haru, we are criticizing the composer of the exercise. HJ is content with the 255 N -- even when standing still is already 365 N per foot...
Criticising the composer is not the point. It is important that HJ^2 understands how to get the correct answer.

haruspex said:
Criticising the composer is not the point. It is important that HJ^2 understands how to get the correct answer.
I understood the concepts; it was just a weird problem that didn't make a lot of sense at first

HJ^2 said:
I understood the concepts; it was just a weird problem that didn't make a lot of sense at first
Ok, but do you understand that the book answer, 255N, is wildly wrong?

haruspex said:
Ok, but do you understand that the book answer, 255N, is wildly wrong?
I think so? Since the numbers here can't really be applied to a real life problem and that some values are hardly considered in the equation. The professor of my class is more concerned about if we know how to calculate the material, I think.

HJ^2 said:
I think so? Since the numbers here can't really be applied to a real life problem and that some values are hardly considered in the equation. The professor of my class is more concerned about if we know how to calculate the material, I think.
No, it's more straightforward than that.
As I wrote in post #19, one way to approach it is to consider the FBD of the athlete after making contact with the ground but still moving. What are the forces acting on the athlete? What is the net acceleration?
BvU's method is sound too, but doesn't make it as clear what happens.

haruspex said:
No, it's more straightforward than that.
As I wrote in post #19, one way to approach it is to consider the FBD of the athlete after making contact with the ground but still moving. What are the forces acting on the athlete? What is the net acceleration?
BvU's method is sound too, but doesn't make it as clear what happens.

Well the forces acting on the athlete would include gravity & normal force,? Net acc. would have to be zero?

HJ^2 said:
Well the forces acting on the athlete would include gravity & normal force,? Net acc. would have to be zero?
The athlete hasn't come to a stop yet, but will do so. You calculated the speed on making contact with the ground, and you are given the time to come to rest.

haruspex said:
The athlete hasn't come to a stop yet, but will do so. You calculated the speed on making contact with the ground, and you are given the time to come to rest.
Well the acceleration of the athlete is 6.8411 and there's a -9.8 force as well; -2.9589?

HJ^2 said:
Well the acceleration of the athlete is 6.8411
Yes.
HJ^2 said:
and there's a -9.8 force as well; -2.9589?
As well? Don't know what you mean by that.
The acceleration is the result of the net force. What equation does that give you?

haruspex said:
Yes.

As well? Don't know what you mean by that.
The acceleration is the result of the net force. What equation does that give you?

Since acceleration is the result of NF; would it not be mass * acc (f=ma)

HJ^2 said:

Since acceleration is the result of NF; would it not be mass * acc (f=ma)
Yes, but expand that net force into a sum of forces.

haruspex said:
Yes, but expand that net force into a sum of forces.
okay; sum of forces, as in, Fnet = m*a of which Fnet covers x / y values like friction, gravity, normal force? Am I missing something?

HJ^2 said:
okay; sum of forces, as in, Fnet = m*a of which Fnet covers x / y values like friction, gravity, normal force? Am I missing something?
Sure, but write out what Fnet is here, in terms of normal force etc. There are only vertical forces involved, so it's easy. Remember, the normal force is what you are asked to find, so it's an unknown at this stage.

haruspex said:
Sure, but write out what Fnet is here, in terms of normal force etc. There are only vertical forces involved, so it's easy. Remember, the normal force is what you are asked to find, so it's an unknown at this stage.
So fnet would equal Fg + Fn = m*a

HJ^2 said:
So fnet would equal Fg + Fn = m*a
Right. The only unknown there is Fn. Solve.

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