What constant force does the floor exert

[haruspex]

Forgot I had already incorporated 9.8 (gravity) already, my mistake.

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.

 

[HJ^2]

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?

 

[haruspex]

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.

 

[HJ^2]

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

 

[haruspex]

So fnet would equal Fg + Fn = m*a

Right. The only unknown there is Fn. Solve.

 

[HJ^2]

Right. The only unknown there is Fn. Solve.

Fn = (m*a) - Fg

With variables is;

Fn = (74.4 * 6.8411) - 9.8 = 499.177 ?

 

[haruspex]

Fn = (m*a) - Fg

With variables is;

Fn = (74.4 * 6.8411) - 9.8 = 499.177 ?

Fg is not 9.8 For one thing, it will be negative (since you have chosen up as positive).

 

[HJ^2]

Fg is not 9.8 For one thing, it will be negative (since you have chosen up as positive).

Fgrav = -9.8*74.4?
If that's right, then;
Fn = (74.4*6.8411) - (-729.12) = 1238.10

 

[haruspex]

Fgrav = -9.8*74.4?
If that's right, then;
Fn = (74.4*6.8411) - (-729.12) = 1238.10

Right (but shared across two feet).
So, do you see where your prof went wrong?

 

[HJ^2]

Right (but shared across two feet).
So, do you see where your prof went wrong?

Absolutely. I'll have to bring it up to her as I'm sure my class mates had similar concerns.

 

[haruspex]

Absolutely. I'll have to bring it up to her as I'm sure my class mates had similar concerns.

Good.
On a side note, if you calculate how far the athlete's mass centre must travel after touch down and before coming to rest, it's about 3m!

 

[HJ^2]

Good.
On a side note, if you calculate how far the athlete's mass centre must travel after touch down and before coming to rest, it's about 3m!

Oh thanks! And thank you so much for your help + everyone else in this thread :)