What constant force does the floor exert

Click For Summary

Homework Help Overview

The discussion revolves around a physics problem involving a person falling from a height and the forces exerted by the ground upon landing. The subject area includes concepts from mechanics, specifically kinematics and dynamics, as participants explore the calculations related to force, acceleration, and velocity during the impact.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the initial conditions of the fall, the calculations for final velocity upon impact, and the forces acting on the person as they come to a stop. There are attempts to clarify the correct application of equations and the interpretation of variables such as acceleration and force.

Discussion Status

The discussion is ongoing, with participants providing various calculations and questioning the validity of the given answer in the problem statement. Some participants have suggested alternative methods and highlighted potential inaccuracies in the problem setup, while others are still working through their understanding of the concepts involved.

Contextual Notes

Participants note discrepancies in the problem's parameters, such as the accuracy of time measurements and the implications of bending knees during the landing. There is also a recognition that the problem may not accurately reflect real-life scenarios, prompting discussions about the assumptions made in the calculations.

  • #31
HJ^2 said:
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.
 
Physics news on Phys.org
  • #32
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?
 
  • #33
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.
 
  • #34
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
 
  • #35
HJ^2 said:
So fnet would equal Fg + Fn = m*a
Right. The only unknown there is Fn. Solve.
 
  • #36
haruspex said:
Right. The only unknown there is Fn. Solve.
Fn = (m*a) - Fg

With variables is;

Fn = (74.4 * 6.8411) - 9.8 = 499.177 ?
 
  • #37
HJ^2 said:
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).
 
  • #38
haruspex said:
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
 
  • #39
HJ^2 said:
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?
 
  • #40
haruspex said:
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.
 
  • #41
HJ^2 said:
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!
 
  • #42
haruspex said:
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 :)
 

Similar threads

Replies
7
Views
10K
  • · Replies 38 ·
2
Replies
38
Views
4K
  • · Replies 2 ·
Replies
2
Views
14K
  • · Replies 2 ·
Replies
2
Views
7K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 1 ·
Replies
1
Views
8K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 1 ·
Replies
1
Views
7K
  • · Replies 9 ·
Replies
9
Views
6K