How Is Average Force Calculated in Ball Bounce Dynamics?

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Homework Help Overview

The discussion revolves around calculating average force in the context of ball bounce dynamics and analyzing forces in a physics problem involving a leg and cast system. The original poster presents a scenario involving a ball that rebounds after falling from a height, seeking to understand the average force exerted during the contact with the ground.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to calculate average force using velocity changes and time of contact but expresses confusion over the resulting value. Some participants suggest checking the signs of the velocities to ensure correct calculations.
  • Another participant provides guidance on analyzing vertical and horizontal components of forces in a different problem involving a leg and cast, indicating a need for further exploration of the problem setup.

Discussion Status

The discussion is ongoing, with participants providing hints and suggestions for further analysis. There is no explicit consensus, but some guidance has been offered regarding the need to consider both vertical and horizontal components in the second problem.

Contextual Notes

Participants note potential issues with the original poster's calculations and assumptions, particularly regarding the signs of velocities. There is also mention of the need for visual aids in the second problem, which may not be readily available.

perfect_piccolo
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Homework Statement



After falling from rest at a height of 29.8 m, a 0.596 kg ball rebounds upward, reaching a height of 23.9 m. If the contact between ball and ground lasted 1.68 ms, what average force was exerted on the ball?


Homework Equations





The Attempt at a Solution



so I figured out that the final velocity before the ball first hits the ground is 24.17 m/s, and the velocity when the ball begins to bounce back up its inital velocity is 21.64 m/s. If the a= vf-vi / t, that gives me 21.64 - 24.17 which is -2.53 / 0.00168 which equals -1505.95, which I think is totally unreasonable...where am I going wrong?
 
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Careful with signs. Downward velocity should be negative. That will give the average acceleration; you still have to calculate the force of ground on ball.
 
ok thanks got it!
 
All right new question :D

The leg and cast in the figure below weigh 232 N (w1). (Assume w3 = 100 N and q = 39.2o). (See pic)


So I know that Ft3 = Fg3 = 100N

[tex]\Sigma[/tex]Fy2 = m2ay2
Ftyx + Fty3 - Fg1 = 0
Ft2 sin [tex]\alpha[/tex] + Ft3sin[tex]\vartheta[/tex] - Fg1 = 0
Ft2 sin [tex]\alpha[/tex] +100sin39.2 - 232 N = 0
Ft2 sin [tex]\alpha[/tex] = 168.797

But now I'm stuck...I would like to solve for Fty2, because I have found that Ftx2 is 77.49, but I'm not sure how to solve fot Fty2 with the information I have
 

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It looks like you analyzed the vertical components, so now analyze the horizontal components.
 
perfect_piccolo said:
All right new question :D

The leg and cast in the figure below weigh 232 N (w1). (Assume w3 = 100 N and q = 39.2o). (See pic)


So I know that Ft3 = Fg3 = 100N

[tex]\Sigma[/tex]Fy2 = m2ay2
Ftyx + Fty3 - Fg1 = 0
Ft2 sin [tex]\alpha[/tex] + Ft3sin[tex]\vartheta[/tex] - Fg1 = 0
Ft2 sin [tex]\alpha[/tex] +100sin39.2 - 232 N = 0
Ft2 sin [tex]\alpha[/tex] = 168.797

But now I'm stuck...I would like to solve for Fty2, because I have found that Ftx2 is 77.49, but I'm not sure how to solve fot Fty2 with the information I have



I would start a new thread for this, it will get much better traffic. Also, if you have an image host like Photobucket, I would upload though that. The PF image hosting has to be approved by a moderator first...which could take all day as they have a lot of them to deal with.

Casey

Edit: But it looks like Doc Al was just waiting to prove me wrong:rolleyes:
 

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