Solve the Bungee Physics Problem: Net Force and Height Calculation"

  • Thread starter Thread starter tandoorichicken
  • Start date Start date
  • Tags Tags
    Bungee Physics
Click For Summary

Homework Help Overview

The problem involves a bungee jump scenario where a jumper, Barbara, falls from a height of 30 meters, attached to a bungee cord with specific properties. The goal is to determine the height at which the net force acting on her becomes zero, considering the forces of gravity and the spring-like behavior of the bungee cord.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the relationship between net force, gravity, and the tension in the bungee cord, questioning why the net force is not zero at the lowest point of the fall. Some explore the dynamics of free fall and the subsequent stretching of the bungee cord.

Discussion Status

The discussion includes various interpretations of the forces at play, with some participants offering mathematical reasoning and equations to describe the motion. There is an ongoing exploration of the conditions under which the net force is zero, with no explicit consensus reached on the final outcome.

Contextual Notes

Participants are working under the assumption that the bungee cord behaves like an ideal spring and are considering the effects of momentum and acceleration during the fall. The problem constraints include the specific measurements of height, weight, and spring constant provided in the original problem statement.

tandoorichicken
Messages
245
Reaction score
0
The problem is as follows:

"Bungee Barbara jumps off a crane platform 30m above the ground. The bungee cord to which she is hooked has an unstretched length of 9.0m We assume the bungee cord behaves like an ideal spring with spring constant k=100 N/m. Barbara has a weight of 600 N. At what height will the net force acting on Barbara be zero?"

I don't understand why this point is not the lowest point in the fall. I mean, wouldn't the net force be zero when the force of gravity and the tension in the spring are balanced, and wouldn't that be when Barbara momentarily comes to rest at the lowest point?
 
Physics news on Phys.org
At the lowest point in the fall, the net force on her is pulling her up.

While she's falling, she builds up momentum that carries her past the equilibrium point.
 
So, how do I find out exactly what the net force is acting on Barbara at the lowest point in the fall?
 
At her lowest point I think her ΣFy = 0, because there is a surface holding her up with the same force she is pulling the Earth with.
 
The net force will be the force from gravity plus the force from the bungee.
 
At her lowest point I think her ÓFy = 0, because there is a surface holding her up with the same force she is pulling the Earth with.
No, when the total force is 0, there is no acceleration so she continues downward (remind me not to bungee jump with you! :smile:). That had already been said. The lowest point comes when her velocity is 0, not acceleration.

If I am reading the problem correctly, Barbara will first "free fall" until she reaches 9 m below the platform. Then the cord will start stretching. For those first 9 m her acceleration is -9.8 m/s2 so her speed after t seconds is -9.8t and her height above the ground will be 30- 4.9 t2. That will be 9 m when 30- 4.9t2= 30-9= 21 or t2= 9/4.9= 1.83 seconds so t= 1.35 seconds At that time her speed will be -9.8(1.35)= -13.3 m/s.

After that first 9 m the cord will act with force 100(21-x) Newtons (21- x is the stretch of the cord in m when the Barbara's height is x meters). The total force on Barbara will be -600+ 100(21-x)= 1500- 100x Newtons. Her mass is 600/9.8= 61.2 kg so her acceleration is
a= (1500- 100x)/61.2= 24.5- 1.63x.
We have the differential equation x"= 24.5- 1.63x or x"+ 1.63x= 24.5

The General solution to that is x(t)= C1Cos(1.28t)+ C2Sin(1.28t)+ 14.8. Taking t= 0 at the point at which the bungee cord starts stretching, x(0)= 21 m and x'(0)= -13.3 m/s.
x(0)= C1+ 14.8= 21 so C1= 6.2.
x'(t)= -1.28C1Sin(1.28t)+ 1.28C2Cos(1.28t)
so x'(0)= 1.28 C2= -13.3 so C2= -10.4.

That is: x(t)= 6.2 Cos(1.28t)- 10.4 Sin(1.28t)+ 14.8 meters.
her speed at time t is x'(t)= -7.94 Sin(1.28t)- -13.3 Cos(1.28t).
Barbara's lowest point will come when her speed is 0: when
-7.94 Sin(1.28t)- 13.3 Cos(1.28t)= 0 or Tan(1.28t)= -13.3/7.94= -1.68.
That means 1.28t= 2.1 or t= 1.6 seconds.
Putting that into x(t), x(1.6)= 6.2Cos(2.1)- 10.4Sin(2.1)+ 14.8= 2.7 meters.
 

Similar threads

Replies
44
Views
8K
Replies
8
Views
3K
  • · Replies 1 ·
Replies
1
Views
6K
  • · Replies 16 ·
Replies
16
Views
5K
Replies
6
Views
4K
  • · Replies 11 ·
Replies
11
Views
4K
  • · Replies 9 ·
Replies
9
Views
4K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 9 ·
Replies
9
Views
7K