Why Is My Bungee Cord Tension Calculation Incorrect?

  • #1
I_Try_Math
65
17
Homework Statement
A bungee cord is essentially a very long rubber band that can stretch up to four times its unstretched length. However, its spring constant varies over its stretch. Take the length of the cord to be along the x-direction and define the stretch x as the length of the cord l minus its un-stretched length l_0. Suppose a particular bungee cord has a spring constant, for 0 <= x <= 4.88, of k_1 = 204 N/m and for x > 4.88, of k_2 = 111 N/m (Recall that the spring constant is the slope of the force F(x) versus its stretch x.) (a) What is the tension in the cord when the stretch is 16.7 m (the maximum desired for a given jump)?
Relevant Equations
F(x) = kx
(a)

x = 16.7

Since x > 4.88, k = k_2 = 111 N/m.

T = F(16.7) = k_2*x = 111*16.7 = 1853.7 N

The answer key says the correct tension is 2320 N. Any help is appreciated, I'm not seeing where my math is wrong.
 
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  • #2
I_Try_Math said:
Homework Statement: A bungee cord is essentially a very long rubber band that can stretch up to four times its unstretched length. However, its spring constant varies over its stretch. Take the length of the cord to be along the x-direction and define the stretch x as the length of the cord l minus its un-stretched length l_0. Suppose a particular bungee cord has a spring constant, for 0 <= x <= 4.88, of k_1 = 204 N/m and for x > 4.88, of k_2 = 111 N/m (Recall that the spring constant is the slope of the force F(x) versus its stretch x.) (a) What is the tension in the cord when the stretch is 16.7 m (the maximum desired for a given jump)?
Relevant Equations: F(x) = kx

(a)

x = 16.7

Since x > 4.88, k = k_2 = 111 N/m.

T = F(16.7) = k_2*x = 111*16.7 = 1853.7 N

The answer key says the correct tension is 2320 N. Any help is appreciated, I'm not seeing where my math is wrong.
If the constant is k at some point, stretching it by a further distance x requires an additional force kx.
Sketching a graph might help.
 
  • #3
haruspex said:
If the constant is k at some point, stretching it by a further distance x requires an additional force kx.
Sketching a graph might help.
I'm having trouble understanding. So the total tension at a stretch, x > 4.88, is not given by F(x) = k_2*x? I can't see where the extra tension comes from.
 
  • #4
I_Try_Math said:
So the total tension at a stretch, x > 4.88, is not given by F(x) = k_2*x?
Right, it is given by ##\Delta F=k_2\Delta x##.
What is the tension at 4.88m? What is it at 4.89m? It can't be less, right?
 
  • #5
I_Try_Math said:
I'm having trouble understanding. So the total tension at a stretch, x > 4.88, is not given by F(x) = k_2*x? I can't see where the extra tension comes from.
The problem drops a very specific hint:
I_Try_Math said:
(Recall that the spring constant is the slope of the force F(x) versus its stretch x.)
But furthermore the force must be continuous. If it were ##k_1 x## in the first region and ##k_2x## in the second, what would happen to the force at the boundary between regions?Also note that the hint can be written ##F’(x) = k## if you are familiar with calculus.
 
  • #6
Orodruin said:
The problem drops a very specific hint:

But furthermore the force must be continuous. If it were ##k_1 x## in the first region and ##k_2x## in the second, what would happen to the force at the boundary between regions?Also note that the hint can be written ##F’(x) = k## if you are familiar with calculus.
Maybe I should reread the chapter because there must be something I'm missing. This example seems different than all others I've seen so far. First, I'm not sure how it is taken as a given that the spring force must be continuous. I mean if you take a spring and stretch it enough perhaps it would break or its force would drop off dramatically. In addition, I'm struggling to see what the relevance is of the boundary between the two regions because a stretch of 16.7 m is significantly distant from it.

If the spring force is given by F(x) = k_2*x for x > 4.88 then F'(x) = k_2, correct? How does that help solve the problem?
 
  • #7
I_Try_Math said:
Maybe I should reread the chapter because there must be something I'm missing. This example seems different than all others I've seen so far. First, I'm not sure how it is taken as a given that the spring force must be continuous. I mean if you take a spring and stretch it enough perhaps it would break or its force would drop off dramatically. In addition, I'm struggling to see what the relevance is of the boundary between the two regions because a stretch of 16.7 m is significantly distant from it.
Yes, the force can exhibit strange behavior in general. Particularly close to the limits of the material strength. However, physical changes are typically continuous (with some striking exceptions). The formulation of the problem itself also indicates that ##F’(x) = k## is what is intended.

I_Try_Math said:
If the spring force is given by F(x) = k_2*x for x > 4.88 then F'(x) = k_2, correct? How does that help solve the problem?
Yes, but it is not the only solution to ##F’(x) = k_2## and it does not give a continuous behaviour for the force.
 
  • #8
I_Try_Math said:
This example seems different than all others I've seen
The only difference will be that k was a 'constant constant' in previous examples.

The question reminds you:
Recall that the spring constant is the slope of the force F(x) versus its stretch x
and it gives you the values of that slope for two adjacent ranges of x.
So draw a line which starts at the origin and has exactly two positive slopes, the first greater than the second. Note, no other slopes, not even vertical.
Since the y axis is force and the x axis is displacement, the area under it out to a displacement x is the work done to stretch it that far.
 
  • #9
haruspex said:
The only difference will be that k was a 'constant constant' in previous examples.

The question reminds you:
Recall that the spring constant is the slope of the force F(x) versus its stretch x
and it gives you the values of that slope for two adjacent ranges of x.
So draw a line which starts at the origin and has exactly two positive slopes, the first greater than the second. Note, no other slopes, not even vertical.
Since the y axis is force and the x axis is displacement, the area under it out to a displacement x is the work done to stretch it that far.
But if it's asking for tension then I need to find to force (N) of the spring, correct? I'm not sure how work factors in.
IMG_20240201_060748.jpg
 
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  • #10
I_Try_Math said:
But if it's asking for tension then I need to find to force (N) of the spring, correct? I'm not sure how work factors in.
View attachment 339575
Now put a vertical line marking a stretch of 4.88 m(for clarity), and then one marking the total stretch of 16.7 m. What is the value of ##F## at 16.7 m?
 
  • #11
erobz said:
Now put a vertical line marking a stretch of 4.88 m(for clarity), and then one marking the total stretch of 16.7 m. What is the value of ##F## at 16.7 m?
phys_forum_q0.jpg

So this is incorrect?
 
  • #12
I_Try_Math said:
View attachment 339577
So this is incorrect?
Its the correct graph, but your calculation is incorrect.

Can you write the equation of the line beyond ##x = 4.88 \rm{m}##?

What is ##y(x)##

1706799868652.png
 
  • #13
erobz said:
Its the correct graph, but your calculation is incorrect.

Can you write the equation of the line beyond ##x = 4.88 \rm{m}##?
If Hooke's Law applies to bungee cord then the force is F(x) = kx, and in this particular case k=k_2=111? I seem to be going in circles and not sure where I'm going wrong. Is the force supposed to be a negative value?
 
  • #14
I_Try_Math said:
If Hooke's Law applies to bungee cord then the force is F(x) = kx, and in this particular case k=k_2=111? I seem to be going in circles and not sure where I'm going wrong. Is the force supposed to be a negative value?
You are evaluating the value at ##x = 16.7## m incorrectly.

The equation of the red dashed line in the graph beyond 4.88 m is not ## y= k_2 x##

There is a discontinuity in the spring constant. It's not really a "Hookean Spring" to the letter of that definition. You are saying the red dashed line passes through the origin, it clearly doesn't.

1706800749236.png
 
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  • #15
erobz said:
You are evaluating the value at ##x = 16.7## m incorrectly.

The equation of the red dashed line in the graph beyond 4.88 m is not ## y= k_2 x##

There is a discontinuity in the spring constant. It's not really a "Hookean Spring" to the letter of that definition. You are saying the red dashed line passes through the origin, it clearly doesn't.

View attachment 339581
You're exactly right. That was confusing. Just got used to that formula always being in the form k*x I guess. Math isn't my strong suit but thanks a bunch for your help.
 
  • #16
I_Try_Math said:
You're exactly right. That was confusing. Just got used to that formula always being in the form k*x I guess. Math isn't my strong suit but thanks a bunch for your help.
It's not your math so much. The definitions can be subtle. The spring behaves in linear manner over each range, which makes one think "Hooks Law". However, the spring "constant" of this spring is not constant- i.e. not "Hooks Law".
 
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