If a chicken & rooster are fighting over a spaghetti string

[Ocata]

Homework Statement

A chicken and rooster are fighting over a 1 kg string of spaghetti. The chicken pulls with 999 N and Rooster pulls with 1000 N. The net force is 1 N, so relative to the ground, will the spaghetti string accelerate at 1m/s^2 (considering both the chicken and rooster continue to grasp the spaghetti string and friction on the ground is equated into the net force of the spaghetti string)?

Homework Equations

F=ma

The Attempt at a Solution

Yes. The string will accelerate at 1m/s^2


1. Homework Statement
Continuing, this unbalanced pull between the chicken and rooster continues for 5 seconds. What is the velocity of the spaghetti string relative to the ground after 5 seconds?

2. Homework Equations
vf = at + v0

The Attempt at a Solution

5m/s^2

Homework Statement

After 5 seconds, the chicken increases its pull to 1000 N. The net force on the spaghetti string is 0. What is the velocity of the spaghetti string relative to the ground at this time? 0 m/s^2 or 1m/s^2 or 5m/s^2?

The Attempt at a Solution

5m/s^2

 

[gneill]

You haven't asked a question. You haven't presented any work or argument that can be critiqued, or pointed out areas where you are having doubts. So what were you hoping to happen as a result of your post? I should point out that we won't just confirm answers that are unsupported by explicit work.

 

[Ocata]

After 5 seconds of pulling with unequal forces, when the chicken and rooster both begin to pull the spaghetti string with equal forces of 1000N, they do so for another 5 seconds.

So, from 0-5 seconds, the net force on the spaghetti is 1N, from 5-10 seconds, the net force is 0 N.

The rooster gets tired at 10 seconds and is now the one pulling at 999N while the chicken continues pulling at 1000N for 10 seconds.

What is the rate of acceleration at 3, 7, 12, 15, 18, 20 seconds? And in what direction is the acceleration occurring at each time?
F/m = a
At 3 seconds: (1000 – 999)/1kg = 1N/1kg = 1m/s^2 to the right
At 7 seconds: (1000-1000)/1kg = 0K/1kg = 0m/s^2 to the right
At 12 seconds: (l999-1000l)/1kg = 1N/1kg = 1m/s^2 to the left
At 15 seconds: (l999-1000)/kg = 1N/1kg = 1m/s^2 to the left
At 18 seconds: (l999-1000l)/kg = 1N/1kg = 1m/s^2 to the left
At 20 seconds: (l999-1000l)/kg = 1N/1kg = 1m/s^2 to the left

At what speed (velocity) is the spaghetti traveling at 3, 7, 12, 15, 18, and 20 seconds?
Vf = at + v0 = (f/m)t + v0

At 3 seconds: Vf = (1N/1kg)(3)
At 7 seconds: Vf = (0N/1kg)(7-5)+5
At 12 seconds: Vf = (-1N/1kg)(12-10)+5=3
At 15 seconds: Vf = (-1kg/1kg)(15-10)+5=0
At 18 seconds: Vf = (-1N/1kg)(18-10)+5=-3
At 20 seconds: Vf = (-1N/1kg)(20-10)+5=-5The spaghetti package says that at 1000N the spaghetti string will emit loud popping sounds near the area of the string where most of the tension is felt (if there is an unequal distribution of tension), then it will turn blue at 1001N, and at 1002, the spaghetti string will break.

What is the tension in the spaghetti string at the times 3,7,12,15 18,20

At 3 seconds the tension is 1N
At 7 seconds the tension is 0N
At 12 seconds the tension is 1N
At 15 seconds the tension is 1N
At 18 seconds the tension is 1N
At 20 seconds the tension is 1N

At 3,7,12,15,18,20 seconds:

What force does the right side of the rope feel due to the rooster?
1000N, 1000N, 999N, 999N, 999N, 999N

What force does the rooster feel by the rope?
1000N, 1000N, 999N, 999N, 999N, 999N

What force does the left side of the rope feel due to the pull of the chicken?
999N, 1000N, 1000N, 1000N, 1000N, 1000N

What force does the chicken feel due to the pull of the rope?
999N, 1000N, 1000N, 1000N, 1000N, 1000N

At what time interval(s) will the spaghetti string most likely emit loud popping sounds and why?

During the first 5 seconds.
Because even though the net tension on the rope is 1N, at least some portion of the rope is experiencing a force of 1000N due to the rooster. I believe the pops will emit from the side closest to the rooster because the force of 1000N are coming from the rooster.
Then it will continue to pop during the 5-10 second interval because the rope is experiencing 1000N from both the rooster and the chicken. I think the pops will emit evenly across the string because the 1000N is coming from both the rooster and the chicken.

------- This is where I get confused.

At 0-5 seconds, will the popping sounds emit evenly because even though unequal forces are acting on the rope, the entire rope experiences 1000N? This doesn't seem intuitive given that the string is apparently pulling on the chicken on the left side with a force of 999N.
So I'm guessing the entire rope is experiencing a varying level of tension from 999N on the right to 1000N on the left. Which assumption is correct? Or are both assumptions incorrect?

Thanks

 

[Ocata]

You haven't asked a question. You haven't presented any work or argument that can be critiqued, or pointed out areas where you are having doubts. So what were you hoping to happen as a result of your post? I should point out that we won't just confirm answers that are unsupported by explicit work.

Hi Gneill,

I created this scenario yesterday randomly and assigned myself some questions about the scenario, which I believe would help me to understand this concept better. Hopefully the questions (and answers) are organized enough to be considered as homework or coursework - regardless of from who it was assigned.

Hope this is acceptable.

Best Regards,
Ocata

 

[gneill]

Hi Gneill,

I created this scenario yesterday randomly and assigned myself some questions about the scenario, which I believe would help me to understand this concept better. Hopefully the questions (and answers) are organized enough to be considered as homework or coursework - regardless of from who it was assigned.

Hope this is acceptable.

Best Regards,
Ocata

Hi Ocata. It's more than acceptable, in fact it's a great way to investigate a subject and improve your understanding

 

[gneill]

What is the tension in the spaghetti string at the times 3,7,12,15 18,20

At 3 seconds the tension is 1N
At 7 seconds the tension is 0N
At 12 seconds the tension is 1N
At 15 seconds the tension is 1N
At 18 seconds the tension is 1N
At 20 seconds the tension is 1N

You'll want to rethink this. The tension is not the same thing as net force on the system of interest (the string). Because the string is not massless and is not a point mass, you can expect the tension to vary all along its length if the system is accelerating (hint: think about the Equivalence Principle), matching the boundary conditions at the ends where the external forces are applied.

 

[Ocata]

You'll want to rethink this. The tension is not the same thing as net force on the system of interest (the string). Because the string is not massless and is not a point mass, you can expect the tension to vary all along its length if the system is accelerating (hint: think about the Equivalence Principle), matching the boundary conditions at the ends where the external forces are applied.

I looked up equivalency principle and it seems to have to do with general relativity, which I'm not sure I'm there yet.

From your hint: the tension is not the net force- I'm realizing I've probably been confusing "tension in the string" with "tension of the string on connected mass." I think the confusion is coming from the idea that, in terms of a string, tension, in plane English, to me, sounds like "tightness of a string." And so when I hear about measuring the tension between two objects, I think I'm supposed to be measuring some value of tightness of a string as opposed to measuring the interaction between the string and the connected mass(es).

Perhaps tension could mean both things? At every point, from the rope on to the rooster, opposing forces within the rope itself, and the rope on to the chicken.

So would it be correct to assume that when the question of finding tension between two masses is asked, the question is implying: What is the force that the string has on the mass at the point the string is in contact with the mass. And in terms of which mass (the left or right side of the string), the question most likely wants to know about the force on the mass that is being pulled in the direction of motion ( as opposed to the side where the mass is pulling the string in the direction of motion.

For instance, within the first 5 seconds, the rooster on the right hand side is applying the greater force and causing an acceleration to the right, so if I'm asked to find the tension between the two masses, would I then evaluate the force of the string/ spaghetti on the chicken? Or would the question need to be more specific? Like, what is the tension of the rope on the rooster or what I the tension that the rope had on the chicken? Or what is the tension on the string 3/4 way toward the rooster?

Or maybe it depends on how the problem is set up? For instance, maybe being asked to find the tension between two masses in which one of the masses is being pulled by another string is inherently different than being asked to find the tension in which two masses are pulling on a string on their own (without another string pulling on the opposite side of one of the masses).

So my next attempt at solving this problem is

At 3 seconds:
The tension is 100N nearest the rooster and 99N nearest the chicken. And then (100N+99N)/2 = 99.5 right in the middle??

 

[gneill]

Tension is property that is internal to a body, so you shouldn't see a phrases like "tension of the string on connected mass" or "tension between two masses" in a properly formed problem. Tension does not have a single direction as forces do. At the boundary of an object where it meets another the tension manifests as a force which has a specific direction and can be represented by a vector.

Forces exerted at the ends of a rope that pull it taught result in tension internal to the rope. At the rope ends, the tension manifests as a force countering the externally applied force forming the Newton's 3rd Law force pair).

The tension will vary along the length of a massive object undergoing acceleration. Invoking the equivalency principle, it is analogous to the way the tension in a hanging chain varies along its length due to gravity. At the bottom of the chain the tension is zero because there is no chain below it pulling it down. At the top of the chain where it's suspended the entire weight of the chain is pulling down.

For your noodle, if one end is being pulled with a force of 100 N and the other with 99 N, the tension proximate to the ends must be 100 N and 99 N respectively. The forces that are manifested by the tensions at the ends must similarly be 100 N and 99 N, and are directed inwards, balancing the externally applied forces which are directed outwards. An imbalance of the applied forces of 1 N means there's a net external force of 1 N accelerating the system as a whole. The equivalence principle then accounts for the change in tension along the length of the noodle. Since in this example the noodle is considered to be of uniform density, the tension must vary uniformly as well. So your prediction of the tension in the center of the noodle being the average of the forces applied at the ends is correct.

 

[Ocata]

Gneill, Thank you for your advice and guidance.

 

[gneill]

Gneill, Thank you for your advice and guidance.

You're welcome