What is the tension of the rope connecting two masses in Yellow Creek?

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Homework Help Overview

The problem involves two masses connected by a rope, specifically a carp and a trout, with a focus on determining the tension in the rope while one mass is being lifted with a specified force. The context is rooted in classical mechanics, particularly Newton's laws of motion.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss free body diagrams (FBDs) and the forces acting on each mass. There is uncertainty about the correct setup of equations and whether the system is accelerating or in equilibrium. Some participants question the relevance of the carp in the tension calculation and suggest that the system may be lifted at a constant force.

Discussion Status

There is ongoing exploration of the equations representing the forces on each mass. Some participants have pointed out potential errors in the original equations and have encouraged re-evaluation of the arithmetic involved in calculating tension. Multiple interpretations of the problem setup are being considered.

Contextual Notes

Participants are grappling with the implications of the upward force applied and the gravitational forces acting on both masses. There is a suggestion that the system may be in a state of constant velocity, leading to discussions about the assumptions made regarding acceleration.

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Homework Statement



Jimmy has caught two fish in Yellow Creek. He has tied the line holding the 4.10 kg steelhead trout to the tail of the 1.45 kg carp. To show the fish to a friend, he lifts upward on the carp with a force of 86.7 N. What is the tension of the rope connecting the steel trout and carp?

m1 = 1.45 kg
m2 = 4.1 kg

Homework Equations



F = ma
Basically Newton's three laws

The Attempt at a Solution



I'm not entirely sure if this is correct or not, but I had a hard time drawing the FBDs and wasn't sure where to really begin. For the carp, I came up with this:

m1a = 86.7 - m1g - m2g - T

and the trout:

m2a = T - m2g
 
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I don't think you even need to worry about the carp. If the carp is what is being held, it won't have an effect on the tension. However, your equation for the trout would then be incorrect. You have the tension and the upward force both in the positive direction and the force of gravity in the negative direction. I think you may have to assume that the system is being lifted up at a constant force, so the acceleration of the system would be zero.
 
Welcome to Physics Forums :smile:

sdiddy said:

The Attempt at a Solution



I'm not entirely sure if this is correct or not, but I had a hard time drawing the FBDs and wasn't sure where to really begin. For the carp, I came up with this:

m1a = 86.7 - m1g - m2g - T

and the trout:

m2a = T - m2g

This is pretty close, I see only one problem. Note that -m2g is the force due to gravity acting on the trout; it should not enter into the forces acting on the carp.

The downward "pull" on the carp, due to the trout, is expressed by the tension force -T.

w3390 said:
I think you may have to assume that the system is being lifted up at a constant force, so the acceleration of the system would be zero.
Compare the 86.7 N upward force with the combined weight of the two fish. Do you still think the acceleration is zero?
 
So what you are saying is:

m1a = 86.7 - m1g - T
m2a = T - m2g

Even then, when I solve for tension, I keep getting the wrong answer. The answer I keep getting is 14.7 N. I still don't know what I'm doing wrong.
 
Your equations are correct, but I am getting a different value for the tension. You probably have a simple arithmetic or algebra mistake somewhere.

Try again, and if the answer is still wrong then post your algebra work and we'll try to find just where the error is.
 

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