Cable car system : Tension in the pull cable

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Homework Help Overview

The discussion revolves around a physics problem involving a cable car system, specifically focusing on the tension in the pull cable as the cars are accelerated up an incline. The problem states the maximum mass of the cars and provides an angle of inclination, along with an acceleration value.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore the relationship between tension in adjacent sections of the pull cable, with some attempting to derive equations based on the forces acting on the cars. There is a discussion about whether to consider the difference in tension between specific sections of the cable and the role of acceleration in the calculations.

Discussion Status

The conversation is ongoing, with participants sharing their interpretations of the problem and questioning the clarity of the exercise's wording. Some have offered alternative formulations and equations, while others have expressed uncertainty about the correct approach to take. There is no explicit consensus yet on the best method to solve the problem.

Contextual Notes

Participants note potential confusion in the problem statement regarding the definition of "adjacent sections" and the significance of the provided acceleration value. There is also mention of a desire for reference materials to clarify the problem further.

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Homework Statement



Figure shows a section of a cable car system. The maximum permissible mass of each car with occupants is 2800 kg. The cars, riding on a support cable are pulled by a second cable attached to the support tower on each car. Assume that the cables are taut and inclined at θ= 35°. What is the difference in tension between adjacent sections of pull cable if the cars are at the maximum permissible mass and are being accelerated up the incline at .81 m/s2.

Homework Equations



FNET=ma

The Attempt at a Solution



From figure,

T2 = T1 +Mg sin θ
T1 = Mg sin θ
Thus, T2 = 2 Mg sin θ

And Tn = n Mg sin θ

Thus, T5 - T2 = (5-2) Mg sin θ = 3*9.8*2800*sin(35°)= 47216.8 N

Is this answer correct?
 

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I read the exercise somewhat differently: adjacent sections of pull cable means you have to take e.g. T2 - T1, not T5 -T2 as you seem to be doing. Then: the pull cable is the one that provides the acceleration, so I would expect the acceleration should appear somewhere in the answer, isn't it ?
 
Oh of course! I thought the same about the T2-T1 thing and then saw this
https://www.physicsforums.com/showthread.php?t=291753
and redid it.
But yeah! The acceleration is in the answer, It skipped my mind soo bad when I redid it with the section things and all. :cry:
So the free body diagram goes to this,
and so the equation for Tn = n(ma+mg sinθ)
And T5-T3= 2(2800)(.80+9.8 sin(35°))
For Tn-Tn-1 = 2800(.80+9.8 sin(35°))

So, i guess we could answer both! But also refer to the question,
Figure shows a section of a cable car system. The maximum permissible mass of each car with occupants is 2800 kg. The cars, riding on a support cable are pulled by a second cable attached to the support tower on each car. Assume that the cables are taut and inclined at θ= 35°. What is the difference in tension between adjacent sections of pull cable if the cars are at the maximum permissible mass and are being accelerated up the incline at .81 m/s2.

But then, there's be a chance to get at the implied meaning if someone has a solution manual of some sort to the Fundamentals of Physics- 8th edition extended by Halliday, Resnick and Walker..o have had it in class or something.. I guess there's no better way to know it than that!
 

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I see. This has nothing to do with physics any more. It is clear you completely understand the situation. The formulation of the exercise is confusing, not intentionally, I suppose, but they could have written "tension between adjacent sections of three cars each" to keep it clear...

They mention an acceleration of .81 m/s2. Is there a reason you write 0.80 ?
 
No! It has to be .81! Clearly saturated with the problem i am now! :-p
 

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