Cable car system : Tension in the pull cable

AI Thread Summary
The discussion revolves around calculating the tension difference in a cable car system with a maximum mass of 2800 kg per car, inclined at 35° and accelerating at 0.81 m/s². Participants clarify the correct approach to find the tension difference between adjacent sections of the pull cable, emphasizing the need to include acceleration in the calculations. There is confusion regarding whether to calculate T2 - T1 or T5 - T2, with a consensus emerging that adjacent sections should be considered. The importance of accurately interpreting the problem statement is highlighted, as it could lead to different solutions. Ultimately, the participants agree on the necessity of precise values and clear communication in physics problems.
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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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