Cable car system : Tension in the pull cable

[Curieuse]

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

F_NET=ma

The Attempt at a Solution

From figure,

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

And T_n = 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?

 

[BvU]

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 ?

 

[Curieuse]

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.
So the free body diagram goes to this,
and so the equation for T_n = n(ma+mg sinθ)
And T₅-T₃= 2(2800)(.80+9.8 sin(35°))
For T_n-T_n-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!

 

[BvU]

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 ?

 

[Curieuse]

No! It has to be .81! Clearly saturated with the problem i am now! :tongue2: