Which solution is correct (Tension)

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The discussion revolves around solving a physics problem involving the tension in cables of a cable car system. Two different methods to calculate the tension differences yield conflicting results. The first method fails to account for the normal force acting on the cable car, while the second method correctly eliminates this force by aligning the axes along the cable's angle. The correct approach shows that the tension difference is influenced by both acceleration and gravitational components. Ultimately, the second method is validated as the accurate solution.
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Homework Statement


A section of a cable car system, showing three cars. They are being pulled by drag cables at 35 degrees above the horizontal. Each car weighs 2800 kg. The are all being accelerated at the same rate of 0.81 m/s/s. The problem asks for the difference between the tensions in the two cables that connect to the middle car.

Homework Equations


F=ma

The Attempt at a Solution



Solution 1:

I started by writing the force equations in X and Y:

Fy = ma sin 35 = T sin 35 - mg
Fx = ma cos 35 = T2 cos 35 - T1 cos 35

Obviously, I don't need the Fy equation at all, since the cos 35 drops out and I have:

ma = T2- T1

Then I thought to try solving it by having the X axis along the 35 degree line of the cables. This gives me:

Fx = ma = T1 - mg sin 35 - T2

with T1 - T2 = m(a + g sin 35)

They both give very different results, and both seem reasonable to me. I have no idea which one is correct or why. If someone could help me out, I would appreciate it.

Thanks,
Sheldon
 
Last edited:
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Hi SheldonG,

This one is correct:

SheldonG said:
Then I thought to try solving it by having the X axis along the 35 degree line of the cables. This gives me:

Fx = ma = T1 - mg sin 35 - T2

with T1 - T2 = m(a + g sin 35)


The reason the first one is not correct is that there is a normal force on the cable car, perpendicular to the cable. With your axes chosen as they were in method 1, the components of this force would need to be included. But for method 2, the normal force had not component along the cable, so it was not needed.
 
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