Tension in cable car system

In summary, the problem involves a cable car system with a maximum permissible mass of 2800kg for each car with occupants. The cars are pulled by a second cable attached to the support tower on each car. The cables are taut and inclined at an angle of 35 degrees. The question asks for the difference in tension between adjacent sections of pull cable when the cars are at maximum mass and being accelerated up the incline at 0.81 m/s². The solution involves using the equation Fnet = ma and finding the difference in tension between car 3 and car 6. The final answer is 143520.54 N.
  • #1
aebrenn
1
0

Homework Statement


Figure 5-60 shows a section of a cable car system. The maximum permissible mass of each car with occupants is 2800kg. 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 an angle theta = 35 degrees. 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/s squared.

the drawing shows three cars in a "section." there is a cable each of the cars is hanging from on along the top and another cable parallel that is the "pull" cable running alone the towers of the cars.


Homework Equations



Fnet = ma

The Attempt at a Solution



I calculated the tension from car 3 and the tension from car 6 (the top car of the second section) and then found the difference between them. Free body diagram for car 3 shows the car with a weight force coming straight down, tension force balancing that up, tension force 35 degrees above the positive x direction and weight force 2mg equal to two cars heading in the direction directly opposite of the tension force (at an angle down). Used trig to find the weight force down at an angle = mg/sin theta

Sums of forces:
Car 3 : sum of fnet = ma

T - 2mg = ma
T - 2(mg/sin theta) = ma
T = ma + 2(mg/sin theta)
T = (2800)(.81) + 2((2800x9.8)/sin 35)
T = 97948.36 N

Car 6 : sum of fnet = ma

T - 5mg = ma
T - 5(mg/sin theta) = ma
T = ma + 5(mg/sin theta)
T = (2800)(.81) + 5((2800x9.8)/sin 35)
T = 241468.9 N

241468.9 - 97948.36 = 143520.54
change in Tension = 143520.54 N
 
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  • #2
The diagram is missing and (IMO) the description is insufficient to properly understand the problem.

Also, there is (at least one) inconsistency: the description of the (missing) diagram says it shows ‘three cars’; but later in the post there is a reference to ‘car 6’.
 

1. What is tension in a cable car system?

Tension in a cable car system refers to the force that is exerted on the cables that support the car. This force is necessary to keep the car stable and prevent it from falling.

2. How is tension created in a cable car system?

Tension is created in a cable car system through the use of a counterweight. The weight of the counterweight creates tension in the cables, which in turn supports the car and allows it to move smoothly.

3. Why is tension important in a cable car system?

Tension is important in a cable car system because it helps to keep the car stable and in motion. Without tension, the car could become unbalanced and potentially cause accidents or malfunctions.

4. How is tension measured in a cable car system?

Tension in a cable car system is typically measured using a tension gauge. This tool measures the amount of force being exerted on the cables and can help ensure that the tension is within a safe and optimal range.

5. Can tension in a cable car system be adjusted?

Yes, tension in a cable car system can be adjusted by changing the weight of the counterweight or by tightening or loosening the cables. This may be necessary if there are changes in weather conditions or if the car is carrying a heavier or lighter load than usual.

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