Tension in cable car system

by aebrenn
Tags: cable, tension
aebrenn is offline
Feb11-09, 09:15 PM
P: 1
1. The problem statement, all variables and given/known data
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.

2. Relevant equations

Fnet = ma

3. 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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