Find Tension Between Sections of Cable?

[GingerBread27]

Figure 5-58 shows a section of an airplane cable-car system. The maximum permissible mass of each car with occupants is 3000 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 angle = 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 0.81 m/s2?

I have no idea what this question is asking or even how to start? Please help.

 

[HobieDude16]

i had this problem... you just have to find what T1-T2 is, and that's equal to
m(Ax+gsin(theta))
Ax equals accel (.81) g is 9.8 (gravity) m is the mass (3000) and theta is your angle (35) so do that, you should get 1.9x10^4

 

[M98Ranger]

To find the tension between sections of the pull cable, we first need to understand the forces acting on the system. In this scenario, the cars are being pulled up the incline at an acceleration of 0.81 m/s2, which means there must be a force acting on them to cause this acceleration. This force is provided by the pull cable attached to the support tower on each car.

Now, since the cables are taut and inclined at an angle of 35°, we can use trigonometry to break down the forces acting on the cable. The weight of the car and its occupants, which is the maximum permissible mass of 3000 kg, can be broken down into two components: one parallel to the incline (mg sinθ) and one perpendicular to the incline (mg cosθ). The force provided by the pull cable will have to counteract the component of weight parallel to the incline (mg sinθ) and provide an additional force to cause the acceleration (ma).

We can use Newton's second law (F=ma) to calculate the tension in the pull cable: T = mg sinθ + ma. Since the cars are being accelerated at 0.81 m/s2, we can substitute the values and get T = (3000 kg)(9.8 m/s2)(sin35°) + (3000 kg)(0.81 m/s2) = 14,145 N.

Now, to find the difference in tension between adjacent sections of the pull cable, we need to consider the forces acting on the cable at different points. As the cable moves up the incline, the weight of the car and its occupants will decrease (since they are being pulled up), and therefore the tension in the cable will also decrease. However, the force required to cause the acceleration (ma) will remain the same. This means that the difference in tension between adjacent sections of the pull cable will be equal to the force required for acceleration (ma).

Therefore, the difference in tension between adjacent sections of the pull cable will be 3000 kg x 0.81 m/s2 = 2430 N. This means that the tension in the cable will decrease by 2430 N as we move up the incline.