Figure 5 57 shows a section of a cable car system the...

Figure $5-57$ shows a section of a cable-car system. The maximum permissible mass of each car with occupants is 2800 $\mathrm{kg}$ . 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 $\theta=35^{\circ} .$ 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 $\mathrm{m} / \mathrm{s}^{2} ?$

Transcript

00:01 Hi there, so for this problem we have a cable car system as is shown in this figure.

00:08 So the maximum permissible mass of each car with occupants is of the value of the mass is 2 ,800 kilograms.

00:24 We also have that the angle that the cables are taught and incline at an angle of 35 degrees.

00:41 And what we need to obtain is the difference in tension between adjacent sections of pull cable if the cars are the maximum permissible mass and are being accelerated up the incline, an acceleration of 0 .81 meters per second.

01:05 So as the problem sets, we need to find the difference between the tensions.

01:11 So we are going to call dot of the temptions, tension one and the other tension two.

01:16 But to do that, we need to draw all of the forces that are adding on this cable car.

01:24 So i'm going to draw the s -adz -its parallel to the forces of the tension.

01:34 So this is the y -axis.

01:38 We have the car in here.

01:41 Just let me draw some.

01:44 Similar to the figure.

01:49 So we have something like this.

01:52 So the forces that are acting on this core cable are attention one and attention two.

02:08 And also we have the length and the weight of the cable card that we just call w and the angle that corresponds to this angle in here that the weight makes with at the vertical.

02:27 So we apply newton's second law to sum all of the forces...

Question

Please give Ace some feedback