A section of an alpne cable-car system has a maximum permissible mass of each car with occupants in 2800kg. The cars riding on a support calbe ar pulled by a second cable attached to each pylon(support tower); assume the cables are straight. What is the difference in tension b/w adjacent sections of pull cable if the cars are at the maxiumum permissible mass and are being accelerated up the 35 degree incline at .81 m/s^2?
What is the equation that would be used to solve this problem?
2046n
To solve this problem, we need to consider the force acting on each car and the tension in the pull cable.
The force acting on each car can be calculated using Newton's second law of motion, which states that force (F) = mass (m) x acceleration (a). In this case, the mass of each car (including occupants) is given as 2800 kg, and the acceleration is given as 0.81 m/s^2. Therefore, the force acting on each car is:
F = m x a
F = 2800 kg x 0.81 m/s^2
F = 2268 N
Since the cable is attached to the pylon (support tower) and pulls the cars, the tension in the cable must be equal to or greater than this force. Let's assume there are two consecutive cars connected to the pull cable.
To find the difference in tension between adjacent sections of the pull cable, we need to consider the forces acting on each section. At the higher section, let's denote the tension as T1, and at the lower section, let's denote the tension as T2.
Since the cars are being accelerated up a 35° incline, the component of weight parallel to the incline (mg sin θ) contributes to the force pulling the cars upwards. The tension in the cable must overcome this force in addition to providing the necessary force to accelerate the cars. The weight of each car (mg) is given by:
mg = 2800 kg x 9.8 m/s^2
≈ 27440 N
Therefore, the force due to the weight of each car parallel to the incline is:
Force_parallel = mg sin θ
= 27440 N x sin 35°
≈ 15686 N
To find the difference in tension, we need to consider the net force acting on each car. At the higher section, the net force will be the sum of the force due to the tension (T1) and the force due to the weight parallel to the incline. At the lower section, the net force will be the sum of the force due to the tension (T2) and opposite of the force due to the weight parallel to the incline, as the direction of the force will be opposite.
Therefore, we have the following equations:
Net force at the higher section = T1 - Force_parallel = m x a
Net force at the lower section = T2 + Force_parallel = m x a
Simplifying these equations, we can solve for the difference in tension (ΔT = T1 - T2):
ΔT = T1 - T2 = 2 x Force_parallel
The equation used to solve this problem is ΔT = 2 x Force_parallel, where ΔT represents the difference in tension between adjacent sections of the pull cable and Force_parallel represents the force due to the weight of each car parallel to the incline.