Solving Tension Question for Figure 5-58: Max Mass, Angle & Acceleration

  • Thread starter Thread starter greenman100
  • Start date Start date
  • Tags Tags
    Tension
AI Thread Summary
The discussion revolves around solving a physics problem related to an airplane cable-car system, specifically calculating the difference in tension between adjacent sections of pull cable. The maximum mass of each car is 3000 kg, and they are being accelerated up an incline at 0.78 m/s² with the cable inclined at 35°. Participants emphasize applying Newton's 2nd Law to analyze the forces acting on the cars, considering both gravitational and acceleration components. The tension formula discussed is T = mg + ma, where m is mass, g is gravitational acceleration, and a is the acceleration. The problem's wording is noted as challenging, but the underlying physics principles are understood.
greenman100
Messages
9
Reaction score
0
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 0.78 m/s^2?


http://timelmore.ipupdater.com/05_55.gif



I don't even know how to get started on this, and neither does anyone else taking the class. The professor is out of town, does anyone have any thoughts or advice? Equations and theory would be nice- I'm not looking for someone to do my homework for me, as I know how to do othre tension problems, I just don't really know what tehy're asking for and how the angle plays into it.

Thanks,
Tim
 
Last edited by a moderator:
Physics news on Phys.org
Here's how I read the problem. The support cable supports the weight of the cars just as a frictionless incline plane would support a mass resting on it: it supplies a normal force. The pull cables, on the other hand, provide a force parallel to the cables that prevents the cars from sliding down and imparts some acceleration to the cars.

So identify all the forces on each car and apply Newton's 2nd Law for components parallel to the cable.
 
Doc Al said:
Here's how I read the problem. The support cable supports the weight of the cars just as a frictionless incline plane would support a mass resting on it: it supplies a normal force. The pull cables, on the other hand, provide a force parallel to the cables that prevents the cars from sliding down and imparts some acceleration to the cars.

So identify all the forces on each car and apply Newton's 2nd Law for components parallel to the cable.


that's what I thought

T=mg+ma

so, (3000*sin(35)*9.81)+(3000*.78)

I could do it, the problem was just worded tough
 
Thread 'Minimum mass of a block'
Here we know that if block B is going to move up or just be at the verge of moving up ##Mg \sin \theta ## will act downwards and maximum static friction will act downwards ## \mu Mg \cos \theta ## Now what im confused by is how will we know " how quickly" block B reaches its maximum static friction value without any numbers, the suggested solution says that when block A is at its maximum extension, then block B will start to move up but with a certain set of values couldn't block A reach...
TL;DR Summary: Find Electric field due to charges between 2 parallel infinite planes using Gauss law at any point Here's the diagram. We have a uniform p (rho) density of charges between 2 infinite planes in the cartesian coordinates system. I used a cube of thickness a that spans from z=-a/2 to z=a/2 as a Gaussian surface, each side of the cube has area A. I know that the field depends only on z since there is translational invariance in x and y directions because the planes are...
Thread 'Calculation of Tensile Forces in Piston-Type Water-Lifting Devices at Elevated Locations'
Figure 1 Overall Structure Diagram Figure 2: Top view of the piston when it is cylindrical A circular opening is created at a height of 5 meters above the water surface. Inside this opening is a sleeve-type piston with a cross-sectional area of 1 square meter. The piston is pulled to the right at a constant speed. The pulling force is(Figure 2): F = ρshg = 1000 × 1 × 5 × 10 = 50,000 N. Figure 3: Modifying the structure to incorporate a fixed internal piston When I modify the piston...
Back
Top