What Is the Tension in San Francisco's Cable Car System?

  • Thread starter Thread starter accobos
  • Start date Start date
  • Tags Tags
    Cable Tension
AI Thread Summary
San Francisco's cable car system operates by pulling cars along tracks with an underground steel cable that moves at 9.5 mph, driven by motors in a central power station. The cable can stretch up to 100 feet over its lifetime, and a tensioning pulley system maintains constant tension. A 2000 kg block is connected to the tensioning pulley, which is crucial for understanding the forces at play. The discussion emphasizes that the system is not accelerating, indicating that the net force equals zero. To analyze the tension in the cable, it is suggested to treat the cart and pulley as a single system and consider the forces acting on it.
accobos
Messages
2
Reaction score
0
The cable cars in San Francisco are pulled along their tracks by an underground steel cable that moves along at 9.5 mph. The cable is driven by large motors at a central power station and extends, via an intricate pulley arrangement, for several miles beneath the city streets. The length of a cable stretches by up to 100 ft during its lifetime. To keep the tension constant, the cable passes around a 1.5-m-diameter "tensioning pulley" that rolls back and forth on rails, as shown in the figure. A 2000 kg block is attached to the tensioning pulley's cart, via a rope and pulley, and is suspended in a deep hole.What is the tension in the cable car's cable?T_{b on a}=T_{a on b}

I just don't even know where to start. I know that it's not accelerating, so Net Force=0 but I still can't figure out where to go from here. If I could just get pointed in the right direction, that'd be great! Thanks so very much!
 

Attachments

  • knight_Figure_08_14.jpg
    knight_Figure_08_14.jpg
    6.2 KB · Views: 812
Physics news on Phys.org
Hint: Treat the cart and pulley as a single system. What forces act on it?
 
The tension of the hanging mass and the normal force of the rail on the cart, and the tension from the rope in the pulley?
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top