1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

A 1130-kg car is held in place by a light cable on a frictionless ramp

  1. Dec 16, 2013 #1
    1. The problem statement, all variables and given/known data

    A 1130-kg car is held in place by a light cable on a very smooth (frictionless) ramp. The cable makes an angle of 31.0° above the surface of the ramp, and the ramp itself rises at 25.0° above the horizontal.

    a). Find the tension in the cable.
    b). How hard does the surface of the ramp push on the car?

    2. Relevant equations

    Newton's Laws of Motion

    3. The attempt at a solution

    From my understanding, there are three arrows, one showing the weight of the car pointing straight downwards through the ramp, another is the normal force arrow drawn perpendicular to the surface of the ramp, and the third is the cable. The problem is, I don't understand where T=mgsin25 came from? How do I know which angle to use? Also, if this is the case, why is there a component of the car's weight parallel to the surface of the ramp? Then what is T?
     
  2. jcsd
  3. Dec 16, 2013 #2

    CAF123

    User Avatar
    Gold Member

    That equation is just the force balance down the slope. What is the component of the weight down the slope?
     
  4. Dec 16, 2013 #3
    I'm not sure, do you mean the component perpendicular to the surface of the ramp that goes downwards, or is there a weight component parallel to the surface of the ramp?
     
  5. Dec 16, 2013 #4
    There is a weight component parallel to the surface. If there was no weight component in the opposite direction of the tension, the car would be pulled by the cable.
     
  6. Dec 16, 2013 #5
    Try drawing the system at an angle, so that the surface of the ramp is flat, horizontal. Then apply your force diagram, with x being left/right and y being up/down. Just don't forget that the force of gravity is now down at an angle to the y axis, equal to the angle of inclination of the ramp.

    Then go from there!
     
  7. Dec 16, 2013 #6
    Oh ok! So the parallel component of the weight equals the horizontal component of the tension force in the cable? Would I then use this horizontal component to find the vertical component of T and then use basic pythagorean theorem to find the tension in the cable?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: A 1130-kg car is held in place by a light cable on a frictionless ramp
  1. Car Ramping (Replies: 8)

  2. Frictionless ramp (Replies: 1)

Loading...