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

In summary, the 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. The tension in the cable is mgsin(31.0°-25.0°), and the surface of the ramp pushes on the car with a force of T=mgsin(31.0°-25.0°).
  • #1
mirs
18
0

Homework Statement



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?

Homework Equations



Newton's Laws of Motion

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?
 
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  • #2
mirs said:
The problem is, I don't understand where T=mgsin25 came from? How do I know which angle to use?
That equation is just the force balance down the slope. What is the component of the weight down the slope?
 
  • #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?
 
  • #4
mirs said:
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?

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.
 
  • #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!
 
  • #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?
 

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

1. What is the weight of the car?

The weight of the car is 1130 kilograms.

2. How is the car being held in place?

The car is being held in place by a light cable.

3. What is the purpose of the cable?

The cable is used to prevent the car from rolling down the ramp.

4. How does the frictionless ramp affect the car?

The frictionless ramp reduces the resistance on the car, allowing it to move more easily.

5. What is the significance of the weight and cable in this scenario?

The weight and cable demonstrate the balance of forces acting on the car, as the cable counteracts the force of gravity pulling the car down the ramp.

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