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

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SUMMARY

The discussion revolves around a physics problem involving a 1130-kg car held in place by a cable on a frictionless ramp inclined at 25.0° to the horizontal, with the cable making an angle of 31.0° above the ramp. Participants clarify the derivation of the equation T = mg sin(25°), which represents the force balance down the slope, and the importance of understanding the weight components acting on the car. The tension in the cable is determined by analyzing the forces acting parallel and perpendicular to the ramp's surface.

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  • Understanding of Newton's Laws of Motion
  • Knowledge of vector components in physics
  • Familiarity with trigonometric functions and their applications in force analysis
  • Ability to construct free-body diagrams
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mirs
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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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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?
 
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?
 
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.
 
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!
 
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?
 

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