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

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Homework Help Overview

The problem involves 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. The discussion centers around understanding the forces acting on the car, particularly the tension in the cable and the normal force exerted by the ramp.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants are discussing the components of forces acting on the car, including weight and tension. Questions arise about the derivation of the equation T=mgsin25 and the appropriate angles to consider. There is also confusion regarding the components of weight acting parallel and perpendicular to the ramp's surface.

Discussion Status

The discussion is ongoing, with participants exploring different interpretations of the forces involved. Some guidance has been offered regarding drawing the system and applying force diagrams, but there is no explicit consensus on the approach to take.

Contextual Notes

Participants are navigating the complexities of force components on an inclined plane, with specific attention to the angles involved and the implications of a frictionless surface. There is an emphasis on understanding the relationships between the forces rather than arriving at a final solution.

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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