Friction force of a car rolling down an incline at constant velocity.

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Homework Help Overview

The problem involves analyzing the forces acting on a car coasting down an incline at a constant velocity, specifically focusing on the resistive forces such as friction and air resistance. The scenario is set on a stretch of road where the elevation decreases by a certain height over a specified horizontal distance.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to resolve the forces acting on the car by breaking them down into components and using trigonometric relationships. They express confusion regarding their diagram and the correct application of trigonometric functions.

Discussion Status

Participants are actively engaging in clarifying the original poster's understanding of the problem setup and the trigonometric relationships involved. Some guidance has been provided regarding the correct interpretation of the triangle formed by the incline.

Contextual Notes

There appears to be a misunderstanding regarding the geometric representation of the incline and the corresponding trigonometric functions to be used. The original poster is encouraged to reconsider their diagram and the definitions of the sides of the triangle.

robbondo
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Homework Statement


At a certain part of your drive from home to school, your car (mass m) will coast in neutral at a constant speed of v if there is no wind. Examination of a topographical map shows that for this stretch of straight highway the elevation decreases an amount h for each segment of road of length L. What is the total resistive force (friction plus air resistance) that acts on your car when it is traveling at velocity v?


Homework Equations





The Attempt at a Solution


well I broke the forces down into components. I drew the diagram as a triangle with height h and length L. So, the angle of the hill is arctan(H/L). so then I found the x component of the downward force should be the hypotenuse, mg times sin of the angle of the hill which is arctan(H/L). Well, with this i got mgsin(arctan(h/L) as the force down the hill. So since the total force is zero the resistance must be equal to it in the negative. Well, I'm wrong. I was told "check your trigenometry". I hope someone can explain to me what the flaws in my reasoning were. Thanks a lot.
 
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robbondo said:
the elevation decreases an amount h for each segment of road of length L.

Read once again.
 
I don't undersand. Is there a mistake in the way that I drew my diagram. If it decreases h for every l then shouldn't that equate to a triangle with the height being H and the length L?
 
nm... thanks radou. I see know that the hypotenuse is length l, not the adjacent side. I should be using arcsin not arctan. THANKS!
 

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