Finding the force up an incline given horizontal force

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Homework Help Overview

The discussion revolves around determining the force acting up an incline when a horizontal force, referred to as a "headwind," is applied. The context involves a skier on an incline and the relationship between the applied force and its components.

Discussion Character

  • Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to use trigonometric relationships to resolve the forces acting on the skier. There is confusion regarding the correct representation of the forces in a triangle, particularly the distinction between the applied force and the wind force. Some participants question the setup of the problem and the interpretation of the forces involved.

Discussion Status

Participants are exploring the relationships between the forces and their components. Some guidance has been offered regarding the correct approach to finding the force component along the incline, but there is still uncertainty about the definitions and roles of the forces involved.

Contextual Notes

There is a mention of confusing wording in the problem statement, which may affect the understanding of the forces at play. The original poster's reference to a horizontal headwind while facing an incline raises questions about the problem's assumptions.

OneObstacle
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Homework Statement


I need help with part of a problem. I am having problems understanding just how to find the force up an incline when a horizontal force ("headwind") is given.


Homework Equations





The Attempt at a Solution


When drawing a triangle to figure it out I draw the hypotenuse as x (force I need to find) and the side adjacent to angle θ as Fapplied. From this, I get cosθ = Fapp/x, therefore x=Fapp/cosθ. However, my online homework listed it as Fapp*cosθ. Did I draw the triangle wrong?
 
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Welcome to PF!

Hi OneObstacle! Welcome to PF! :smile:
OneObstacle said:
… how to find the force up an incline when a horizontal force ("headwind") is given.

When drawing a triangle to figure it out I draw the hypotenuse as x (force I need to find) and the side adjacent to angle θ as Fapplied. From this, I get cosθ = Fapp/x, therefore x=Fapp/cosθ. However, my online homework listed it as Fapp*cosθ. Did I draw the triangle wrong?

I don't understand which of x and Fapp is the force to be applied, and which is the wind. :confused:

(I also don't understand how you can have a horizontal headwind if you're facing up an incline. :redface:)
 
The original problem is a skier heading down an incline (sorry, confusing wording). Fapp is the headwind heading into the slope, and Force x is what I randomly assigned the force of the wind acting on the skier.
 
Hi OneObstacle! :smile:

(just got up :zzz:)
OneObstacle said:
The original problem is a skier heading down an incline (sorry, confusing wording). Fapp is the headwind heading into the slope, and Force x is what I randomly assigned the force of the wind acting on the skier.

ohh! … you mean Fx is the component of the wind force in the x direction (ie up the slope)?

ok, then your book is correct …

don't start drawing triangles, just use this rule:

to find the component of a (complete) force, we always multiply by cos of the angle between

Fx = Fappcosθ :smile:
 

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