Finding the force up an incline given horizontal force

[OneObstacle]

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?

 

[tiny-tim]

Welcome to PF!

Hi OneObstacle! Welcome to PF!

… 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 F_app is the force to be applied, and which is the wind.

(I also don't understand how you can have a horizontal headwind if you're facing up an incline. )

 

[OneObstacle]

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.

 

[tiny-tim]

Hi OneObstacle!

(just got up :zzz:)

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 F_x 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

F_x = F_appcosθ ​