Finding the force up an incline given horizontal force

In summary, the problem involves finding the force up an incline when a horizontal force is given. The original problem involves a skier heading down an incline, with the horizontal force acting as a headwind into the slope. The force to be found, Fx, is the component of the wind force in the x direction and can be calculated using the formula Fx = Fappcosθ.
  • #1
OneObstacle
6
0

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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  • #2
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:)
 
  • #3
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.
 
  • #4
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:
 
  • #5


Your approach using trigonometry is correct. However, the difference in notation may be due to the use of different reference angles. In your approach, you are using the angle θ as the reference angle, which is measured from the horizontal. In this case, the force up the incline would be given by x = Fapp/cosθ. However, some textbooks or online resources may use the angle α as the reference angle, which is measured from the incline itself. In this case, the force up the incline would be given by x = Fapp*cosα. Both approaches are correct, it just depends on how the angle is defined in the problem. Make sure to double check the given angle and use the correct trigonometric function accordingly.
 
  • #6



It seems like you have the right idea in terms of using trigonometry to solve for the force up an incline. However, the key is to make sure you are using the correct trigonometric function based on the information given in the problem.

If the horizontal force (Fapp) is given, then the force up the incline (Fup) would be the opposite side of the triangle, not the adjacent side. In this case, you would use sine instead of cosine. So the correct equation would be Fup = Fapp/sinθ.

Also, make sure to pay attention to the units of the given forces. If the horizontal force is given in Newtons (N), then the force up the incline should also be in N. If the horizontal force is given in pounds (lbs), then the force up the incline should also be in lbs.

In terms of your online homework listing the equation as Fapp*cosθ, it is possible that they may have used a different convention or notation. It would be best to double check with your instructor or classmates to clarify the correct equation to use in this case.
 

1. What is the formula for finding the force up an incline given horizontal force?

The formula for finding the force up an incline given horizontal force is F = Fh / sinθ, where F is the force up the incline, Fh is the horizontal force applied, and θ is the angle of the incline.

2. How is the force up an incline related to the angle of the incline?

The force up the incline is directly related to the angle of the incline. As the angle increases, the force required to move an object up the incline also increases.

3. Can the force up an incline be greater than the horizontal force applied?

Yes, the force up the incline can be greater than the horizontal force applied. This can happen when the angle of the incline is steep, requiring more force to overcome the incline's resistance.

4. How does friction affect the force up an incline?

Friction can decrease the force up the incline because it acts in the opposite direction of the applied force. This means that more force is needed to overcome friction and move an object up the incline.

5. Is there a maximum force that can be applied up an incline?

There is no maximum force that can be applied up an incline. However, there may be physical limitations or constraints that prevent a certain amount of force from being applied. For example, the weight of the object being moved or the strength of the person applying the force may limit the maximum force that can be applied.

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