# Block on Incline

#### PierceJ

1. Homework Statement
http://i.minus.com/iVbsy2FomprcM.PNG [Broken]
http://i.minus.com/iVbsy2FomprcM.PNG [Broken]

2. Homework Equations
F=ma

3. The Attempt at a Solution
I'm struggling with the normal force and I'm not sure if the component of the force F is the cosine or sine. I see examples where its sine but that doesnt make any sense to me. Isn't the x direction supposed to be cosine?

Fn = mgsinθ
81.2cosθ - Fnμ = 9.23
-Fnμ = -65.897

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#### PhanthomJay

Homework Helper
Gold Member
1. Homework Statement
http://i.minus.com/iVbsy2FomprcM.PNG [Broken]
http://i.minus.com/iVbsy2FomprcM.PNG [Broken]

2. Homework Equations
F=ma

3. The Attempt at a Solution
I'm struggling with the normal force and I'm not sure if the component of the force F is the cosine or sine. I see examples where its sine but that doesnt make any sense to me. Isn't the x direction supposed to be cosine?

Fn = mgsinθ
81.2cosθ - Fnμ = 9.23
-Fnμ = -65.897
For simplicity of the solution, the x axis is chosen to be the axis parallel to the incline, and the y axis is the axis perpendicular to the incline. In this manner, the applied force F is already in the in the x direction, and when you draw your FBD for the forces (you are missing at least one), a bit of geometry/trig and newton 1 in the chosen y direction will you give you the relationship between the normal force and weight. Since the applied force is already in the x direction and the normal force in the y direction, it is the weight force that needs to be broken up into its x and y components before applying Newton's Laws. And no, the x axis is not always the cos, it depends on what angle you are working with and the trig properties of a right triangle.

Last edited by a moderator:

#### PierceJ

Okay.

So the normal force is equal to mgsinθ?

#### BvU

No. If you aren't sure, then you look at a limiting case: let $\theta$ go to zero, and you know that then the normal force is equal to mg. Forces the choice between $mg\sin\theta$ and $mg\cos\theta$ to be the latter because cos(0) = 1.
The normal force is $\perp$ the incline. In your FBD you should easily see that the angle between mg and the normal force is $\theta$ and not ${\pi\over 2}-\theta$. So cosine to project mg on the perpendicular.