Solve Block on Incline HW: F=ma & Fn=mgsinθ

[PierceJ]

Homework Statement

http://i.minus.com/iVbsy2FomprcM.PNG
http://i.minus.com/iVbsy2FomprcM.PNG

Homework Equations

F=ma

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 doesn't make any sense to me. Isn't the x direction supposed to be cosine?

F_n = mgsinθ
81.2cosθ - F_nμ = 9.23
-F_nμ = -65.897

 

[PhanthomJay]

Homework Statement

http://i.minus.com/iVbsy2FomprcM.PNG
http://i.minus.com/iVbsy2FomprcM.PNG

Homework Equations

F=ma

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 doesn't make any sense to me. Isn't the x direction supposed to be cosine?

F_n = mgsinθ
81.2cosθ - F_nμ = 9.23
-F_nμ = -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.

 

[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.

Don't associate x with the one and y with the other. Sometimes it's like this, other times it is the other way.