How Do You Calculate Forces on a Crate on an Inclined Plane?

[Nikstykal]

Homework Statement

Crate of mass m is pushed horizontally with force F up an inclined plane of θ from ground. The plane has a friction coefficient of μ.

a) draw free body diagram of crate
b) derive expression for normal force
c) derive expression for acceleration

*** Use ONLY m, g, F, μ, θ ****

Homework Equations

ΣF = ma
|a| = √(a_x²+a_y²)

The Attempt at a Solution

[/B]
Included is a picture of the problem with my free body diagram.

I set the normal as the y-axis so that when solving ΣFy=ma_y I can set a_y=0.

ΣF_x=ma_x=-μN+Fcosθ-mgsinθ
ΣF_y=ma_y=N-Fsinθ-mgcosθ

a_y=0 so N = Fsinθ+mgcosθ (part b)

Since a_y=0, a=√a_x² = a_x. Substituting N into a_x:
ma_x=-μ(Fsinθ+mgcosθ)+Fcosθ-mgsinθ

is this correct?

 

[mfb]

Looks correct.

 

[CWatters]

Looks correct to me as well although I would have drawn the fbd at the same angle. Nothing wrong with yours it just look odd.

 

[Nikstykal]

Looks correct to me as well although I would have drawn the fbd at the same angle. Nothing wrong with yours it just look odd.

If you do that then you will have two components of acceleration (in the x AND y direction) and so your definition of N would involve more variables than the ones they allow, or at least would involve a lot more manipulation to isolate N.

 

[haruspex]

If you do that then you will have two components of acceleration (in the x AND y direction) and so your definition of N would involve more variables than the ones they allow, or at least would involve a lot more manipulation to isolate N.

mfb was only commenting on the orientation of the FBD. You can draw it with the same orientation as the physical arrangement but still choose the X and Y axes as parallel and normal to the plane.

 

[Nikstykal]

Oh... my bad