- #1

Freemark

- 14

- 0

## Homework Statement

A crate is being pushed up a hill with friction. Given: m1 = 131 kg, θ = 23°, [itex]\mu[/itex]s = 0.15, and [itex]\mu[/itex]k = 0.08. If the crate isn't being pushed anymore, what will be the magnitude of the acceleration of the crate sliding downhill?

What force must be exerted to move the crate uphill at constant speed?

What force must be exerted to move the crate downhill at constant speed?

## Homework Equations

[itex]\Sigma[/itex]Fx = m*a

[itex]\Sigma[/itex]Fy = m*a

## The Attempt at a Solution

I drew a free body diagram of the crate, m1. The crate is diagonal since it's on an incline, so I made my coordinate system correlate with that. the only diagonal I had was Mg, so I found the x and y components of it. So Mxg = MgSinθ and Myg = MgCosθ.

Overall the forces acting upon the box were Mg, N, and f (friction). So for [itex]\Sigma[/itex]Fx = M*a:

-MgSinθ - f = m*ax

-MgSinθ - [itex]\mu[/itex]s*N = M*ax

[itex]\frac{-MgSinθ - ms*N}{M}[/itex]=ax

For [itex]\Sigma[/itex]Fy = M*a:

N-MgCosθ=M*ay (there's no vertical a, so the right side is 0)

N = MgCosθ

Since I'm trying to find horizontal ax, I'll sub in MgCosθ in for N.

Now I have [itex]\frac{-MgSinθ - ms*MgCosθ}{M}[/itex]=ax

I can factor out the M...

-gSinθ - [itex]\mu[/itex]s*gCosθ =ax (the M's cancel on top/bottom)

Bring out a g and -1 from both...

-g(Sinθ + [itex]\mu[/itex]s*Cosθ) =ax

Plug in and get my answer for part 1? The real question I guess is if my math was right and if I'm using [itex]\mu[/itex]k instead of [itex]\mu[/itex]s. I have no idea how to start the other 2 parts.