- #1

Occidomortem

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## Homework Statement

A 250-kg crate is on a rough ramp, inclined at 30° above the horizontal. The coefficient of kinetic friction between the crate and ramp is 0.22. A horizontal force of 5000 N is applied to the crate, pushing it up the ramp. What is the acceleration of the crate?

Force (horizontal)= 5000N

U_K= 0.22

m= 250kg

a (upward) = ?

## Homework Equations

a=Fnet/m

F_k=U_kn

## The Attempt at a Solution

The first thing I did was find the net force. In the upward direction the force is equal to cos(30)5000N, The force of gravity acting downward on the incline is sin(30)mg, the kinetic friction also pointing downward of the incline is F_k=.22n, where normal force is equal to cos(30)mg from the incline plus the force from applying the horizontal force sin(30)5000, n=cos(30)mg+sin(30)5000

So I got an upward force on the incline to be 4330.127N. and the downward for from gravity to be 1225N. The downward force from friction= 0.22(cos(30)mg+sin(30)5000) = 1016.79.

the Fnet upward is = 4330.127N-1225N-1016.79N = 2088.337

a=F/m

a=2088.337/250kg

a=8.35m/s^2, this however is wrong. I think I messed up with my normal force somewhere.

So I guess my main question is how does a horizontal force affect the normal force on an incline. I apologize for the bad formatting/english I'm not used to typing these equations ect.