Crate pushed down an incline.

[Student3.41]

Homework Statement

A 219-kg crate rests on a surface that is inclined above the horizontal at an angle of 20.1°. A horizontal force (magnitude = 501 N and parallel to the ground, not the incline) is required to start the crate moving down the incline. What is the coefficient of static friction between the crate and the incline?

Homework Equations

F=ma

The Attempt at a Solution

Made the coordinate system in the direction of the acceleration.

Drew FBD. F_N in the +y direction. mgCOS20.1 in the -y direction.
F_x(501N) in the +x direction. mgSIN21.0 in the +x direction and F_k in the -x direction.

So,

F_y=ma_y ---> a=0
F_N=mgcos20.1

F_x=ma_x
-F_k+F_x+mgsin20.1 = ma
-F_k+501N+770N=219(kg)a
-U_kmgcos20.1+1270N = 219(kg)a
-U_k2010(N)=219(kg)a-1270N

This is where I am stuck I am unable to solve for the kinetic coefficient because I have two variables. How do I get a_x?

 

[Student3.41]

Is the acceleration 501N/219kg since this is the required force to move the crate.

 

[joej24]

First of all, since we are looking for the coeff of static friction, not kinetic (U k).
You were correct in saying that F net y = 0 and F net x = 0. There is no need to worry about acceleration since these equal zero.

F net y = n - mg cos (theta) - F sin (theta) = 0
The force that pushes it horizontally can be divided into components.

F net x = F cos (theta) - static friction - mg sin (theta) = 0

solve for n, plug it into the equation for the forces in the x direction and only static friction's coeff will be left.
static friction = (static coeff) * normal