Finding the coefficient of kinetic friction

[riseofphoenix]

A 3.40 kg block is pushed along the ceiling with a constant applied force of 85.0 N that acts at an angle of 55.0° with the horizontal, as in the figure below. The block accelerates to the right at 6.00 m/s². Determine the coefficient of kinetic friction between block and ceiling.

Ugghhh so... this is what I did...

F = 85.0 N
m = 3.40 kg
a = 6.00 m/s²
f_k = μ_kn

1) Forces in the x-direction:
F cos θ
f_k (kinetic friction)

Forces in the y-direction:
F sin θ
n (normal force)

2) Sum of forces

ƩF_x = F cos θ + f_k = 0
ƩF_y = F sin θ + n = 0
ƩF_y = F sin θ + n = 0
ƩF_y = n = -F sin θ

3) ƩF_x = F cos θ + μ_kn = 0
3) ƩF_x = F cos θ + μ_k(-F sin θ) = 0
3) ƩF_x = F cos θ + μ_k(-F sin θ) = 0
3) ƩF_x = 85 cos 55 - μ_k(85 sin 55) = 0
3) ƩF_x = 85 cos 55 = μ_k(85 sin 55)
3) ƩF_x = (85 cos 55)/(85 sin 55) = μ_k
3) ƩF_x = 0.700 = μ_k

So I got this wrong, the answer is actually 0.781 (?)

I don't know what to do!

 

[vela]

You forgot a force.

 

[howie8594]

You have only considered the upward force that is applied to the block and the normal force that the wall exerts downward on the block. You forgot to include the force of gravity in the equation.