Determining coefficient of kinetic friction help

[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!

 

[azizlwl]

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
-----------------------
It is accelerating, net Fx not zero.
The force applied is pressing the block to the ceiling. The weight is pulling it away.
I got this value - 0.78093138

 

[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
-----------------------
It is accelerating, net Fx not zero.
The force applied is pressing the block to the ceiling. The weight is pulling it away.
I got this value - 0.78093138

So is it supposed to be this?

ƩFx = F cos θ + μ_kn = 6

85 cos 55 + μ_k(-85 sin 55) = 6
85 cos 55 - μ_k(85 sin 55) = 6
85 cos 55 = 6 + μ_k(85 sin 55)
48.75 - 6 = 69.62μ_k
42.75/69.62 = μ_k
0.614 = μ_k

-____-

I didn't get that answer!
Can you please show me what you did?

 

[riseofphoenix]

Force of kinetic friction:
f_k = μ_kn

Horizontal force of acceleration in the x-direction/Value of the bottom leg of the triangle:
F_{acceleration-x} = F_a(cos θ)
F_{acceleration-x} = 85(cos 55)
F_{acceleration-x} = 48.75 N

Normal force n (value of the upright, vertical leg of the triangle):
n = F_{acceleration-y}(sin θ) + mg
n = 85(sin 55) + (3.40 kg)(6.00 m/s²)
n = 69.62 + 20.4
n = 90.02 N

f_k = μ_kn
48.75 = μ_k(90.02)
48.75/90.02 = μ_k
0.541 = μ_k

aksjflkjadsfjds -____-
Help?

 

[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
-----------------------
It is accelerating, net Fx not zero.
The force applied is pressing the block to the ceiling. The weight is pulling it away.
I got this value - 0.78093138

You were completely wrong.
Net force doesn't equal 6.
Thanks for nothing.

 

[azizlwl]

You were completely wrong.
Net force doesn't equal 6.
Thanks for nothing.

I not sure whether you read the question.
I just copy and paste the original question you posted and just underlined it.

 

[azizlwl]

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.

Here your original question and i underlined it.