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How do I determine the wretched coefficient of kinetic friction? It's impossible

  • #1
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.

p4-60.gif



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

F = 85.0 N
m = 3.40 kg
a = 6.00 m/s2
fk = μkn

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

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

2) Sum of forces

ƩFx = F cos θ + fk = 0
ƩFy = F sin θ + n = 0
ƩFy = F sin θ + n = 0
ƩFy = n = -F sin θ

3) ƩFx = F cos θ + μkn = 0
3) ƩFx = F cos θ + μk(-F sin θ) = 0
3) ƩFx = F cos θ + μk(-F sin θ) = 0
3) ƩFx = 85 cos 55 - μk(85 sin 55) = 0
3) ƩFx = 85 cos 55 = μk(85 sin 55)
3) ƩFx = (85 cos 55)/(85 sin 55) = μk
3) ƩFx = 0.700 = μk

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

I don't know what to do!
 

Answers and Replies

  • #2
And THEN, i tried doing this:

ƩFx = F cos θ + μkn = 6

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

-____-

I didn't get that answer!!!
Can you please show me how to get to that answer please?
 
Last edited:
  • #3
gneill
Mentor
20,793
2,773
I don't see where you've taken gravity into account. Won't it affect the normal force?
 

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