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Finding the coefficient of kinetic friction

  • #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
vela
Staff Emeritus
Science Advisor
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You forgot a force.
 
  • #3
46
0
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.
 

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