Kinetic Friction Problem

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Homework Statement



A child goes down a playground slide with an acceleration of 1.26 m/s^2. Find the coefficient of the kinetic friction between the child and the slide if the slide is inclined at an angle of 33 degrees below horizontal.


Homework Equations



F=MA

Kinetic Friction = (Coefficient)N

The Attempt at a Solution



I'm assuming I need to find the X and Y components using 33 degrees and gravity?

(-9.81Cosine(-33) = -8.66
(-9.81)Sin(-33) = 4.6

Now I'm pretty lost. I'm having a hard time understanding how f=ma and Fk = (μk)N relate to each other.

I think I can plug in MA into N? Fk = (uk)(ma) ?
 

Answers and Replies

  • #2
PhanthomJay
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Please draw a free body diagram and identify and show the forces acting on the child. Choose the x axis parallel to the incline and the y axis perpendicular to the incline. The child's weight acts down, vertically. There is also a normal force that acts perpendicular to the incline pushing against the child. Then there is the friction force which acts where and in which direction? The friction force is uN.
You will need to break up the weight into its x and y components. Then apply newton 1 in the y direction to solve for N. Then apply newton 2 , F_net = ma, in the x direction to solve for the friction coefficient. What is the net force in the x direction? You should do a www. search on inclined planes with friction.
 
  • #3
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Please draw a free body diagram and identify and show the forces acting on the child. Choose the x axis parallel to the incline and the y axis perpendicular to the incline. The child's weight acts down, vertically. There is also a normal force that acts perpendicular to the incline pushing against the child. Then there is the friction force which acts where and in which direction? The friction force is uN.
You will need to break up the weight into its x and y components. Then apply newton 1 in the y direction to solve for N. Then apply newton 2 , F_net = ma, in the x direction to solve for the friction coefficient. What is the net force in the x direction? You should do a www. search on inclined planes with friction.

I think I made some progress on it by doing some other questions, I have this so far.

may = (uf)(max)

may-1.26/max = uf

4.6-1.26 / 8.21 = uf = .38

or is it... 1.28 = uk (N)

where N = x component + y component ( -8.22+5.34)

1.28 / -2.877 = -.45? Or maybe without the negative
 
Last edited:
  • #4
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5.05 - 8.41mu = 1.36
-8.41mu=-3.69
mu = -3.69 / -8.41 = .433

Can anyone confirm if this is right or not? Unfortunately I don't have the answers since I'm practicing all the even questions in the book :D
 

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