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Calculating coefficient of kinetic friction

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[SOLVED] Calculating coefficient of kinetic friction

Homework Statement



An applied force of 450 N {forward} is needed to drag a 1000-kg crate at constant speed across a horizontal, rough floor. Calculate the coefficient of kinetic friction for the crate on the floor.

Fapp=450 N {forward}
m=1000 kg
a=0
g=9.81 m/s^2

Homework Equations


Fnet=F(f(kinetic)) + F(app) + F(g)
Fnet=ma
Ffkinetic=-(mu)kFn
Fg=-mg


The Attempt at a Solution



Fnet=-(mu)kFn + Fapp + Fg
Since Fnet = ma
ma=-(mu)kFn + Fapp + Fg
0=-(mu)k1000 + 450 - 1000(9.81)
(mu)k1000=-9360
(mu)k=-9.36

The answer is supposed to be 4.59 x 10^-2, which makes a whole lot more sense than -9.36. I'm completely lost. The text only has one example and it's completely different from this question. tia!
 

Answers and Replies

  • #2
nrqed
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Homework Statement



An applied force of 450 N {forward} is needed to drag a 1000-kg crate at constant speed across a horizontal, rough floor. Calculate the coefficient of kinetic friction for the crate on the floor.

Fapp=450 N {forward}
m=1000 kg
a=0
g=9.81 m/s^2

Homework Equations


Fnet=F(f(kinetic)) + F(app) + F(g)
Fnet=ma
Ffkinetic=-(mu)kFn
Fg=-mg


The Attempt at a Solution



Fnet=-(mu)kFn + Fapp + Fg
Since Fnet = ma
ma=-(mu)kFn + Fapp + Fg
0=-(mu)k1000 + 450 - 1000(9.81)
(mu)k1000=-9360
(mu)k=-9.36

The answer is supposed to be 4.59 x 10^-2, which makes a whole lot more sense than -9.36. I'm completely lost. The text only has one example and it's completely different from this question. tia!

You are mixing x and y components!!! You must apply Newton's second law separately to the x and y components! So you have two equations :

[tex] \sum F_x = m a_x [/tex]

and

[tex] \sum F_y = m a_y [/tex]

Draw a FBD, identify the forces along x and along y, write down their x and y components (including the correct signs) and plug in the above equations.
 
  • #3
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Thank you. I looked at the text again and retried writing my information making sure I was dealing with just the horizontal components (I was looking at the equations for slopes and combining the wrong things), using from the text this formula:

F(net_h) = F(app) + F(f_kinetic)

Since there is no acceleration Fnet is 0:

0 = F(app) + F(f_kinetic)
F(f_kinetic) = -F(app)

Since F(f_kinetic) = -(mu)k_F(N)

-(mu)k_F(n) = -F(app)
(mu)k = F(app)/F(N)

Since F(N) is mg:

(mu)k = 450/(1000*9.81)
= 4.59 * 10^-2
 

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