Friction and Incline

  • Thread starter jstiel
  • Start date
  • #1
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Homework Statement


The coefficient of static friction between the 3.00kg crate and teh 35.0 degree incline is 0.300. What minimum force perependicular to the incline must be applied to the crate to prevent it from sliding?

Gravity = 9.80 m/s2

Homework Equations


Sigma Fx = Fcos(theta) + fs - wsin(theta) = 0
Sigma Fy = n - Fsin(theta) - wcos(theta) = 0

The Attempt at a Solution



w = (3.00kg)(9.80m/s) = 29.4N
n = .574F + 24.1N
fs = .300(.574F + 24.1N) = .1722F + 7.23N
Fx = .819F + 1.72F + 7.23N 0 16.9N = 0
.991F - 9.67N = 0
F = 9.75
n = .574(9.75) + 24.0
n = 29.6N

Problem that I am running into is that the solution to the question is actually 32.1N and I can't see where I went wrong. Help would be much appreciated. Thank you.
 

Answers and Replies

  • #2
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Anyone?
 
  • #3
alphysicist
Homework Helper
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Hi jstiel,

Homework Statement


The coefficient of static friction between the 3.00kg crate and teh 35.0 degree incline is 0.300. What minimum force perependicular to the incline must be applied to the crate to prevent it from sliding?

Gravity = 9.80 m/s2

Homework Equations


Sigma Fx = Fcos(theta) + fs - wsin(theta) = 0
Sigma Fy = n - Fsin(theta) - wcos(theta) = 0

I don't believe the Fcos(theta) and Fsin(theta) are correct in these equations. What do you know about the direction of the applied force? Do you get the right answer?
 

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