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Coeff of friction

  • Thread starter cugirl
  • Start date
  • #1
17
0

Homework Statement


A force of 10N is applied downward at an angle of 30 deg with respect to the horizontal on a block. There is friction between the block and the floor and the block remains stationary. M=2kg
(a) What is the magnitude of the normal force acting on the block?

(b) The coefficient of friction between the floor and block is μ = 0.8. What is the magnitude of the frictional force acting on the block?

Homework Equations


Fn = mg + 10sin30
Fn(mu) = Fk


The Attempt at a Solution


I got part A -- Fn = mg + 10sin30 = 19.6 + 5.0 = 24.6

But for part B, I thought it would be 24.6*.8 = 19.7, but according to the answer key, it should be 8.7.
 
Last edited:

Answers and Replies

  • #2
674
2
The block is stationary on the floor so the sum of the horizontal forces must equal 0. So draw the free body diagram along with the horizontal forces acting on it.
 
  • #3
17
0
So, Fμ = 10 cos 30 ?
 
  • #4
674
2
Correct.
 

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