Forces on a Block and Friction on a Surface

  • #1
JessicaHelena
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3

Homework Statement


A 2.5 kg block is initially at rest on a horizontal surface. A horizontal force
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of magnitude 5.8 N and a vertical force
Parrowitalic.gif
are then applied to the block. The coefficients of friction for the block and surface are μs = 0.43 and μk = 0.24.
(a) Determine the magnitude of the frictional force acting on the block if the magnitude of
Parrowitalic.gif
is 8.0 N.


Homework Equations


F_friction = Mu*F_normal

The Attempt at a Solution


I really need help with this one, because I don't understand why I get a wrong answer.
Firstly, the net force in the y direction is 0. So F_g = P + F_N.
Then F_N = F_g - P = 2.5 * 9.8 - 8 = 16.5 N
Then since (I believe) the block is stationary, we use μs = 0.43, and so F_f = 0.43 * 16.5 =7.095.
Satisfied with my answer, I put this in the webassign, but it won't accept this as the correct answer. Where did I go wrong?
 

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Answers and Replies

  • #2
TSny
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The formula ##f_s = \mu_s N## gives the maximum possible value for the static friction force. (It would be better to write the formula as ##f_s^{\rm max} = \mu_s N##.)

The static friction force in your situation need not be at its maximum possible value.
 
  • #3
JessicaHelena
188
3
@TSny — so does that mean the WebAssign's looking for an answer that's like say, 7, rather than 7.095?

update: no, what I said above isn't right bc I checked... but how else should I interpret your hint?
 
  • #4
haruspex
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It would be better to write the formula as ##f_s^{\rm max} = \mu_s N##.
The form I use is ##|F_s|≤\mu_s|N|##. This emphasises that the direction of the frictional force is not dictated by the equation.
 
  • #5
JessicaHelena
188
3
@haruspex @TSny —Oh so since the maximum frictional force is 7.095N, but the other horizontal force being applied is 5.8 N, would the answer be 5.8 N?
 
  • #6
haruspex
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how else should I interpret your hint?
The relevant equation you quoted is incorrect. See either of the forms in posts 2 and 4.
Consider this: if the applied horizontal force is 5.8N and the frictional force in the other direction is 7N, what will happen to the block?
 
  • #8
JessicaHelena
188
3
@haruspex @TSny — thank you! Can I ask another question though?

(c) Determine the magnitude of the frictional force acting on the block if the magnitude of
Parrowitalic.gif
is 12.0 N.

So now that yields a maximum friction of 0.43*(2.5*9.8 - 12) = 5.375 N, which is less than the horizontal force F=5.8N. Then it's going to move right off the bat, right? How should I answer this?
 

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  • #9
haruspex
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Then it's going to move right off the bat, right?
Yes, but I am uncertain how to answer this part. It depends whether it means just before it starts moving or after it has started moving. My guess is they want the second.
 

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