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Friction Force between the box and the floor?

  • #1

Homework Statement


A heavy box (200.0 kg) is sitting still on a warehouse floor, despite the efforts of a worker to push it (with a force of 300.0 N). If the coefficient of static friction between the box and the floor is 0.6, what is the friction force between the box and the floor? What is the frictional force between the box and the floor if another worker puts another box (also 200.0 kg) on top of the first box while the first worker continues to pull with 300.0 N?


Homework Equations


Not sure


The Attempt at a Solution


ok, so this problem deals with static friction. What is the equation for the friction force? I am assuming for the first part of the problem, you plug in the force and mass. Also, If you simply double the the weight in kg, doesnt the force quadruple?
 

Answers and Replies

  • #2
Draw a free body diagram for this problem and include all of the relavent forces.

Friction force equals normal force times coefficiant of friction.
 
  • #3
so would that be 9.8 m/s2 times 0.6?
 
  • #5
so 200kg(9.8) = 1960.. times that by 0.6 and you get 1176?
 
  • #6
tiny-tim
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Welcome to PF!

Hi IneedHelp:-)! Welcome to PF! :wink:
Also, If you simply double the the weight in kg, doesnt the force quadruple?
No, Fr = µN = µmg, so if you double m, then you double the maximum static friction force also. :smile:
 
  • #7
so 200kg(9.8) = 1960.. times that by 0.6 and you get 1176?
Yes! There is your answer for the first part. :biggrin:
 
  • #8
PhanthomJay
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so 200kg(9.8) = 1960.. times that by 0.6 and you get 1176?
The problem is asking for the friction force between the box and floor when the worker pushes on it with a 300 N force (presumably in a horizontal direction), and it doesn't move. Using one of Newton's laws will help to solve for it.
 
  • #9
ok so if you double the mass, then you double 1176 and that is the second part of the answer correct?
 
  • #10
PhanthomJay
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ok so if you double the mass, then you double 1176 and that is the second part of the answer correct?
No. You are not trying to determine in part 1 or 2 the max available static friction force. You are trying to determine the actual friction force. Hint: Use Newton 1.
 
  • #11
ah so would that mean the force is 0, bc there is no acceleration? thats the only thing i can come up with out of the first law clue you are giving me
 
  • #12
tiny-tim
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ah so would that mean the force is 0, bc there is no acceleration? thats the only thing i can come up with out of the first law clue you are giving me
No, it means the net force (the sum of all the forces) is 0. :wink:

So what are all the forces? :smile:
 

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