Friction Help finding the coefficent of static friction

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Homework Help Overview

The discussion revolves around determining the coefficients of static and kinetic friction for a box being moved across a concrete floor. The original poster presents a scenario involving a force applied to initiate movement and subsequent acceleration.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the relationship between the applied force and static friction, with some suggesting that the maximum static friction force equals the applied force. Questions arise about how to approach the second part of the problem regarding kinetic friction and the necessary force for acceleration.

Discussion Status

There is an ongoing exploration of the concepts involved in the problem, with participants providing guidance on how to relate the forces and acceleration. Some participants express confusion about the setup and calculations, indicating a lack of consensus on the next steps.

Contextual Notes

The original poster mentions difficulty in understanding the problem setup and expresses frustration with the lack of explanation from their teacher. There is also a reference to the need for clarification on the role of friction in the context of the applied forces.

Jazzybelle4
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Homework Statement


A force of 48.0 N is required to start a 5.0 kg box moving across a horizontal concrete floor. (a) what is the coefficient of static friction between the box and the floor?
(b) if the 48.0N force continues the box accelerates at 0.70m/sec2. what is the coefficient of kinetic friction


I have no idea how to set up up problem (a), all i know is the Ffr=us(mg)
 
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You assume the force that moves the box is only just enough to overcome static friction and thus you can set the max force of static friction equal to the force applied to the box.
 
so then the equation set up would look like:
Ffr max= us(mg)
48N=us(mg)
 
Yes. You can then find the coefficient.
 
I found the coefficient it is .98 now i do not know what to do at all for the second part of the question (b)
 
What force would be required to accelerate the box at that rate without any friction?
 
any force about 48.0? there will always be friction. the coefficient needs to be smaller at .98
 
What about Newton's second law. You can work out the force required to make a mass the same as the box accelerate at the value you're given. How does that compare to the force you're having to use?
 
second law is
F=ma
48=m(.70)
I am so lost... my teacher doesn't explain anything
 
  • #10
You're finding the force required to accelerate a 5 kg mass at 0.7 ms[su[]-2[/sup]. You don't need the 48 in there. Hang in there all will become clear. :wink:
 

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