Calculating Frictional Force: A Problem-Solving Guide

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Homework Help Overview

The discussion revolves around calculating the frictional force acting on a crate resting on a rough horizontal surface when a horizontal force is applied. The problem involves understanding the coefficients of static and kinetic friction and determining whether the applied force is sufficient to overcome static friction.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the application of static and kinetic friction coefficients, questioning the validity of using both in the calculations. There are inquiries about how to determine if the applied force is sufficient to move the crate.

Discussion Status

The discussion is ongoing, with participants providing hints and guidance on the definitions and implications of static friction. There is an exploration of the conditions under which the crate would remain stationary versus when it would begin to move.

Contextual Notes

Participants are navigating the definitions of friction coefficients and the conditions for static versus kinetic friction, indicating a need for clarity on these concepts. The original poster's calculations are being scrutinized for correctness in the context of the problem.

gracemir
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A 40-N crate rests on a rough horizontal floor. A 12-N horizonatl force is then applied to it. If the coefficients of friction are s = 0.5 and k = 0.4, the magnitude of the frictional force on the crate is:

what i did was :
40*0.5=20 , 40*0.4=16 20+16=36 40/12=3.3 36/3=12

how to solve this problem? could someone help?
thank you.
 
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Welcome to PF!

Hi gracemir! Welcome to PF! :smile:

(have a mu: µ :wink:)
gracemir said:
… what i did was :
40*0.5=20 , 40*0.4=16 20+16=36

Nooo … you never use both µs and µk

it's always either one or the other! :wink:

Hint: first decide whether 12N is enough to move the crate, then decide what the friction force is. :smile:
 


how to decide whether 12N is enough to move or not?

tiny-tim said:
Hi gracemir! Welcome to PF! :smile:

(have a mu: µ :wink:)


Nooo … you never use both µs and µk

it's always either one or the other! :wink:

Hint: first decide whether 12N is enough to move the crate, then decide what the friction force is. :smile:
 
gracemir said:
how to decide whether 12N is enough to move or not?

oh come on … think …

what is the definition of µs ?​
 
the definition of mu s is F sub s / N?

tiny-tim said:
oh come on … think …

what is the definition of µs ?​
 
gracemir said:
the definition of mu s is F sub s / N?

No.

µs is the maximum possible value of Fs/N.

To find the actual value of Fs, we can use the fact that the acceleration (and velocity) is zero, so all the forces must add to zero. :smile:
 

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