Calculating Frictional Force: A Problem-Solving Guide

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To calculate the frictional force on a 40-N crate with a 12-N applied force, it's essential to distinguish between static and kinetic friction coefficients. The static friction coefficient (µs = 0.5) indicates the maximum frictional force before movement, while the kinetic friction coefficient (µk = 0.4) applies once the crate is in motion. The initial step is to determine if the 12-N force is sufficient to overcome static friction, which requires calculating the maximum static friction force (20 N). If the applied force is less than this maximum, the crate does not move, and the frictional force equals the applied force. Understanding these principles is crucial for accurately solving the problem.
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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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