Calculating Frictional Force: A Problem-Solving Guide

[gracemir]

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.

 

[tiny-tim]

Welcome to PF!

Hi gracemir! Welcome to PF!

(have a mu: µ )

… what i did was :
40*0.5=20 , 40*0.4=16 20+16=36

Nooo … you never use both µ_s and µ_k …

it's always either one or the other!

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

 

[gracemir]

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

Hi gracemir! Welcome to PF!

(have a mu: µ )


Nooo … you never use both µ_s and µ_k …

it's always either one or the other!

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

 

[tiny-tim]

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

oh come on … think …

what is the definition of µ_s ?​

 

[gracemir]

the definition of mu s is F sub s / N?

oh come on … think …

what is the definition of µ_s ?​

 

[tiny-tim]

the definition of mu s is F sub s / N?

No.

µ_s is the maximum possible value of F_s/N.

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