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Static and kinetic friction of a crate

  • Thread starter dcangulo
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7
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a dockworker loading crates on a ship finds that a 20kg. crate, intially at rest on a horizontal surface, requires a 75-N horizontal force to set it in motion. However, after the crate is in motion, a horizontal force of 60-N is required to keep it moving with a constant speed. Find the coefficients of static and kinetic friction between crate and floor.

how do i find the coefficients? there is a table in the book with coefficients but it doesnt have one with wood on floor.
 
327
1
K + U = K_o + U_o

what is k,u,k_o,u_o ?
 

Doc Al

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how do i find the coefficients? there is a table in the book with coefficients but it doesnt have one with wood on floor.
You're supposed to solve for the coefficients using the information given in the problem, not look them up in some table.

Hint: How does static and kinetic friction relate to the normal force?
 
7
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[tex]\mu[/tex]=f/N

where N is the normal force and f would be the 20kg?
 
Last edited:

Doc Al

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[tex]\mu[/tex]=f/N
Good. But realize there's a difference between static and kinetic friction.

where N is the normal force and f would be the 20kg?
N is the normal force. What does the normal force equal? Hint: It's related to the 20 kg mass.

f is the friction force. Hint: The static and kinetic friction forces are given in the problem statement.
 
You need to solve for the two values of friction. Static friction being the frictional force that one must "overcome" to put the object into motion, and kinetic friction being the frictional force that must be "overcome" to sustain the objects motion.
 
7
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ok so what i did was find the normal force of the static friction which i believe was N=mg which is N=20(9.8) and then i did the little [tex]\mu[/tex]=f/N with [tex]\mu[/tex]=75/196

and that found me, what i believe to be the static friction.

Now how do i find kinetic friction coefficient??
 

Doc Al

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44,803
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ok so what i did was find the normal force of the static friction which i believe was N=mg which is N=20(9.8) and then i did the little [tex]\mu[/tex]=f/N with [tex]\mu[/tex]=75/196

and that found me, what i believe to be the static friction.
Good! The normal force is just the weight of the crate. The maximum static friction force was given as 75 N. You combined those facts to find the static friction coefficient.

Now how do i find kinetic friction coefficient??
Exactly the same way. What's the kinetic friction force? Does the normal force change?
 
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i think kinetic friction force is 60... or do i subtract 75-60 and that'll give me the kinetic friction??

this is what i did

75-f[tex]_{}s[/tex]=0
f[tex]_{}s[/tex]=75

this is the only thing i can think of
 

Doc Al

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44,803
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i think kinetic friction force is 60...
Right.
or do i subtract 75-60 and that'll give me the kinetic friction??
No. Kinetic and static friction are two different things, so don't combine them.

But you calculate the coefficient of friction in the exact same manner for both:
μ = F/N, where F is the friction force and N is the normal force.
 

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