# Static Friction and Frictional Force Ranking Task

William2

## Homework Statement

Below are six crates at rest on level surfaces. The crates have different masses and the frictional coefficients [given as (μs, μk) ] between the crates and the surfaces differ. The same external force is applied to each crate, but none of the crates move.

Rank the crates on the basis of the frictional force acting on them.
Rank from largest to smallest. To rank items as equivalent, overlap them.

Items:
(mass in kg, μs, μk)
A. 250, 0.2, 0.1
B. 1500, 0.3, 0.1
C. 500, 0.6, 0.3
D. 750, 0.6, 0.5
E. 600, 0.8, 0.5
F. 750, 0.4, 0.3

fs ≤ μs * m * g

## The Attempt at a Solution

I used the equation fs = μs * m* g and got the answer wrong. Now I am just completely confused. Any help will be appreciated, thanks.

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## Homework Equations

fs μs * m * g

Note the less than or equal sign. The static force of friction could be less than μs*m *g. So, you'll need to bring in another idea to determine what the actual force of friction would be for each block. Hint: Newton's 2nd law.

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William2
F=ma

there is no acceleration because the crates do not move, correct? so force would be 0 for all of them?

still a little confused.

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Homework Helper
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The "F" on the left side of the second law represents the net force (i.e., the sum of all the forces acting on the object.)

William2
Ok, sooo

F - μkmg = m * a

there is no acceleration so

F = μkmg

So am I supposed to use μk instead of μs?

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No, there is not kinetic friction in this problem since the blocks don't slide. You don't need to use any specific formula for friction. If I push horizontally on a block sitting on a horizontal surface with 5 N of force and it doesn't move, what is the value of the friction force?

William2
No, there is not kinetic friction in this problem since the blocks don't slide. You don't need to use any specific formula for friction. If I push horizontally on a block sitting on a horizontal surface with 5 N of force and it doesn't move, what is the value of the friction force?

5 N.

how would I figure out the horizontal force in this case?

is this equation correct?

F - μs(m * g) = 0 ; where F is the horizontal force

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5 N.
Right, 5N. That makes the net force zero which agrees with the second law since we know a = 0.
how would I figure out the horizontal force in this case?

is this equation correct?

F - μs(m * g) = 0 ; where F is the horizontal force

In your problem, you don't need to figure out the horizontal applied force. Whatever it is, it's the same applied force for each crate. So, how would the friction forces compare for the various crates? (Again, μs*m * g is not relevant here since it represents only the maximum possible static friction force and you cannot assume that the forces of friction are at the maximum value.)

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William2
so...
would it simply be ranked based on the static frictional coefficient?

the only other known variable is mass.

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Let F represent the applied force (assumed horizontal) and let fs represent the friction force. How would you express the net horizontal force using these symbols?

William2
Let F represent the applied force (assumed horizontal) and let fs represent the friction force. How would you express the net horizontal force using these symbols?

Net Force = F - fs

i think

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Good. What does the net force have to equal according to the 2nd law?

William2
Good. What does the net force have to equal according to the 2nd law?

it has to equal ma

so... F - fs = m * a

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Yes. And what is the value of the acceleration for the crates?

William2
Yes. And what is the value of the acceleration for the crates?

0, i think.

a = F/m ; and net Force, F, is 0

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Well, we know a = 0 because we are told that the crates don't move. So, if a = 0, what does the formula F - fs = ma tell you about the amount of friction force?

William2
Well, we know a = 0 because we are told that the crates don't move. So, if a = 0, what does the formula F - fs = ma tell you about the amount of friction force?

if a = 0 , then F = fs ; which was already determined

this seems like a loop to me

Homework Helper
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Good. So, you should now be able to answer the question. Remember, the problem states that each crate has the same applied force.

William2
OMG, WOW!

IT WAS SO EASY! They are all the same?!

I LOVE YOU!

Homework Helper
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Hey hey, now . Anyway, good work. azizlwl
Since all blocks are not moving, the frictional force equal to applied force.
The static frictional force is not constant from zero to maximum μs.
Thus all the frictional forces are equal.