Solving Physics Problem: Find Coefficient of Friction

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The discussion revolves around solving two physics problems related to friction. In the first problem, a 25 kg block has a maximum static friction force of 55 N, but an external force of 25 N results in a net frictional force of 30 N, indicating the block remains stationary. The second problem involves a 5 kg block where a 10 N force is applied, leading to confusion about the coefficient of friction, which is calculated incorrectly as 5. The correct understanding is that the coefficient of static friction should yield a value between 0 and 1, and the actual frictional force can vary up to the maximum static friction. Clarifications about the equilibrium state and the nature of friction forces were appreciated by the participants.
Eiano
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Hello all,
I'm new to this site and think it's a great idea to have a Physics site to help with, b/c we all know physics is the devil... (j/k) :)

Anyway I have this problem which I think i figured out but want to be sure.



Suppose a block of mass 25 kg rests on a horizontal surface and the coefficient of static friction is .22

a). What is the maximum possible Fs that could act on the block.

This is what I got, Fs=m*g*Us
x=(25)(10)(.22) = 55N

b). What is the avtual Fs that acts on the block if an external force of 25N acts horizontally on the block.

ME:
55-25= 30M


And this problem:

2). A 5kg block rests on a horizontal plane a force of 10N applied horizontally causes the block to more horizontally at a constant velocity. What is the coefficient of friction b/w the block and the plane, assume G is 9.8 m/s^2.

My answer doesn't make sense because the coefficient is not between 0 and 1.0 :rolleyes:

Anyway:

ME:
(9.8)(5)/10= about 5

Any ideas?

Thanks in advance!
 
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For 1b) the block is not moving. What must be true of the sum of the forces acting on the block?

#2 is upside down.
 
Last edited:
Okay, so for 1b.)

since it's at equilibrium, the sum of the forces must be equal. (err, must equal 0)?

Fs= (uS)(N)?
Fs= (.22)(25)
Fs=5.5?

Is this good or am I still missing something.

and thanks for correcting my error on 2. :blushing:
 
Last edited:
Eiano said:
Okay, so for 1b.)

since it's at equilibrium, the sum of the forces must be equal. (err, must equal 0)?

Fs= (uS)(N)?
Fs= (.22)(25)
Fs=5.5?

Is this good or am I still missing something.

and thanks for correcting my error on 2. :blushing:

The coefficient of static friction tells you the maximum possible force of friction acting on a stationary object. The actual frictional force can be any value between zero and the maximum. It is often less than the maximum, and that is true in this case.
 
OlderDan said:
The coefficient of static friction tells you the maximum possible force of friction acting on a stationary object. The actual frictional force can be any value between zero and the maximum. It is often less than the maximum, and that is true in this case.

thanks, i read the book and didn't get this concept, thanks again for clearing it up.
 
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