Calculating Coefficient of Friction for Sliding Book on Floor

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Homework Help Overview

The problem involves calculating the coefficient of friction for a book sliding on the floor, given its mass, the applied force, and the resulting acceleration. The subject area pertains to dynamics and friction in physics.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the relationship between applied force, frictional force, and acceleration, with some attempting to derive the coefficient of friction using equations of motion.
  • There are questions about the relevance of the applied force and the interpretation of negative acceleration.
  • Some participants express confusion regarding the terminology of acceleration versus deceleration.

Discussion Status

The discussion is ongoing, with various interpretations of the problem being explored. Some participants offer insights into the relationship between forces and acceleration, while others question assumptions about the forces involved. There is no explicit consensus on the correct approach or terminology.

Contextual Notes

Participants note potential confusion arising from the initial conditions of the problem, including the direction of forces and the nature of acceleration. The problem's setup may lead to differing interpretations of the forces acting on the book.

rajames429
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I am not a student, but once took physics in college more that 20 years ago.

I have the following problem that I am having trouble with:

A book has a mass of 400 grams. When you slide the book against the floor with a force of 5 N, it accelerated at a rate of -1.5 m/s^2.

What would the coefficient of friction be between the book and the floor?

So, my solution seems to be:

Where F is force
N is normal force
FF is force of friction
u is coefficient of friction

I know F = ma, N = -mg, and FF = uN, so that FF = -umg.

Is this just a matter of setting ma = -umg, then a = -ug, and then u = -a/g ?

So given that g = 9.8 m/s^2:

I get an answer of 0.15, is this correct?

Is this the correct way to approach this problem?

Is the 5 N force just unneeded information here?

Thank you for your help
 
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yea you've got the right idea except -a is not 1.5 it's 11, it's acceleration caused by the force of friction

u= -(11)/9.8 = 1.22
 
Yea, I think the 5N part is suppose to be a trap to get people to think that a force of 5N is applied continuously so they use Fa + Ff = Fnet
 
Assuming you're designating the direction of the 5 N force to be positive, how could the acceleration possibly be negative?

Forgive me if I'm missing something obvious; it is getting late.
 
Well, not quite, the force applied is not the same as the force of friction because there is an acceleration. The only time the force of friction is the same as the force applied is when acceleration is zero.

The reason they give you an initial force is so you can find the -acceleration due to the force of friction. Basically this is what you do:
F=ma( your variable is a) as you will see the acceleration is different than that of the actual acceleration.
Now subtract this value from the actual acceleration to find the -acceleration due to friction. You can now find the force due to friction again using F=ma you have now found the force of friction and can solve for the coefficient of friction.
(for convenience I changed the axis and made the initial acceleration +.)
 
Last edited:
trajan22 said:
Well, not quite, the force applied is not the same as the force of friction because there is an acceleration. The only time the force of friction is the same as the force applied is when acceleration is zero.

The reason they give you an initial force is so you can find the deceleration due to the force of friction. Basically this is what you do:
F=ma( your variable is a) as you will see the acceleration is different than that of the actual acceleration.
Now subtract this value from the actual acceleration to find the deceleration due to friction. You can now find the force due to friction again using F=ma you have now found the force of friction and can solve for the coefficient of friction.
(for convenience I changed the axis and made the initial acceleration + but I don't think it will change the problem anyway.

there's no such thing as deceleration
 
What do you mean there is no deceleration? If the force applied produces a larger acceleration than is present. Then that means in effect the force of friction must be slowing the object down or decelerating it.

Is there a better way you could put it? Sometimes I'm a little confusing.
 
Last edited:
trajan22 said:
What do you mean there is no deceleration? If the force applied produces a larger acceleration than is present. Then that means in effect the force of friction must be slowing the object down or decelerating it.

acceleration can be positive or negative, deceleration is a misnomer
 
Ok Ill concede. But I was just trying to avoid directional confusion. Since the problem starts with an - direction for acceleration, and to simplify it I did the problem assuming a positive acceleration.

Edit* there it should be fixed.
 

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