Coefficient of friction for a shoe sole

[chrispat]

Homework Statement

A person of mass 60kg standing on a horizontal floor attempts to push a large wheeled crate of mass 800kg, which can roll on the same horizontal floor with coefficient of rolling friction equal to ur=0.03. What is the minimum coefficient of friction required between the soles of the person's shoes and the horizontal floor if the person is to succeed in keeping the crate rolling at a constant speed without slipping herself?

I don't understand what force is moving the person forward but I have attempted a solution

Homework Equations

For the crate:

Ff<---crate--------->Fpush

Fnet=Fpush-Ff=0

For the shoe shoes:

Ff<---ss---------->Fpush

Fnet=Fpush-Ff=0

The Attempt at a Solution

For the crate:

Ff=(ur)(m)(g)
=(0.03)(800kg)(9.81m/s^2)=235N

For the shoe soles:

Fnet=(235N)-(us)(60kg)(9.81m/s^2)
us=235/(60)(9.81)=0.399

 

[LowlyPion]

Homework Statement

A person of mass 60kg standing on a horizontal floor attempts to push a large wheeled crate of mass 800kg, which can roll on the same horizontal floor with coefficient of rolling friction equal to ur=0.03. What is the minimum coefficient of friction required between the soles of the person's shoes and the horizontal floor if the person is to succeed in keeping the crate rolling at a constant speed without slipping herself?

I don't understand what force is moving the person forward but I have attempted a solution

Homework Equations

For the crate:

Ff<---crate--------->Fpush

Fnet=Fpush-Ff=0

For the shoe shoes:

Ff<---ss---------->Fpush

Fnet=Fpush-Ff=0

The Attempt at a Solution

For the crate:
Ff=(ur)(m)(g)
=(0.03)(800kg)(9.81m/s^2)=235N

For the shoe soles:
Fnet=(235N)-(us)(60kg)(9.81m/s^2)
us=235/(60)(9.81)=0.399

Welcome to PF.

Looks to me like you understood what needed to happen.
Your method looks good. I won't check the math.