Coefficient of Friction (Styrofoam on Water)

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Homework Statement



fgrav = mg
fgrav = (0.0821)(9.8) = 0.80458 N

fnet = ma = (0.0821)(0.0096) = 0.0096 m/s^2

fnorm = fgrav

Homework Equations



kinetic mew = force of friction / fnorm

The Attempt at a Solution



Required: Ffr

kinetic mew = force of friction / fnorm
kinetic mew = force of friction / 0.80458 N

I need to find the friction of the object when it is on the water. The object is made out of styrofoam. I'm completely stumped and I tried looking around internet using the google search engine to find the coefficient of friction of two surfaces.
 
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dashed said:

Homework Statement




fnet = ma = (0.0821)(0.0096) = 0.0096 m/s^2

This does not make much sense...mathematically, it is not true. (0.0821)(0.0096) = 0.00079.

Also...what forces are acting on this styrofoam contributing to fnet? is friction the only force and the styrofoam is actually slowing down? Or are there other forces involved?

If it is the first case, then 0.00079 = force of friction and "mu" can be calculated as per the equation you show. If it is the second case, the force of friction can be calculated using the fnet equation and the values of the other forces involved.
 
Galileo's Ghost said:
This does not make much sense...mathematically, it is not true. (0.0821)(0.0096) = 0.00079.

Also...what forces are acting on this styrofoam contributing to fnet? is friction the only force and the styrofoam is actually slowing down? Or are there other forces involved?

If it is the first case, then 0.00079 = force of friction and "mu" can be calculated as per the equation you show. If it is the second case, the force of friction can be calculated using the fnet equation and the values of the other forces involved.

there is Fa which is opposite of Ffr. However at the moment Fa is unknown.