Finding the force to break static friction

[nevererdofit]

Homework Statement

Samantha and Rebekah each push on a stack of doughnuts equally. If the coefficient of static friction is .28 and the box has a mass of 18kg, what force does each push individually to break static friction? Once it's moving, the coefficient of kinetic friction is .17, how much will the doughnuts be accelerated onto the floor?

mass of object = 18kg
coefficient of friction (static) = .28
coefficient of friction (kinetic) = .17[/B]

Homework Equations

The only one I know that is obvious to use is f = ma, not sure how to apply it to this context[/B]

The Attempt at a Solution

On the x and y plane, I know the force diagram reads normal force traveling up, force of gravity traveling down, Applied force traveling right, and force of friction traveling left.

 

[paisiello2]

Do you know the formula for static and kinetic friction?

 

[nevererdofit]

Do you know the formula for static and kinetic friction?

I was never taught the formulas for each type of friction no

 

[paisiello2]

You need to know them:
http://en.wikipedia.org/wiki/Friction#Static_friction

 

[nevererdofit]

So after trying this is how I attempt it
(I assume gravity as 10m/s/s)

I know normal force = mg so fn = (18)(10) fn = 180N
then
f = (.28)x(180)
f = 50.4N
then I assume you divide the answer in two as it asks for the individual force, so
f = 25.2N

to find the acceleration you just use f = ma substituting your found force
25.2 = (18)a
a = 1.4m/s/s

am I correct?

 

[paisiello2]

Not quite. Is the box sliding or not?

 

[nevererdofit]

Not quite. Is the box sliding or not?

they both push on the stack of doughnuts, so I believe it is sliding. Really though you are just trying to find the force that was needed to break the static force, or when it wasn't in motion. Then the second part of the question just wanted the acceleration of their push

 

[paisiello2]

Yes, so when you found the first answer did you assume the box was sliding or not?

 

[nevererdofit]

yes I did assume the box was sliding as there was force being emitted on it

 

[paisiello2]

Well, you are contradicting yourself because you also said there was no motion. Sliding= motion.

 

[billy_joule]

So after trying this is how I attempt it
(I assume gravity as 10m/s/s)

I know normal force = mg so fn = (18)(10) fn = 180N
then
f = (.28)x(180)
f = 50.4N
then I assume you divide the answer in two as it asks for the individual force, so
f = 25.2N

That looks correct.

to find the acceleration you just use f = ma substituting your found force
25.2 = (18)a
a = 1.4m/s/s

am I correct?

A couple problems with this part. You've ignore kinetic friction force and only used one of the applied forces.