# Maximum frictional force experienced and the acceleration

1. Apr 22, 2013

### gerard.caleb

1. The problem statement, all variables and given/known data

2. A block of mass 35 kg is on a flat floor with a static friction coefficient of 0.36 and a kinetic friction coefficient of 0.29. A person attempts to slide the box from rest in a forward direction. Find the maximum frictional force experienced and the acceleration of the box once it is moving if:
a) the person pushes horizontally with 120 N.

m=35kg
μs=0.36
μk=0.29

2. Relevant equations
Fnet=m*a
Fnorm=g*m
μk=Fnet/Fnorm

3. The attempt at a solution

*Disclaimer: I'm taking an online class with very few sample questions so I will not be held liable for any injuries that occur due to facepalming at my attempts to solve these*
Fn=m*g
Fn=35kg*9.8kg/N
Fn= 343N

F=Fn*μs
F=343N*0.36
F=123.48N

So 123.48N is the maximum friction force experienced?
And the acceleration can be found from:

We know uK=0.29
And uK=Fnet/Fnorm

So 0.29=Fnet/343N
0.29*343N=Fnet
99.47N=Fnet

We know Fnet=m*a, so:
99.47N=35kg*a
a=2.842m/s^2

So acceleration is 2.842m/s^2?

I am 20000% percent sure that I have these wrong because I didn't include the applied force given in the question at all (among other reasons).
I have a few other questions similar to this but involving angles and different applied forces, I'm sure once I understand how to do this one I can do the others, but I thought the most basic one would be a good starting place for someone here to point out my errors.

Thank you!

2. Apr 22, 2013

### BruceW

You've done a good attempt. But it is not a totally correct answer. You are correct that 123 N is the max static frictional force experienced. And you have calculated the correct force due to kinetic friction. But this is not the net force, since there is also the applied force (as you said).

3. Apr 22, 2013

### CWatters

Unless I've made a mistake I believe you need to have a rethink. Try these questions in order..

What's the force required to overcome static friction and start it moving?
How much force does the man actually apply?
Bearing in mind newtons third law...what's the maximium frictional force?

4. Apr 22, 2013

### CWatters

I dissagree :-)

5. Apr 22, 2013

### gerard.caleb

Thank you so much for your response! I'm sorry for the late reply. Thank you to the other person who responded as well, I'd respond individually but I'm not sure if that's proper conduct on this board.

Ffr = μs*Fn
Ffr = 343N * 0.36
Ffr= 123.48N

As he only applied 120N, he doesn't overcome this and therefore the block doesn't move (wow, I really wish I thought this out before I wrote all that!)

I'm still a little confused about maximum frictional force though, I've been reading and a few things have said the same thing as this: http://www.sfu.ca/~boal/101lecs/101lec6.pdf
"The maximum force of static friction, Fmax, is Fmax= µs*N,where µs is the coefficient of static friction." I'm confused what the difference between maximum friction force and maximum force of static friction is.

6. Apr 22, 2013

### BruceW

Oh yeah, duh! Thanks, CWatters. I was misled by the question using the word 'maximum', and I guess gerard was too. So, the question says: "Find the maximum frictional force experienced" But what they mean is find the actual frictional force experienced. I hope this helps, gerard.

7. Apr 23, 2013

### CWatters

Yes what Bruce said, it's the difference between...

1) The maximum frictional force that will occur if enough force is applied to move the box and
2) The actual maximum frictional force in this particular set up.

We've worked out that the box doesn't move so draw a free body diagram and then ask... What do the forces sum to if the box doesn't move. The hint I gave earlier was to remember newtons third law.