Homework Help: Two blocks stacked on one another

1. Feb 19, 2006

pureouchies4717

question:

The coefficient of static friction is 0.60 between the two blocks in figure. The coefficient of kinetic friction between the lower block and the floor is 0.20. Force F_vec causes both blocks to cross a distance of 5.0 m, starting from rest.What is the least amount of time in which this motion can be completed without the top block sliding on the lower block?

this is a very tricky question

force of friction for lower block: Ff= 13.72N

Fsmax= 23.52

so the force being exerted has to have a max of 9.8N

F=ma
9.8= 7a
a=1.4 m/s^2

i have no idea how i can get the time from this

Last edited: Feb 19, 2006
2. Feb 19, 2006

Hootenanny

Staff Emeritus
could you post some of your working / thoughts?

3. Feb 19, 2006

pureouchies4717

so here, i tried to incorporate it into the kinematics equations

http://www.glenbrook.k12.il.us/GBSSCI/PHYS/Class/1DKin/U1L6a1.gif [Broken]

5=.5(a)t^2
and got a time of 2.67s, which is wrong

Last edited by a moderator: May 2, 2017
4. Feb 19, 2006

Integral

Staff Emeritus
Where did the friction enter into the problem?

You have jumped a step to far in your choice of equations. Start from F=ma, you should have something in your text relating the friction to this fundamental starting point. You have 2 friction problems to do here. The first will tell you the maximum acceleration you can have without the top block slipping. Then that acceleration and the friction between the floor and the bottom block will give you the time you need.

5. Feb 19, 2006

pureouchies4717

thanks for the response

so

a=fmax/m
= 5.88m/s^2
d= .5at^2
5= .5(5.88)(t^2)
t= 1.304 (wrong)

Last edited: Feb 19, 2006
6. Feb 19, 2006

Hootenanny

Staff Emeritus
Consider the maximum force applicable to the blocks : $F = 0.6 \times 4 = 2.4$. Now apply this to the body of two block (m = 7) to give you the maximum force, which will allow you to calc. the acceleration.

7. Feb 19, 2006

pureouchies4717

guys, thanks alot for helping

ok so f=ma

2.4=7a
a= .349 m/s^2

d=.5at^2
t= 5.4s (this comes out to the wrong answer)

sorry to bother you guys..

8. Feb 19, 2006

Hootenanny

Staff Emeritus
You forgot to factor in the force of friction during movement.

9. Feb 19, 2006

pureouchies4717

ah yes! thanks sooooo much

it came out to 1.75

:!!)

10. Oct 1, 2006

kmt271

Hey, I have the same question but with different values, I'm trying to do it using this method and it just won't work.
My values are:
Box on top m=2.73kg
Box on bottom m=1.41kg
Co-efficient static = 0.641
Co-efficient kinetic = 0.116
Distance = 4.25m

My work:
Force max = fs = 0.641(9.8)(1.41)
= 8.857N
Force of kinetic friction = 0.116(9.8)(1.41+2.73)
= 4.706N
Fnet = 8.857 - 4.706
= 4.15N

4.15 = 4.14a
a = 1.00m/s/s
d = 0.5(1)t^2
t = 2.91 s - Wrong!

Can anyone spot the bugger?

11. Oct 21, 2010

pointguard 2

"You forgot to factor in the force of friction during movement. "

12. Oct 6, 2011

OleWik

Yeah, bump! hehe, id like to know the last "formula" there. Thanks

13. Oct 6, 2011

Staff: Mentor

Check this.

14. Oct 6, 2011

Staff: Mentor

Where do you mean "there"?

15. Oct 21, 2012

jdsp23

How did you get this ? :S I've tried everything.
Please explain to me how you did this.

16. Oct 21, 2012

jdsp23

How did you get to this? How did you factor the kinetic friction ?
i don't understand what just happened...

17. Oct 21, 2012

Staff: Mentor

Hi jdsp23. Welcome to physics forums.

The message to which you refer has a date on it indicating it was posted in 2006. Its author, nick727, has not logged in to his physics forums account here for 3 years. It's likely he no longer reads this forum, so your expectations should not be high of receiving a reply from him.

18. Oct 28, 2012

stagename

And what is it is asking to cross a distance "d" but without providing one?