Two blocks stacked on one another

In summary, the author is asking for help with a difficult question regarding the motion of two blocks. The coefficient of static friction is 0.60 between the two blocks, while 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. The author asks for help with solving a problem in which the force of friction must be taken into account.
  • #1
pureouchies4717
99
0
hey guys, i was wondering if you could please help me

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?
knight_Figure_08_32.jpg


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:
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  • #2
could you post some of your working / thoughts?
 
  • #3
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:
  • #4
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
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)

:frown:
 
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  • #6
Consider the maximum force applicable to the blocks : [itex] F = 0.6 \times 4 = 2.4 [/itex]. 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
guys, thanks a lot 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
You forgot to factor in the force of friction during movement.
 
  • #9
ah yes! thanks sooooo much

it came out to 1.75

:!)
 
  • #10
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
"You forgot to factor in the force of friction during movement. "

Can you please explain this?
 
  • #12
Yeah, bump! hehe, id like to know the last "formula" there. Thanks
 
  • #13
kmt271 said:
My work:
Force max = fs = 0.641(9.8)(1.41)
Check this.
 
  • #14
OleWik said:
Yeah, bump! hehe, id like to know the last "formula" there. Thanks
Where do you mean "there"?
 
  • #15
nick727kcin said:
ah yes! thanks sooooo much

it came out to 1.75

:!)

How did you get this ? :S I've tried everything.
Please explain to me how you did this.
 
  • #16
nick727kcin said:
ah yes! thanks sooooo much

it came out to 1.75

:!)
How did you get to this? How did you factor the kinetic friction ?
i don't understand what just happened...
 
  • #17
nick727 said:
ah yes! thanks sooooo much

it came out to 1.75

jdsp23 said:
How did you get this ? :S I've tried everything.
Please explain to me how you did this.

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

What is the concept of two blocks stacked on one another?

The concept of two blocks stacked on one another is a physical arrangement where one block is placed on top of another block, creating a vertical structure. This arrangement is commonly used in construction, engineering, and physics experiments.

What is the purpose of stacking two blocks on one another?

The purpose of stacking two blocks on one another is to create a stable structure with increased height. This is useful in building tall structures, such as skyscrapers, and in experiments where a higher platform is needed.

How do the properties of the blocks affect the stability of the stack?

The stability of the stack is affected by the size, shape, and weight of the blocks. Larger, heavier blocks with flat surfaces are more stable than smaller, lighter blocks with irregular shapes. The distribution of weight on the stack also plays a role in its stability.

What is the maximum height that two blocks can be stacked before it becomes unstable?

The maximum height of a stack of two blocks depends on the size and weight of the blocks, as well as the distribution of weight on the stack. Generally, the stack will become unstable when the center of gravity is no longer located above the base of the stack.

How can the stability of a stack of two blocks be increased?

The stability of a stack of two blocks can be increased by using blocks with flat, even surfaces and placing them directly on top of each other in a balanced manner. Adding a third block in the middle of the stack can also help distribute weight and increase stability.

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