Friction between objects

  • #1
dawn_pingpong
50
0

Homework Statement



Please see attached

Homework Equations



Ff=μFn,
F=ma

The Attempt at a Solution


(a) The blocks can have the same acceleration due to static friction.
consider the 2 blocks on the left: Ff=fmg

then consider the 2 blocks on the right: Ff is also fmg? I am rather confused at this point... how does M come into the equation? because the answer for part (a) is 2(M+m)/(M+2m)f. There is no g anywhere in the equation, though from the equation, friction is to the normal, which is mg?

THank you!
 

Attachments

  • friction.png
    friction.png
    30.6 KB · Views: 402

Answers and Replies

  • #2
azizlwl
1,065
10
The frictional force is given as fs. It is equivalent to mgμ.
It is a system of 4 objects in unison, one identical acceleration.
 
  • #3
dawn_pingpong
50
0
yeah okay, but how to do the question? Can give me a head start? Thanks!
 
  • #4
azizlwl
1,065
10
The static friction force created by small mass m on top of the right side will pull both itself and the 2 blocks behind it.
As a body of 4 blocks there is only one the accelerations value.
 
  • #5
dawn_pingpong
50
0
sorry, I don't really get what you mean, please elaborate... So for the small mass on the right, f value must be (m+M)f?
 
  • #6
CAF123
Gold Member
2,950
88
If the blocks are all to move with the same acceleration then the small mass [itex] m [/itex] relative to the large mass [itex] M [/itex] must be stationary. I suggest drawing a free body diagram for the [itex] m, M [/itex] system, being careful to get the direction of the frictional force correct.
 
  • #7
kushan
256
0
Net external force applied on a system always equals to total mass of system into its acceleration .
 
  • #8
azizlwl
1,065
10
F=2(M+m)a ...(1)
fs=(2m+M)a ...(2)
 
  • #9
dawn_pingpong
50
0
Okay, I think I know...

So the whole thing F=2(M+m)a ...(1) (as stated by azizlwl)
the greatest friction is between the blocks on the left, so fs=(2m+M)a ...(2) (thanks azizlwl!), because the bottom block on the right need not be considered. So F/fs=2(M+m)/(2m+M), and F=2(M+m)f/(2m+M)

Then for the 2nd part, F=2(M+m)a ...(1) still,
then f is only for the block M on the right, thus f=Ma,
by the same reasoning, F=2(M+m)f/M.

Yay thanks guys! Any tips for force analysis (and consider which few objects as a system?)
 

Suggested for: Friction between objects

Replies
11
Views
353
  • Last Post
Replies
12
Views
315
Replies
30
Views
602
  • Last Post
Replies
30
Views
853
Replies
23
Views
225
Replies
5
Views
383
Replies
10
Views
5K
Top