# Friction between objects

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

The frictional force is given as fs. It is equivalent to mgμ.
It is a system of 4 objects in unison, one identical acceleration.

yeah okay, but how to do the question? Can give me a head start? Thanks!

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.

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?

CAF123
Gold Member
If the blocks are all to move with the same acceleration then the small mass $m$ relative to the large mass $M$ must be stationary. I suggest drawing a free body diagram for the $m, M$ system, being careful to get the direction of the frictional force correct.

Net external force applied on a system always equals to total mass of system into its acceleration .

F=2(M+m)a .........(1)
fs=(2m+M)a ........(2)

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?)