Maximizing Friction: A Force Analysis Approach to Solving Problems

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The discussion revolves around solving a physics problem involving friction and acceleration of a system of blocks. Participants clarify the equations for frictional force and acceleration, emphasizing that the blocks can move with the same acceleration due to static friction. The key equations derived include F=2(M+m)a and fs=(2m+M)a, highlighting the relationship between mass, friction, and acceleration. A suggestion is made to draw a free body diagram to better understand the forces at play. Overall, the conversation focuses on understanding the dynamics of the system and the role of friction in determining motion.
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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!
 

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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?
 
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?)
 
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