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## Homework Statement

Kleppner and Kolenkow "An Introduction to Mechanics (2nd ed.)" prob. 3.2:

Mass M

_{A}= 4 kg rests on top of mass M

_{B}= 5 kg that rests on a frictionless table. The coefficient of friction between the two blocks is such that the blocks just start to slip when the horizontal force F applied to the lower block is 27 N. Suppose that now a horizontal force is applied to the upper block. What is its maximum value for the blocks to slide without slipping relative to each other?

Hint: if F = 30 N, M

_{A}= 5 kg, M

_{B}= 6 kg, then F' = 25 N

## Homework Equations

##F = ma## and ##f = μmg##

## The Attempt at a Solution

So I start of finding μ with the first system where block A is just about to slide and only friction acts on it (I assume the acceleration for both blocks is equal). The two equations with the two unknowns:

## f=\mu M_A g=M_Aa_1## and ##F-f=F-\mu M_A g = M_Ba_1## which solving for ##\mu## gives ##μ=\frac{F}{(M_A+M_B)g}##

Now the second system where block A has a force applied to it making the force of friction act against it. Since it it just about to slip I assume friction should be at its maximum (i.e. ##f=-\mu M_Ag##). I assume the accelerations are equal as well. The equations are:

##F'-f=M_Aa_2## and ##F+f=M_Ba_2## solving for ##F'## (by substituting ##a_2##) yields:

##F'=f+ M_A(\frac{F+f}{M_B})=\mu M_Ag+ M_A(\frac{F+\mu M_Ag}{M_B})## finally substituting ##\mu## and simplifying:

##F'=\frac{2M_A}{M_B}F## which does not agree with the hint. Any help is welcome.

Thanks in advanced.