What Is the Maximum Force on the Upper Block to Prevent Slipping?

[Born]

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.

 

[Tanya Sharma]

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.

Nice work ! Your work is correct under the assumption that F continues to act when F' is applied on the upper block This is not the case here .

On first attempt I got the same answer as yours :) .I also interpreted the problem as you did .But looking at the hint ,it seems that the statement needs to be interpreted somewhat differently.

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?

This means that now F' is the only horizontal force acting on the system of two blocks and F=27N ceases to act on the lower block . In this case you should get $F'=\frac{M_A}{M_B}F$ which agrees with the hint.

 

[Born]

Got it. Thanks! :)