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This is a question I found in my old high-school textbook(I'm revising the topics for a course).Two blocks A and B of masses $1kg$ and $2kg$ are placed on a fixed triangular wedge by a massless in-extensible string as shown. enter image description here

The pulley is massless and friction-less. The coefficient of friction between block A and wedge is $\mu _1 = 2/3$ and that between B and wedge is $\mu _2 = 1/3$ (let's neglect the difference between static and kinetic friction).The inclines are at $45 ^\circ $ with the horizontal.We need to find the frictional force exerted on each block,and the tension in the string.

I worked out the problem,and I thoroughly checked my steps.Since the maximum resistive(friction) force is higher than the components of weights of the bodies along the inclines,there should be a state of equilibrium.I got the equations: $$T= \frac {10}{\sqrt2}+f_A \space .........(1)$$ $$T+f_B=\frac {20}{\sqrt 2} \space .........(2)$$ $$=> f_A +f_B = \frac {10}{\sqrt 2} \space ....(3); f_A-f_B=2T-\frac{30}{\sqrt 2} \space ....(4)$$ $$max(f_A)=max(f_B)=\frac{10\sqrt2}{3}\space .........(5)$$ (Considering $g=10 m/s^2$) Where $f_A$ and $f_B$ are frictional forces experienced by blocks $ A$ and $B$. There seems to be no way to get exact values of $f_A$ and $f_B$.Still,the book says(only numerical answers are given in the end,no complete solution) that $$f_A=\frac {10}{3\sqrt2} , f_B=\frac{10\sqrt2}{3} and \space T= \frac{40}{3\sqrt2} $$ All I can see by looking at the book's answers is that the author simply guessed that let there exist limiting(max) friction between $B$ and the wedge,and he might have put the value $f_B= max(f_B)$ and solved.In that case,why not the friction between $A$ and the wedge be limiting and the answers be $$f_A=\frac{10\sqrt2}{3} , f_B=\frac {10}{3\sqrt2} and \space T= \frac{50}{3\sqrt2} $$ But when we look at eq$(3)$,we see that neither of the frictions needs to be limiting,their sum needs to satisfy $\frac{10}{\sqrt2}$. Any way to justify the book's answers?

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up vote 1 down vote accepted

Firstly, block B is the heavier of the two and coefficient of friction is lower on B's side. If either of the blocks moved, it has to be B. Given that B has a tendency to move first and its side has lower coefficient of friction, friction on that side would reach a limiting value before A.

This is what the author must have assumed.

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Friction of B must be at its maximum otherwise there would not be tension in the string. Imagine there was no tension in the string first, and increase the tension till the equilibrium can be reached. Or you can think of the same problem but on a plane instead of a wedge.

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"there would not be tension"?Consider the case of friction on A being at its maximum,the tension would exist,right?Moreover,even if the surfaces were frictionless,tension would have existed. Talking about increasing tension while keeping $F_B$ constant at its max doesn't have to be the key condition for achieving equilibrium,right?We can have,say,some optimum values for both the frictions,such that they satisfy all the equations above. – scienceauror Sep 22 '13 at 16:00

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