Q.Ten identical masses (m each) are connected one below the other with 10 strings. Holding the topmost string, the system is accelerated upwards with acceleration g/2. What is the tension in the 6th string from the top (the topmost string being the first string)? [Ans (as printed in the book): 6mg]

[!TLDR] Using the natural reading (string 1 = topmost, holding all 10 masses; string n holds the 11-n masses below it), string 6 holds 5 masses, giving tension $T_6=5m\left(g+\frac{g}{2}\right)=7.5mg$; the book's printed answer of 6mg appears to assume a different string-num …

The system of 10 identical masses (each m), connected by 10 strings and accelerated upward at $a=\frac{g}{2}$, has an effective per-mass 'weight' during acceleration of $m\left(g+\frac{g}{2}\right)=\frac{3mg}{2}$ (since accelerating upward increases the tension needed, exactly like the apparent-weight-gain case of 4.5.3). Reading the string numbering as stated -- 'topmost string being the first string' -- the most natural interpretation is: string 1 (topmost, held by the external hand/support) carries the full weight of all 10 masses below it; string 2 (between the 1st and 2nd mass) carries 9 masses; and in general string n carries $(11-n)$ masses below it. On this reading, string 6 carries $(11-6)=5$ masses, so its tension is $T_6=5\times\frac{3mg}{2}=7.5mg$. …

Identify how many of the 10 masses hang below the 6th string under the stated numbering convention, then apply $T=(\text{ …

Miscounting how many masses hang below a given string number, since the problem's numbering convention (whether the topmost support itself counts a …