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Question
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 (Topmost string being the first string)?
Solution
Consider the 6th string from the top. The number of masses below the 6th string is 5. Thus, FBD for the 6th mass is given in the following figure.
The force equation for that mass is,
5 ma = T – 5 mg
As, a = `"g"/2`
T = `"5mg"/2 + "5mg"`
`= 5 "mg"(1/2 + 1)`
`= (15"mg")/2`
= 7.5 mg
Tension in the 6th string is 7.5 mg.
Notes
In textbook answer given is incorrect.
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