Question

Consider the arrangement shown. Table top is rough and coefficient of friction between table top and block 2m is μ. Pulley and string are ideal.

Minimum value of μ at table top to keep system at equilibrium, is
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Answer 1

`\blue{\bold{\underline{\underline{Answer:}}}}`

• 
  `\green{\tt{\mu=0.5}}`

`\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}\\`

`\green{\underline{\bold{Given :}}} \\\\ \tt: \implies Mass \: of \: blocks = m \: and \: 2 m \\ \\ \red{\underline{\bold{To \: Find :}}} \\ \\ \tt: \implies Minimum \: value \: of \: \mu \: so \: that \: system \: will \: be \: in \: equilibrium =?`

`\blue{\underline{\bold{Calculation :}}}\\`

`\bold{For \: block \: of \: mass \: m}\\ \\ \tt: \implies mg - T= ma \\ \\ \tt \because \: a = 0 { \: m/s}^(2) \\ \\ \tt: \implies T = mg - - - - - (1) \\ \\ \bold{For \: block \: of \: mass \: 2m} \\\\ \tt: \implies T = fr \:\:\:\:\:\:(for\:equilibrium) \\ \\ \tt: \implies T= \mu N \\ \\ \tt: \implies t = \mu \: 2mg \\ \\ \tt: \implies mg = \mu \: 2mg \\ \\ \tt: \implies \mu = (mg)/(2mg) \\ \\ \green{\tt: \implies \mu = 0.5} \\ \\ \green{ \tt \therefore Minimum \: coefficient \: of \: friction \: is \: 0.5 }`