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
​

Answer 1

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

• $\green{\tt{\mu=0.5}}$

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

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

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

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