Question Number 38419 by Rio Mike last updated on 25/Jun/18
Answered by tanmay.chaudhury50@gmail.com last updated on 25/Jun/18
$${m}_{{B}} {g}−{T}−{friction}={m}_{{B}} {a} \\ $$$${T}−\mu{m}_{{A}} {g}={m}_{{A}} {a} \\ $$$${as}\:{per}\:{question}\:{block}\:{B}\:{hangs}\:{freely}\:{so} \\ $$$${friction}\:{between}\:{wooden}\:{big}\:{block}\:{and}\:{mass} \\ $$$${m}_{{B}} \:{is}\:{considered}\:{zero} \\ $$$${m}_{{B}} {g}−{T}={m}_{{B}} {a} \\ $$$${T}−\mu{m}_{{A}} {g}={m}_{{A}} {a} \\ $$$${adding} \\ $$$${m}_{{B}} {g}−\mu{m}_{{A}} {g}={a}\left({m}_{{A}} +{m}_{{B}} \right) \\ $$$$\mathrm{3}×\mathrm{10}−\mu×\mathrm{2}×\mathrm{10}=\mathrm{4}\left(\mathrm{3}+\mathrm{2}\right) \\ $$$$\mathrm{30}−\mathrm{20}=\mathrm{20}\mu \\ $$$$\mu=\frac{\mathrm{10}}{\mathrm{20}}=\mathrm{0}.\mathrm{5} \\ $$$${T}={m}_{{B}} {g}−{m}_{{B}} {a}=\mathrm{3}\left(\mathrm{10}−\mathrm{4}\right)=\mathrm{18}{N} \\ $$