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Question Number 24482 by Tinkutara last updated on 18/Nov/17

Neglecting friction and mass of pulleys,  what is the acceleration of mass B?

$$\mathrm{Neglecting}\:\mathrm{friction}\:\mathrm{and}\:\mathrm{mass}\:\mathrm{of}\:\mathrm{pulleys}, \\ $$$$\mathrm{what}\:\mathrm{is}\:\mathrm{the}\:\mathrm{acceleration}\:\mathrm{of}\:\mathrm{mass}\:{B}? \\ $$

Commented by ajfour last updated on 18/Nov/17

Let tension in string on which  B hangs be T.  a_B  = 2a_A          ....(i)  mg−T = m(2a_A )         .....(ii)  2T−mg = ma_A              .....(iii)  ⇒   adding 2×(ii) and (iii),  mg = 5ma_A   ⇒   a_B  = 2a_A  = ((2g)/5) .

$${Let}\:{tension}\:{in}\:{string}\:{on}\:{which} \\ $$$${B}\:{hangs}\:{be}\:{T}. \\ $$$${a}_{{B}} \:=\:\mathrm{2}{a}_{{A}} \:\:\:\:\:\:\:\:\:....\left({i}\right) \\ $$$${mg}−{T}\:=\:{m}\left(\mathrm{2}{a}_{{A}} \right)\:\:\:\:\:\:\:\:\:.....\left({ii}\right) \\ $$$$\mathrm{2}{T}−{mg}\:=\:{ma}_{{A}} \:\:\:\:\:\:\:\:\:\:\:\:\:.....\left({iii}\right) \\ $$$$\Rightarrow\:\:\:{adding}\:\mathrm{2}×\left({ii}\right)\:{and}\:\left({iii}\right), \\ $$$${mg}\:=\:\mathrm{5}{ma}_{{A}} \\ $$$$\Rightarrow\:\:\:{a}_{{B}} \:=\:\mathrm{2}{a}_{{A}} \:=\:\frac{\mathrm{2}{g}}{\mathrm{5}}\:. \\ $$

Commented by Tinkutara last updated on 18/Nov/17

Commented by mrW1 last updated on 18/Nov/17

a_B =2a_A     m_B g−T_B =m_B a_B   ⇒T_B =m_B (g−a_B )  T_A −m_A g=m_A a_A   ⇒T_A =m_A (a_A +g)=m_A ((a_B /2)+g)    T_A =2T_B   m_A ((a_B /2)+g)=2m_B (g−a_B )  (m_A +4m_B )a_B =2(2m_B −m_A )g  ⇒a_B =((2(2m_B −m_A ))/(m_A +4m_B ))×g  with m_A =m_B =m  ⇒a_B =((2(2−1))/(1+4))×g=0.4g    if m_A =2m_B   ⇒a_B =0, i.e. no motion.  for m_A >2m_B   ⇒a_B <0, i.e. block B moves upwards.

$${a}_{{B}} =\mathrm{2}{a}_{{A}} \\ $$$$ \\ $$$${m}_{{B}} {g}−{T}_{{B}} ={m}_{{B}} {a}_{{B}} \\ $$$$\Rightarrow{T}_{{B}} ={m}_{{B}} \left({g}−{a}_{{B}} \right) \\ $$$${T}_{{A}} −{m}_{{A}} {g}={m}_{{A}} {a}_{{A}} \\ $$$$\Rightarrow{T}_{{A}} ={m}_{{A}} \left({a}_{{A}} +{g}\right)={m}_{{A}} \left(\frac{{a}_{{B}} }{\mathrm{2}}+{g}\right) \\ $$$$ \\ $$$${T}_{{A}} =\mathrm{2}{T}_{{B}} \\ $$$${m}_{{A}} \left(\frac{{a}_{{B}} }{\mathrm{2}}+{g}\right)=\mathrm{2}{m}_{{B}} \left({g}−{a}_{{B}} \right) \\ $$$$\left({m}_{{A}} +\mathrm{4}{m}_{{B}} \right){a}_{{B}} =\mathrm{2}\left(\mathrm{2}{m}_{{B}} −{m}_{{A}} \right){g} \\ $$$$\Rightarrow{a}_{{B}} =\frac{\mathrm{2}\left(\mathrm{2}{m}_{{B}} −{m}_{{A}} \right)}{{m}_{{A}} +\mathrm{4}{m}_{{B}} }×{g} \\ $$$${with}\:{m}_{{A}} ={m}_{{B}} ={m} \\ $$$$\Rightarrow{a}_{{B}} =\frac{\mathrm{2}\left(\mathrm{2}−\mathrm{1}\right)}{\mathrm{1}+\mathrm{4}}×{g}=\mathrm{0}.\mathrm{4}{g} \\ $$$$ \\ $$$${if}\:{m}_{{A}} =\mathrm{2}{m}_{{B}} \\ $$$$\Rightarrow{a}_{{B}} =\mathrm{0},\:{i}.{e}.\:{no}\:{motion}. \\ $$$${for}\:{m}_{{A}} >\mathrm{2}{m}_{{B}} \\ $$$$\Rightarrow{a}_{{B}} <\mathrm{0},\:{i}.{e}.\:{block}\:{B}\:{moves}\:{upwards}. \\ $$

Commented by Tinkutara last updated on 19/Nov/17

Thank you very much Sirs!

$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{Sirs}! \\ $$

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