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Question Number 224466 by Jubr last updated on 13/Sep/25
Commented by fantastic last updated on 14/Sep/25
the acceleration a=((4m_1 m_2 +m_0 (m_1 −m_2 ))/(4m_1 m_2 +m_0 (m_1 +m_2 )))g
$${the}\:{acceleration}\: \\ $$$${a}=\frac{\mathrm{4}{m}_{\mathrm{1}} {m}_{\mathrm{2}} +{m}_{\mathrm{0}} \left({m}_{\mathrm{1}} −{m}_{\mathrm{2}} \right)}{\mathrm{4}{m}_{\mathrm{1}} {m}_{\mathrm{2}} +{m}_{\mathrm{0}} \left({m}_{\mathrm{1}} +{m}_{\mathrm{2}} \right)}{g} \\ $$
Commented by fantastic last updated on 14/Sep/25
the answer in your book is with respect to m_0
$${the}\:{answer}\:{in}\:{your}\:{book} \\ $$$${is}\:{with}\:{respect}\:{to}\:{m}_{\mathrm{0}} \\ $$
Answered by mr W last updated on 14/Sep/25
Commented by mr W last updated on 14/Sep/25
T_1 =2T_2 T_1 =M_0 a ⇒T_2 =((M_0 a)/2) M_1 g−T_2 =M_1 (a+Δa) M_1 g−((M_0 a)/2)=M_1 (a+Δa) ⇒(2M_1 +M_0 )a+2M_1 Δa=2M_1 g ...(i) M_2 g−T_2 =M_2 (a−Δa) M_2 g−((M_0 a)/2)=M_2 (a−Δa) ⇒(2M_2 +M_0 )a−2M_2 Δa=2M_2 g ...(ii) solving (i) and (ii) we get ⇒a=((4M_1 M_2 g)/(4M_1 M_2 +M_0 (M_1 +M_2 ))) acc. of M_1 is a+Δa=g−((M_0 a)/(2M_1 )) =g−((2M_0 M_2 g)/(4M_1 M_2 +M_0 (M_1 +M_2 ))) =((4M_1 M_2 +M_0 (M_1 −M_2 ))/(4M_1 M_2 +M_0 (M_1 +M_2 )))g acc. of M_2 is a−Δa=g−((M_0 a)/(2M_2 )) =g−((2M_0 M_1 g)/(4M_1 M_2 +M_0 (M_1 +M_2 ))) =((4M_1 M_2 −M_0 (M_1 −M_2 ))/(4M_1 M_2 +M_0 (M_1 +M_2 )))g
$${T}_{\mathrm{1}} =\mathrm{2}{T}_{\mathrm{2}} \\ $$$${T}_{\mathrm{1}} ={M}_{\mathrm{0}} {a} \\ $$$$\Rightarrow{T}_{\mathrm{2}} =\frac{{M}_{\mathrm{0}} {a}}{\mathrm{2}} \\ $$$${M}_{\mathrm{1}} {g}−{T}_{\mathrm{2}} ={M}_{\mathrm{1}} \left({a}+\Delta{a}\right) \\ $$$${M}_{\mathrm{1}} {g}−\frac{{M}_{\mathrm{0}} {a}}{\mathrm{2}}={M}_{\mathrm{1}} \left({a}+\Delta{a}\right) \\ $$$$\Rightarrow\left(\mathrm{2}{M}_{\mathrm{1}} +{M}_{\mathrm{0}} \right){a}+\mathrm{2}{M}_{\mathrm{1}} \Delta{a}=\mathrm{2}{M}_{\mathrm{1}} {g}\:\:\:…\left({i}\right) \\ $$$${M}_{\mathrm{2}} {g}−{T}_{\mathrm{2}} ={M}_{\mathrm{2}} \left({a}−\Delta{a}\right) \\ $$$${M}_{\mathrm{2}} {g}−\frac{{M}_{\mathrm{0}} {a}}{\mathrm{2}}={M}_{\mathrm{2}} \left({a}−\Delta{a}\right) \\ $$$$\Rightarrow\left(\mathrm{2}{M}_{\mathrm{2}} +{M}_{\mathrm{0}} \right){a}−\mathrm{2}{M}_{\mathrm{2}} \Delta{a}=\mathrm{2}{M}_{\mathrm{2}} {g}\:\:\:…\left({ii}\right) \\ $$$${solving}\:\left({i}\right)\:{and}\:\left({ii}\right)\:{we}\:{get} \\ $$$$\Rightarrow{a}=\frac{\mathrm{4}{M}_{\mathrm{1}} {M}_{\mathrm{2}} {g}}{\mathrm{4}{M}_{\mathrm{1}} {M}_{\mathrm{2}} +{M}_{\mathrm{0}} \left({M}_{\mathrm{1}} +{M}_{\mathrm{2}} \right)} \\ $$$${acc}.\:{of}\:{M}_{\mathrm{1}} \:{is} \\ $$$${a}+\Delta{a}={g}−\frac{{M}_{\mathrm{0}} {a}}{\mathrm{2}{M}_{\mathrm{1}} } \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:={g}−\frac{\mathrm{2}{M}_{\mathrm{0}} {M}_{\mathrm{2}} {g}}{\mathrm{4}{M}_{\mathrm{1}} {M}_{\mathrm{2}} +{M}_{\mathrm{0}} \left({M}_{\mathrm{1}} +{M}_{\mathrm{2}} \right)} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{4}{M}_{\mathrm{1}} {M}_{\mathrm{2}} +{M}_{\mathrm{0}} \left({M}_{\mathrm{1}} −{M}_{\mathrm{2}} \right)}{\mathrm{4}{M}_{\mathrm{1}} {M}_{\mathrm{2}} +{M}_{\mathrm{0}} \left({M}_{\mathrm{1}} +{M}_{\mathrm{2}} \right)}{g} \\ $$$${acc}.\:{of}\:{M}_{\mathrm{2}} \:{is} \\ $$$${a}−\Delta{a}={g}−\frac{{M}_{\mathrm{0}} {a}}{\mathrm{2}{M}_{\mathrm{2}} } \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:={g}−\frac{\mathrm{2}{M}_{\mathrm{0}} {M}_{\mathrm{1}} {g}}{\mathrm{4}{M}_{\mathrm{1}} {M}_{\mathrm{2}} +{M}_{\mathrm{0}} \left({M}_{\mathrm{1}} +{M}_{\mathrm{2}} \right)} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{4}{M}_{\mathrm{1}} {M}_{\mathrm{2}} −{M}_{\mathrm{0}} \left({M}_{\mathrm{1}} −{M}_{\mathrm{2}} \right)}{\mathrm{4}{M}_{\mathrm{1}} {M}_{\mathrm{2}} +{M}_{\mathrm{0}} \left({M}_{\mathrm{1}} +{M}_{\mathrm{2}} \right)}{g} \\ $$
Commented by fantastic last updated on 14/Sep/25
yes
$${yes} \\ $$
Commented by mr W last updated on 14/Sep/25
bonus question: M_3 ≠0
$${bonus}\:{question}:\:{M}_{\mathrm{3}} \neq\mathrm{0} \\ $$
Commented by mr W last updated on 14/Sep/25
Commented by mr W last updated on 16/Sep/25
T_1 =M_0 a M_3 g+2T_2 −T_1 =M_3 a 2T_2 =(M_0 +M_3 )a−M_3 g ⇒T_2 =(((M_0 +M_3 )a−M_3 g)/2) M_1 g−T_2 =M_1 (a+Δa) M_1 g−(((M_0 +M_3 )a−M_3 g)/2)=M_1 (a+Δa) ⇒(M_0 +M_3 +2M_1 )a+2M_1 Δa=(2M_1 +M_3 )g ...(i) M_2 g−T_2 =M_2 (a−Δa) M_2 g−(((M_0 +M_3 )a−M_3 g)/2)=M_2 (a−Δa) ⇒(M_0 +M_3 +2M_2 )a−2M_2 Δa=(2M_2 +M_3 )g ...(ii) solving (i) and (ii) we get ⇒a=(([4M_1 M_2 +(M_1 +M_2 )M_3 ]g)/(4M_1 M_2 +(M_0 +M_3 )(M_1 +M_2 ))) acc. of M_1 is a+Δa=g−(((M_0 +M_3 )a−M_3 g)/(2M_1 )) =(1+(M_3 /(2M_1 )))g−(((M_0 +M_3 )[4M_1 M_2 +(M_1 +M_2 )M_3 ]g)/(2M_1 [4M_1 M_2 +(M_0 +M_3 )(M_1 +M_2 )])) =(([4M_1 M_2 +M_0 (M_1 −M_2 )+M_3 (M_1 +M_2 )]g)/(4M_1 M_2 +(M_0 +M_3 )(M_1 +M_2 ))) similarly acc. of M_2 is a−Δa=(([4M_1 M_2 −M_0 (M_1 −M_2 )+M_3 (M_1 +M_2 )]g)/(4M_1 M_2 +(M_0 +M_3 )(M_1 +M_2 )))
$${T}_{\mathrm{1}} ={M}_{\mathrm{0}} {a} \\ $$$${M}_{\mathrm{3}} {g}+\mathrm{2}{T}_{\mathrm{2}} −{T}_{\mathrm{1}} ={M}_{\mathrm{3}} {a} \\ $$$$\mathrm{2}{T}_{\mathrm{2}} =\left({M}_{\mathrm{0}} +{M}_{\mathrm{3}} \right){a}−{M}_{\mathrm{3}} {g} \\ $$$$\Rightarrow{T}_{\mathrm{2}} =\frac{\left({M}_{\mathrm{0}} +{M}_{\mathrm{3}} \right){a}−{M}_{\mathrm{3}} {g}}{\mathrm{2}} \\ $$$${M}_{\mathrm{1}} {g}−{T}_{\mathrm{2}} ={M}_{\mathrm{1}} \left({a}+\Delta{a}\right) \\ $$$${M}_{\mathrm{1}} {g}−\frac{\left({M}_{\mathrm{0}} +{M}_{\mathrm{3}} \right){a}−{M}_{\mathrm{3}} {g}}{\mathrm{2}}={M}_{\mathrm{1}} \left({a}+\Delta{a}\right) \\ $$$$\Rightarrow\left({M}_{\mathrm{0}} +{M}_{\mathrm{3}} +\mathrm{2}{M}_{\mathrm{1}} \right){a}+\mathrm{2}{M}_{\mathrm{1}} \Delta{a}=\left(\mathrm{2}{M}_{\mathrm{1}} +{M}_{\mathrm{3}} \right){g}\:\:\:…\left({i}\right) \\ $$$${M}_{\mathrm{2}} {g}−{T}_{\mathrm{2}} ={M}_{\mathrm{2}} \left({a}−\Delta{a}\right) \\ $$$${M}_{\mathrm{2}} {g}−\frac{\left({M}_{\mathrm{0}} +{M}_{\mathrm{3}} \right){a}−{M}_{\mathrm{3}} {g}}{\mathrm{2}}={M}_{\mathrm{2}} \left({a}−\Delta{a}\right) \\ $$$$\Rightarrow\left({M}_{\mathrm{0}} +{M}_{\mathrm{3}} +\mathrm{2}{M}_{\mathrm{2}} \right){a}−\mathrm{2}{M}_{\mathrm{2}} \Delta{a}=\left(\mathrm{2}{M}_{\mathrm{2}} +{M}_{\mathrm{3}} \right){g}\:\:\:…\left({ii}\right) \\ $$$${solving}\:\left({i}\right)\:{and}\:\left({ii}\right)\:{we}\:{get} \\ $$$$\Rightarrow{a}=\frac{\left[\mathrm{4}{M}_{\mathrm{1}} {M}_{\mathrm{2}} +\left({M}_{\mathrm{1}} +{M}_{\mathrm{2}} \right){M}_{\mathrm{3}} \right]{g}}{\mathrm{4}{M}_{\mathrm{1}} {M}_{\mathrm{2}} +\left({M}_{\mathrm{0}} +{M}_{\mathrm{3}} \right)\left({M}_{\mathrm{1}} +{M}_{\mathrm{2}} \right)} \\ $$$${acc}.\:{of}\:{M}_{\mathrm{1}} \:{is} \\ $$$${a}+\Delta{a}={g}−\frac{\left({M}_{\mathrm{0}} +{M}_{\mathrm{3}} \right){a}−{M}_{\mathrm{3}} {g}}{\mathrm{2}{M}_{\mathrm{1}} } \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:=\left(\mathrm{1}+\frac{{M}_{\mathrm{3}} }{\mathrm{2}{M}_{\mathrm{1}} }\right){g}−\frac{\left({M}_{\mathrm{0}} +{M}_{\mathrm{3}} \right)\left[\mathrm{4}{M}_{\mathrm{1}} {M}_{\mathrm{2}} +\left({M}_{\mathrm{1}} +{M}_{\mathrm{2}} \right){M}_{\mathrm{3}} \right]{g}}{\mathrm{2}{M}_{\mathrm{1}} \left[\mathrm{4}{M}_{\mathrm{1}} {M}_{\mathrm{2}} +\left({M}_{\mathrm{0}} +{M}_{\mathrm{3}} \right)\left({M}_{\mathrm{1}} +{M}_{\mathrm{2}} \right)\right]} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\left[\mathrm{4}{M}_{\mathrm{1}} {M}_{\mathrm{2}} +{M}_{\mathrm{0}} \left({M}_{\mathrm{1}} −{M}_{\mathrm{2}} \right)+{M}_{\mathrm{3}} \left({M}_{\mathrm{1}} +{M}_{\mathrm{2}} \right)\right]{g}}{\mathrm{4}{M}_{\mathrm{1}} {M}_{\mathrm{2}} +\left({M}_{\mathrm{0}} +{M}_{\mathrm{3}} \right)\left({M}_{\mathrm{1}} +{M}_{\mathrm{2}} \right)} \\ $$$${similarly} \\ $$$${acc}.\:{of}\:{M}_{\mathrm{2}} \:{is} \\ $$$${a}−\Delta{a}=\frac{\left[\mathrm{4}{M}_{\mathrm{1}} {M}_{\mathrm{2}} −{M}_{\mathrm{0}} \left({M}_{\mathrm{1}} −{M}_{\mathrm{2}} \right)+{M}_{\mathrm{3}} \left({M}_{\mathrm{1}} +{M}_{\mathrm{2}} \right)\right]{g}}{\mathrm{4}{M}_{\mathrm{1}} {M}_{\mathrm{2}} +\left({M}_{\mathrm{0}} +{M}_{\mathrm{3}} \right)\left({M}_{\mathrm{1}} +{M}_{\mathrm{2}} \right)} \\ $$

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