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Question Number 217640 by Tawa11 last updated on 17/Mar/25
Commented by mr W last updated on 17/Mar/25
F should also be given to get a concrete result.
$${F}\:{should}\:{also}\:{be}\:{given}\:{to}\:{get}\:{a} \\ $$$${concrete}\:{result}. \\ $$
Answered by mahdipoor last updated on 17/Mar/25
for B : (ΣF=ma) (→ ≡ +) F−T_1 =M_1 a_1 =10a_1 (1) for A : T_2 =M_2 a_2 =5a_2 (2) for C : T_3 =M_3 a_3 =2a_3 (3) for pully : a^− =a_1 , { ((a^− +θ^(..) r=a_1 +θ^(..) r=a_2 (4))),((a^− −θ^(..) r=a_1 −θ^(..) r=a_3 (5))) :} (↻ ≡ +) ΣF=ma^− ⇒ T_1 −T_2 −T_3 =2a_1 (6) ΣM^− =I^− θ^(..) ⇒ (T_3 −T_2 )r=((mr^2 )/2)θ^(..) ⇒T_3 −T_2 =rθ^(..) (7) 7 unkown (3T+3a+θ^(..) r) and 7 eq ⇒ a_1 =... a_2 =... a_3 =...
$${for}\:{B}\:\::\:\:\:\left(\Sigma{F}={ma}\right)\:\:\:\left(\rightarrow\:\equiv\:+\right) \\ $$$${F}−{T}_{\mathrm{1}} ={M}_{\mathrm{1}} {a}_{\mathrm{1}} =\mathrm{10}{a}_{\mathrm{1}} \:\:\:\:\left(\mathrm{1}\right) \\ $$$${for}\:{A}\:: \\ $$$${T}_{\mathrm{2}} ={M}_{\mathrm{2}} {a}_{\mathrm{2}} =\mathrm{5}{a}_{\mathrm{2}} \:\:\:\:\:\left(\mathrm{2}\right) \\ $$$${for}\:{C}\:: \\ $$$${T}_{\mathrm{3}} ={M}_{\mathrm{3}} {a}_{\mathrm{3}} =\mathrm{2}{a}_{\mathrm{3}} \:\:\:\:\left(\mathrm{3}\right) \\ $$$${for}\:{pully}\:: \\ $$$$\overset{−} {{a}}={a}_{\mathrm{1}} \:\:,\:\begin{cases}{\overset{−} {{a}}+\overset{..} {\theta}{r}={a}_{\mathrm{1}} +\overset{..} {\theta}{r}={a}_{\mathrm{2}} \:\:\:\:\left(\mathrm{4}\right)}\\{\overset{−} {{a}}−\overset{..} {\theta}{r}={a}_{\mathrm{1}} −\overset{..} {\theta}{r}={a}_{\mathrm{3}} \:\:\:\left(\mathrm{5}\right)}\end{cases}\:\left(\circlearrowright\:\equiv\:+\right) \\ $$$$\Sigma{F}={m}\overset{−} {{a}}\:\Rightarrow\:{T}_{\mathrm{1}} −{T}_{\mathrm{2}} −{T}_{\mathrm{3}} =\mathrm{2}{a}_{\mathrm{1}} \:\:\:\:\:\left(\mathrm{6}\right) \\ $$$$\Sigma\overset{−} {{M}}=\overset{−} {{I}}\overset{..} {\theta}\:\Rightarrow\:\left({T}_{\mathrm{3}} −{T}_{\mathrm{2}} \right){r}=\frac{{mr}^{\mathrm{2}} }{\mathrm{2}}\overset{..} {\theta}\:\Rightarrow{T}_{\mathrm{3}} −{T}_{\mathrm{2}} ={r}\overset{..} {\theta}\:\:\:\left(\mathrm{7}\right) \\ $$$$\mathrm{7}\:{unkown}\:\left(\mathrm{3}{T}+\mathrm{3}{a}+\overset{..} {\theta}{r}\right)\:{and}\:\mathrm{7}\:{eq}\:\Rightarrow \\ $$$${a}_{\mathrm{1}} =…\:\:\:\:\:\:\:\:\:\:{a}_{\mathrm{2}} =…\:\:\:\:\:\:\:\:\:\:\:{a}_{\mathrm{3}} =… \\ $$
Commented by Tawa11 last updated on 17/Mar/25
Thanks sir. I appreciate.
$$\mathrm{Thanks}\:\mathrm{sir}. \\ $$$$\mathrm{I}\:\mathrm{appreciate}. \\ $$
Answered by mr W last updated on 17/Mar/25
Commented by mr W last updated on 17/Mar/25
α=angular acc. of pulley (↶) A_1 =acc. of M_1 and m A_2 =acc. of M_2 =A_1 +αr A_3 =acc. of M_3 =A_1 −αr F_2 =M_2 A_2 =M_2 (A_1 +αr) F_3 =M_3 A_3 =M_3 (A_1 −αr) F−F_2 −F_3 =(M_1 +m)A_1 F=(M_1 +m)A_1 +M_2 (A_1 +αr)+M_3 (A_1 −αr) ⇒(M_1 +m+M_2 +M_3 )A_1 +(M_2 −M_3 )αr=F ...(i) F_3 r−F_2 r=Iα=((mr^2 α)/2) M_3 (A_1 −αr)−M_2 (A_1 +αr)=((mrα)/2) ⇒(M_2 −M_3 )A_1 +(M_2 +M_3 +(m/2))αr=0 ...(ii) from (ii): αr=−(((M_2 −M_3 )A_1 )/((M_2 +M_3 +(m/2)))) this into (i): [M_1 +m+M_2 +M_3 −(((M_2 −M_3 )^2 )/(M_2 +M_3 +(m/2)))]A_1 =F ⇒A_1 =(F/(M_1 +m+M_2 +M_3 −(((M_2 −M_3 )^2 )/(M_2 +M_3 +(m/2))))) ⇒A_2 =(1−((M_2 −M_3 )/(M_2 +M_3 +(m/2))))(F/(M_1 +m+M_2 +M_3 −(((M_2 −M_3 )^2 )/(M_2 +M_3 +(m/2))))) ⇒A_3 =(1+((M_2 −M_3 )/(M_2 +M_3 +(m/2))))(F/(M_1 +m+M_2 +M_3 −(((M_2 −M_3 )^2 )/(M_2 +M_3 +(m/2)))))
$$\alpha={angular}\:{acc}.\:{of}\:{pulley}\:\left(\curvearrowleft\right) \\ $$$${A}_{\mathrm{1}} ={acc}.\:{of}\:{M}_{\mathrm{1}} \:{and}\:{m} \\ $$$${A}_{\mathrm{2}} ={acc}.\:{of}\:{M}_{\mathrm{2}} ={A}_{\mathrm{1}} +\alpha{r} \\ $$$${A}_{\mathrm{3}} ={acc}.\:{of}\:{M}_{\mathrm{3}} ={A}_{\mathrm{1}} −\alpha{r} \\ $$$$ \\ $$$${F}_{\mathrm{2}} ={M}_{\mathrm{2}} {A}_{\mathrm{2}} ={M}_{\mathrm{2}} \left({A}_{\mathrm{1}} +\alpha{r}\right) \\ $$$${F}_{\mathrm{3}} ={M}_{\mathrm{3}} {A}_{\mathrm{3}} ={M}_{\mathrm{3}} \left({A}_{\mathrm{1}} −\alpha{r}\right) \\ $$$${F}−{F}_{\mathrm{2}} −{F}_{\mathrm{3}} =\left({M}_{\mathrm{1}} +{m}\right){A}_{\mathrm{1}} \\ $$$${F}=\left({M}_{\mathrm{1}} +{m}\right){A}_{\mathrm{1}} +{M}_{\mathrm{2}} \left({A}_{\mathrm{1}} +\alpha{r}\right)+{M}_{\mathrm{3}} \left({A}_{\mathrm{1}} −\alpha{r}\right) \\ $$$$\Rightarrow\left({M}_{\mathrm{1}} +{m}+{M}_{\mathrm{2}} +{M}_{\mathrm{3}} \right){A}_{\mathrm{1}} +\left({M}_{\mathrm{2}} −{M}_{\mathrm{3}} \right)\alpha{r}={F}\:\:\:…\left({i}\right) \\ $$$$ \\ $$$${F}_{\mathrm{3}} {r}−{F}_{\mathrm{2}} {r}={I}\alpha=\frac{{mr}^{\mathrm{2}} \alpha}{\mathrm{2}} \\ $$$${M}_{\mathrm{3}} \left({A}_{\mathrm{1}} −\alpha{r}\right)−{M}_{\mathrm{2}} \left({A}_{\mathrm{1}} +\alpha{r}\right)=\frac{{mr}\alpha}{\mathrm{2}} \\ $$$$\Rightarrow\left({M}_{\mathrm{2}} −{M}_{\mathrm{3}} \right){A}_{\mathrm{1}} +\left({M}_{\mathrm{2}} +{M}_{\mathrm{3}} +\frac{{m}}{\mathrm{2}}\right)\alpha{r}=\mathrm{0}\:\:\:…\left({ii}\right) \\ $$$${from}\:\left({ii}\right): \\ $$$$\alpha{r}=−\frac{\left({M}_{\mathrm{2}} −{M}_{\mathrm{3}} \right){A}_{\mathrm{1}} }{\left({M}_{\mathrm{2}} +{M}_{\mathrm{3}} +\frac{{m}}{\mathrm{2}}\right)} \\ $$$${this}\:{into}\:\left({i}\right): \\ $$$$\left[{M}_{\mathrm{1}} +{m}+{M}_{\mathrm{2}} +{M}_{\mathrm{3}} −\frac{\left({M}_{\mathrm{2}} −{M}_{\mathrm{3}} \right)^{\mathrm{2}} }{{M}_{\mathrm{2}} +{M}_{\mathrm{3}} +\frac{{m}}{\mathrm{2}}}\right]{A}_{\mathrm{1}} ={F} \\ $$$$\Rightarrow{A}_{\mathrm{1}} =\frac{{F}}{{M}_{\mathrm{1}} +{m}+{M}_{\mathrm{2}} +{M}_{\mathrm{3}} −\frac{\left({M}_{\mathrm{2}} −{M}_{\mathrm{3}} \right)^{\mathrm{2}} }{{M}_{\mathrm{2}} +{M}_{\mathrm{3}} +\frac{{m}}{\mathrm{2}}}} \\ $$$$\Rightarrow{A}_{\mathrm{2}} =\left(\mathrm{1}−\frac{{M}_{\mathrm{2}} −{M}_{\mathrm{3}} }{{M}_{\mathrm{2}} +{M}_{\mathrm{3}} +\frac{{m}}{\mathrm{2}}}\right)\frac{{F}}{{M}_{\mathrm{1}} +{m}+{M}_{\mathrm{2}} +{M}_{\mathrm{3}} −\frac{\left({M}_{\mathrm{2}} −{M}_{\mathrm{3}} \right)^{\mathrm{2}} }{{M}_{\mathrm{2}} +{M}_{\mathrm{3}} +\frac{{m}}{\mathrm{2}}}} \\ $$$$\Rightarrow{A}_{\mathrm{3}} =\left(\mathrm{1}+\frac{{M}_{\mathrm{2}} −{M}_{\mathrm{3}} }{{M}_{\mathrm{2}} +{M}_{\mathrm{3}} +\frac{{m}}{\mathrm{2}}}\right)\frac{{F}}{{M}_{\mathrm{1}} +{m}+{M}_{\mathrm{2}} +{M}_{\mathrm{3}} −\frac{\left({M}_{\mathrm{2}} −{M}_{\mathrm{3}} \right)^{\mathrm{2}} }{{M}_{\mathrm{2}} +{M}_{\mathrm{3}} +\frac{{m}}{\mathrm{2}}}} \\ $$
Commented by Tawa11 last updated on 17/Mar/25
Great sir. Weldone. I appreciate sir.
$$\mathrm{Great}\:\mathrm{sir}. \\ $$$$\mathrm{Weldone}. \\ $$$$\mathrm{I}\:\mathrm{appreciate}\:\mathrm{sir}. \\ $$
Commented by Tawa11 last updated on 17/Mar/25
Commented by Tawa11 last updated on 17/Mar/25
Sir, why is it (M + m)A_1 why not mA_1 . I am just asking to learn why sir.
$$\mathrm{Sir},\:\mathrm{why}\:\mathrm{is}\:\mathrm{it}\:\:\left(\mathrm{M}\:+\:\mathrm{m}\right)\mathrm{A}_{\mathrm{1}} \\ $$$$\mathrm{why}\:\mathrm{not}\:\:\mathrm{mA}_{\mathrm{1}} . \\ $$$$\mathrm{I}\:\mathrm{am}\:\mathrm{just}\:\mathrm{asking}\:\mathrm{to}\:\mathrm{learn}\:\mathrm{why}\:\mathrm{sir}. \\ $$
Commented by mr W last updated on 17/Mar/25
m and M_1 have the same acc. A_1 , is this clear? we have F−F_1 =M_1 A_1 ...(i) F_1 −F_2 −F_3 =mA_1 ..(ii) is this clear? we can also treat m and M_1 as one object which has acc. A_1 , then we have F−F_2 −F_3 =(M_1 +m)A_1 certainly you can get the same by adding (i)+(ii).
$${m}\:{and}\:{M}_{\mathrm{1}} \:{have}\:{the}\:{same}\:{acc}.\:{A}_{\mathrm{1}} , \\ $$$${is}\:{this}\:{clear}? \\ $$$${we}\:{have} \\ $$$${F}−{F}_{\mathrm{1}} ={M}_{\mathrm{1}} {A}_{\mathrm{1}} \:\:\:\:\:…\left({i}\right) \\ $$$${F}_{\mathrm{1}} −{F}_{\mathrm{2}} −{F}_{\mathrm{3}} ={mA}_{\mathrm{1}} \:\:\:..\left({ii}\right) \\ $$$${is}\:{this}\:{clear}? \\ $$$${we}\:{can}\:{also}\:{treat}\:{m}\:{and}\:{M}_{\mathrm{1}} \:{as}\:{one} \\ $$$${object}\:{which}\:{has}\:{acc}.\:{A}_{\mathrm{1}} ,\:{then}\:{we}\: \\ $$$${have} \\ $$$${F}−{F}_{\mathrm{2}} −{F}_{\mathrm{3}} =\left({M}_{\mathrm{1}} +{m}\right){A}_{\mathrm{1}} \\ $$$${certainly}\:{you}\:{can}\:{get}\:{the}\:{same}\:{by} \\ $$$${adding}\:\left({i}\right)+\left({ii}\right). \\ $$
Commented by Tawa11 last updated on 17/Mar/25
I understand now. Thank you sir.
$$\mathrm{I}\:\mathrm{understand}\:\mathrm{now}. \\ $$$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{sir}. \\ $$

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