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Question-222995

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Question Number 222995 by Jubr last updated on 12/Jul/25
Answered by mr W last updated on 12/Jul/25
T=tension string around the pulley A=acceleration of block M a=acceleration of block m assume there is no friction. a=2A a=(T/m) A=((F−2T)/M) (T/m)=2×((F−2T)/M) ⇒T=((2F)/(4+(M/m))) ⇒a=((2F)/(4m+M)) ⇒A=(F/(4m+M))
$${T}={tension}\:{string}\:{around}\:{the}\:{pulley} \\ $$$${A}={acceleration}\:{of}\:{block}\:{M} \\ $$$${a}={acceleration}\:{of}\:{block}\:{m} \\ $$$${assume}\:{there}\:{is}\:{no}\:{friction}. \\ $$$${a}=\mathrm{2}{A} \\ $$$${a}=\frac{{T}}{{m}} \\ $$$${A}=\frac{{F}−\mathrm{2}{T}}{{M}} \\ $$$$\frac{{T}}{{m}}=\mathrm{2}×\frac{{F}−\mathrm{2}{T}}{{M}} \\ $$$$\Rightarrow{T}=\frac{\mathrm{2}{F}}{\mathrm{4}+\frac{{M}}{{m}}} \\ $$$$\Rightarrow{a}=\frac{\mathrm{2}{F}}{\mathrm{4}{m}+{M}} \\ $$$$\Rightarrow{A}=\frac{{F}}{\mathrm{4}{m}+{M}} \\ $$
Commented by Tawa11 last updated on 12/Jul/25
Sir, I tried this and got. a = ((2F − 4mg)/(4m + M)) and A = ((F − 2mg)/(4m + M)) But I noticed if g = 0, I will get your answer. Did you neglect gravity?
$$\mathrm{Sir}, \\ $$$$\mathrm{I}\:\mathrm{tried}\:\mathrm{this}\:\mathrm{and}\:\mathrm{got}. \\ $$$$\:\:\:\:\:\:\mathrm{a}\:\:=\:\:\:\frac{\mathrm{2F}\:\:−\:\:\mathrm{4mg}}{\mathrm{4m}\:\:+\:\:\mathrm{M}}\:\:\:\:\:\:\mathrm{and}\:\:\:\:\:\mathrm{A}\:\:=\:\:\frac{\mathrm{F}\:\:−\:\:\mathrm{2mg}}{\mathrm{4m}\:\:+\:\:\mathrm{M}} \\ $$$$\mathrm{But}\:\mathrm{I}\:\mathrm{noticed}\:\mathrm{if}\:\:\mathrm{g}\:\:=\:\:\mathrm{0},\:\:\mathrm{I}\:\mathrm{will}\:\mathrm{get}\:\mathrm{your}\:\mathrm{answer}. \\ $$$$\mathrm{Did}\:\mathrm{you}\:\mathrm{neglect}\:\mathrm{gravity}? \\ $$
Commented by Tawa11 last updated on 12/Jul/25
Or since the string are massless. we will neglect g?
$$\mathrm{Or}\:\mathrm{since}\:\mathrm{the}\:\mathrm{string}\:\mathrm{are}\:\mathrm{massless}. \\ $$$$\mathrm{we}\:\mathrm{will}\:\mathrm{neglect}\:\:\mathrm{g}? \\ $$
Commented by mr W last updated on 12/Jul/25
the motion is in horizontal direction, what does the gravity have to do with it? show how you did.
$${the}\:{motion}\:{is}\:{in}\:{horizontal}\:{direction}, \\ $$$${what}\:{does}\:{the}\:{gravity}\:{have}\:{to}\:{do}\: \\ $$$${with}\:{it}? \\ $$$${show}\:{how}\:{you}\:{did}. \\ $$
Commented by Tawa11 last updated on 12/Jul/25
I understand now sir.
$$\mathrm{I}\:\mathrm{understand}\:\mathrm{now}\:\mathrm{sir}. \\ $$
Commented by mr W last updated on 12/Jul/25
show how you solved. show why a=2A. assume the friction coefficient between blocks and ground is μ.
$${show}\:{how}\:{you}\:{solved}.\:{show}\:{why} \\ $$$${a}=\mathrm{2}{A}. \\ $$$${assume}\:{the}\:{friction}\:{coefficient}\: \\ $$$${between}\:{blocks}\:{and}\:{ground}\:{is}\:\mu. \\ $$
Commented by Tawa11 last updated on 12/Jul/25
Commented by Tawa11 last updated on 12/Jul/25
L = x + (x − y) = 2x − y (dL/dt) = 2 (dx/dt) − (dy/dt) (d^2 L/dt^2 ) = 2 (d^2 x/dt^2 ) − (d^2 y/dt^2 ) (d^2 L/dt^2 ) = 2A − a = 0 ∴ a = 2A
$$\mathrm{L}\:\:=\:\:\mathrm{x}\:\:+\:\:\left(\mathrm{x}\:\:−\:\:\mathrm{y}\right)\:\:=\:\:\:\mathrm{2x}\:\:\:−\:\:\:\mathrm{y} \\ $$$$\frac{\mathrm{dL}}{\mathrm{dt}}\:\:=\:\:\mathrm{2}\:\frac{\mathrm{dx}}{\mathrm{dt}}\:\:−\:\:\frac{\mathrm{dy}}{\mathrm{dt}} \\ $$$$\frac{\mathrm{d}^{\mathrm{2}} \mathrm{L}}{\mathrm{dt}^{\mathrm{2}} }\:\:=\:\:\mathrm{2}\:\frac{\mathrm{d}^{\mathrm{2}} \mathrm{x}}{\mathrm{dt}^{\mathrm{2}} }\:\:−\:\:\frac{\mathrm{d}^{\mathrm{2}} \mathrm{y}}{\mathrm{dt}^{\mathrm{2}} } \\ $$$$\:\:\frac{\mathrm{d}^{\mathrm{2}} \mathrm{L}}{\mathrm{dt}^{\mathrm{2}} }\:\:\:\:=\:\:\:\mathrm{2A}\:\:−\:\:\mathrm{a}\:\:=\:\:\mathrm{0} \\ $$$$\therefore\:\:\:\mathrm{a}\:\:=\:\:\mathrm{2A}\:\:\:\:\:\: \\ $$
Commented by mr W last updated on 12/Jul/25
that′s right. but the diagram is not correct.
$${that}'{s}\:{right}.\:{but}\:{the}\:{diagram}\:{is}\:{not} \\ $$$${correct}. \\ $$
Commented by Tawa11 last updated on 12/Jul/25
Ohhh. I drew how I think it. Please correct the diagram sir.
$$\mathrm{Ohhh}. \\ $$$$\mathrm{I}\:\mathrm{drew}\:\mathrm{how}\:\mathrm{I}\:\mathrm{think}\:\mathrm{it}. \\ $$$$\mathrm{Please}\:\mathrm{correct}\:\mathrm{the}\:\mathrm{diagram}\:\mathrm{sir}. \\ $$
Commented by mr W last updated on 12/Jul/25
Commented by Tawa11 last updated on 12/Jul/25
Ohh, thanks sir.
$$\mathrm{Ohh},\:\mathrm{thanks}\:\mathrm{sir}. \\ $$
Commented by Tawa11 last updated on 12/Jul/25
Where I put my y before is x − y
$$\mathrm{Where}\:\mathrm{I}\:\mathrm{put}\:\mathrm{my}\:\:\mathrm{y}\:\:\mathrm{before}\:\mathrm{is}\:\:\:\mathrm{x}\:\:−\:\:\mathrm{y} \\ $$
Commented by mr W last updated on 12/Jul/25
the same question, but the block M is connected with the pulley with a spring of natural length l and spring constant k.
$${the}\:{same}\:{question},\:{but}\:{the}\:{block}\:{M} \\ $$$${is}\:{connected}\:{with}\:{the}\:{pulley}\:{with}\:{a} \\ $$$${spring}\:{of}\:{natural}\:{length}\:\boldsymbol{{l}}\:{and}\:{spring} \\ $$$${constant}\:\boldsymbol{{k}}. \\ $$
Commented by mr W last updated on 12/Jul/25
Commented by Jubr last updated on 14/Jul/25
Thank you sir
$${Thank}\:{you}\:{sir} \\ $$
Commented by mr W last updated on 20/Jul/25
at time t=0: force in spring is zero. A=(F/M) a=0 then the tension in spring increases, A ↓, a↑ at sometime t=t_1 , the tension in spring reaches its final value T=((2F)/(4+(M/m))) and A=(F/(4m+M)), a=((2F)/(4m+F)) that means the spring doesn′t affect the final accelerations of the blocks. it only affects their motion from t=0 to t=t_1 . a=0 ↗ ((2F)/(4m+M)) A=(F/M) ↘ (F/(4m+M))
$${at}\:{time}\:{t}=\mathrm{0}: \\ $$$${force}\:{in}\:{spring}\:{is}\:{zero}. \\ $$$${A}=\frac{{F}}{{M}} \\ $$$${a}=\mathrm{0} \\ $$$${then}\:{the}\:{tension}\:{in}\:{spring}\:{increases}, \\ $$$${A}\:\downarrow,\:{a}\uparrow \\ $$$${at}\:{sometime}\:{t}={t}_{\mathrm{1}} ,\:{the}\:{tension}\:{in} \\ $$$${spring}\:{reaches}\:{its}\:{final}\:{value} \\ $$$${T}=\frac{\mathrm{2}{F}}{\mathrm{4}+\frac{{M}}{{m}}}\:{and}\: \\ $$$${A}=\frac{{F}}{\mathrm{4}{m}+{M}},\:{a}=\frac{\mathrm{2}{F}}{\mathrm{4}{m}+{F}} \\ $$$${that}\:{means}\:{the}\:{spring}\:{doesn}'{t} \\ $$$${affect}\:{the}\:{final}\:{accelerations}\:{of}\:{the} \\ $$$${blocks}.\:{it}\:{only}\:{affects}\:{their}\:{motion} \\ $$$${from}\:{t}=\mathrm{0}\:{to}\:{t}={t}_{\mathrm{1}} . \\ $$$${a}=\mathrm{0}\:\nearrow\:\:\frac{\mathrm{2}{F}}{\mathrm{4}{m}+{M}} \\ $$$${A}=\frac{{F}}{{M}}\:\searrow\:\frac{{F}}{\mathrm{4}{m}+{M}} \\ $$

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