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Question Number 21713 by Tinkutara last updated on 01/Oct/17

$$\mathrm{A}\mathrm{constant}\mathrm{force}F=20\mathrm{N}\mathrm{acts}\mathrm{on}\mathrm{a}$$$$\mathrm{block}\mathrm{of}\mathrm{mass}2\mathrm{kg}\mathrm{which}\mathrm{is}\mathrm{connected}\mathrm{to}$$$$\mathrm{two}\mathrm{blocks}\mathrm{of}\mathrm{masses}{m}_1=1\mathrm{kg}\mathrm{and}$$$$m_2=2\mathrm{kg}.\mathrm{Calculate}\mathrm{the}\mathrm{accelerations}$$$$\mathrm{produced}\mathrm{in}\mathrm{all}\mathrm{the}\mathrm{three}\mathrm{blocks}.\mathrm{Assume}$$$$\mathrm{pulleys}\mathrm{are}\mathrm{frictionless}\mathrm{and}\mathrm{weightless}.$$

Commented by Tinkutara last updated on 01/Oct/17

Commented by mrW1 last updated on 01/Oct/17

$$\mathrm{F}-\mathrm{T}=\mathrm{Ma}$$$$2\mathrm{T}-{\mathrm{m}}_1\mathrm{g}={\mathrm{m}}_1{\mathrm{a}}_1$$$$\mathrm{T}-{\mathrm{m}}_2\mathrm{g}={\mathrm{m}}_2{\mathrm{a}}_2$$$${\mathrm{a}}_1=\frac{1}{2}(\mathrm{a}-{\mathrm{a}}_2)$$$$$$$$\mathrm{T}=\mathrm{F}-\mathrm{M}$$$$2(\mathrm{F}-\mathrm{Ma})={\mathrm{m}}_1(\mathrm{g}+{\mathrm{a}}_1)$$$$\Rightarrow {\mathrm{a}}_1=\frac{2(\mathrm{F}-\mathrm{Ma})}{{\mathrm{m}}_1}-\mathrm{g}$$$$\mathrm{F}-\mathrm{Ma}={\mathrm{m}}_2(\mathrm{g}+{\mathrm{a}}_2)$$$$\Rightarrow {\mathrm{a}}_2=\frac{\mathrm{F}-\mathrm{Ma}}{{\mathrm{m}}_2}-\mathrm{g}$$$$\frac{4(\mathrm{F}-\mathrm{Ma})}{{\mathrm{m}}_1}-2\mathrm{g}=\mathrm{a}-\frac{\mathrm{F}-\mathrm{Ma}}{{\mathrm{m}}_2}+\mathrm{g}$$$$\frac{4\mathrm{F}}{{\mathrm{m}}_1}+\frac{\mathrm{F}}{{\mathrm{m}}_2}-3\mathrm{g}=(1+\frac{4\mathrm{M}}{{\mathrm{m}}_1}+\frac{\mathrm{M}}{{\mathrm{m}}_2})\mathrm{a}$$$$\Rightarrow \mathrm{a}=\frac{({\mathrm{m}}_1+{4\mathrm{m}}_2)\mathrm{F}-{3\mathrm{m}}_1{\mathrm{m}}_2\mathrm{g}}{({\mathrm{m}}_1+{4\mathrm{m}}_2)\mathrm{M}+{\mathrm{m}}_1{\mathrm{m}}_2}$$$$=\frac{(1+4\times 2)\times 20-3\times 1\times 2\times 9.8}{(1+4\times 2)\times 2+1\times 2}$$$$=\frac{180-6\times 9.8}{20}$$$$=6.06\mathrm{m}/{\mathrm{s}}^2(\leftarrow )$$$$\Rightarrow {\mathrm{a}}_1=\frac{2(20-2\times 6.06)}{1}-9.8=5.96\mathrm{m}/{\mathrm{s}}^2(\uparrow )$$$$\Rightarrow {\mathrm{a}}_2=\frac{20-2\times 6.06}{2}-9.8=-5.86\mathrm{m}/{\mathrm{s}}^2(\downarrow )$$

Commented by Tinkutara last updated on 01/Oct/17

$$\mathrm{I}{\mathrm{don}}^{\prime }\mathrm{t}\mathrm{think}\mathrm{all}\mathrm{blocks}\mathrm{have}\mathrm{same}$$$$\mathrm{accelerations}.\mathrm{If}\mathrm{you}\mathrm{take}g=9.8\mathrm{m}/{\mathrm{s}}^2,$$$$\mathrm{then}\mathrm{I}\mathrm{can}\mathrm{check}\mathrm{your}\mathrm{answer}\mathrm{from}\mathrm{book}.$$

Commented by mrW1 last updated on 02/Oct/17

$$\mathrm{they}\mathrm{have}\mathrm{the}\mathrm{same}\mathrm{value}\mathrm{of}\mathrm{acceleration}$$$$\mathrm{just}\mathrm{because}\mathrm{of}\mathrm{these}\mathrm{specific}\mathrm{parameters}.$$

Commented by Tinkutara last updated on 02/Oct/17

$$\mathrm{Can}\mathrm{you}\mathrm{explain}\mathrm{why}{\mathrm{a}}_1=\frac{1}{2}(\mathrm{a}-{\mathrm{a}}_2)?$$

Commented by mrW1 last updated on 02/Oct/17

$$\mathrm{since}\mathrm{the}\mathrm{rope}\mathrm{is}\mathrm{inextensible},\mathrm{the}$$$$\mathrm{motion}\mathrm{of}\mathrm{mass}{\mathrm{m}}_1\mathrm{is}\mathrm{dependent}\mathrm{from}$$$$\mathrm{the}\mathrm{motion}\mathrm{of}\mathrm{the}\mathrm{other}\mathrm{two}\mathrm{masses}.$$$$\mathrm{position}\mathrm{of}\mathrm{mass}\mathrm{M}=\mathrm{x}(+\mathrm{ve}\mathrm{if}\leftarrow )$$$$\mathrm{position}\mathrm{of}\mathrm{mass}{\mathrm{m}}_2={\mathrm{x}}_2(+\mathrm{ve}\mathrm{if}\uparrow )$$$$\mathrm{position}\mathrm{of}\mathrm{mass}{\mathrm{m}}_1={\mathrm{x}}_1(+\mathrm{ve}\mathrm{if}\uparrow )$$$$\mathrm{we}\mathrm{get}{\mathrm{x}}_1=\frac{\mathrm{x}}{2}-\frac{{\mathrm{x}}_2}{2}$$$$\mathrm{a}=\frac{{\mathrm{d}}^2\mathrm{x}}{{\mathrm{dt}}^2}$$$${\mathrm{a}}_1=\frac{{\mathrm{d}}^2{\mathrm{x}}_1}{{\mathrm{dt}}^2}$$$${\mathrm{a}}_2=\frac{{\mathrm{d}}^2{\mathrm{x}}_2}{{\mathrm{dt}}^2}$$$$\Rightarrow {\mathrm{a}}_1=\frac{1}{2}(\mathrm{a}-{\mathrm{a}}_2)$$

Commented by Tinkutara last updated on 02/Oct/17

$$\mathrm{Thank}\mathrm{you}\mathrm{very}\mathrm{much}\mathrm{Sir}!$$

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