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A-constant-force-F-20-N-acts-on-a-block-of-mass-2-kg-which-is-connected-to-two-blocks-of-masses-m-1-1-kg-and-m-2-2-kg-Calculate-the-accelerations-produced-in-all-the-three-blocks-Assume-pull

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Question Number 21713 by Tinkutara last updated on 01/Oct/17
A constant force F = 20 N acts on a block of mass 2 kg which is connected to two blocks of masses m_1 = 1 kg and m_2 = 2 kg. Calculate the accelerations produced in all the three blocks. Assume pulleys are frictionless and weightless.
$$\mathrm{A}\:\mathrm{constant}\:\mathrm{force}\:{F}\:=\:\mathrm{20}\:\mathrm{N}\:\mathrm{acts}\:\mathrm{on}\:\mathrm{a} \\ $$$$\mathrm{block}\:\mathrm{of}\:\mathrm{mass}\:\mathrm{2}\:\mathrm{kg}\:\mathrm{which}\:\mathrm{is}\:\mathrm{connected}\:\mathrm{to} \\ $$$$\mathrm{two}\:\mathrm{blocks}\:\mathrm{of}\:\mathrm{masses}\:{m}_{\mathrm{1}} \:=\:\mathrm{1}\:\mathrm{kg}\:\mathrm{and} \\ $$$${m}_{\mathrm{2}} \:=\:\mathrm{2}\:\mathrm{kg}.\:\mathrm{Calculate}\:\mathrm{the}\:\mathrm{accelerations} \\ $$$$\mathrm{produced}\:\mathrm{in}\:\mathrm{all}\:\mathrm{the}\:\mathrm{three}\:\mathrm{blocks}.\:\mathrm{Assume} \\ $$$$\mathrm{pulleys}\:\mathrm{are}\:\mathrm{frictionless}\:\mathrm{and}\:\mathrm{weightless}. \\ $$
Commented by Tinkutara last updated on 01/Oct/17
Commented by mrW1 last updated on 01/Oct/17
F−T=Ma 2T−m_1 g=m_1 a_1 T−m_2 g=m_2 a_2 a_1 =(1/2)(a−a_2 ) T=F−M 2(F−Ma)=m_1 (g+a_1 ) ⇒a_1 =((2(F−Ma))/m_1 )−g F−Ma=m_2 (g+a_2 ) ⇒a_2 =((F−Ma)/m_2 )−g ((4(F−Ma))/m_1 )−2g=a−((F−Ma)/m_2 )+g ((4F)/m_1 )+(F/m_2 )−3g=(1+((4M)/m_1 )+(M/m_2 ))a ⇒a=(((m_1 +4m_2 )F−3m_1 m_2 g)/((m_1 +4m_2 )M+m_1 m_2 )) =(((1+4×2)×20−3×1×2×9.8)/((1+4×2)×2+1×2)) =((180−6×9.8)/(20)) =6.06 m/s^2 (←) ⇒a_1 =((2(20−2×6.06))/1)−9.8=5.96 m/s^2 (↑) ⇒a_2 =((20−2×6.06)/2)−9.8=−5.86 m/s^2 (↓)
$$\mathrm{F}−\mathrm{T}=\mathrm{Ma} \\ $$$$\mathrm{2T}−\mathrm{m}_{\mathrm{1}} \mathrm{g}=\mathrm{m}_{\mathrm{1}} \mathrm{a}_{\mathrm{1}} \\ $$$$\mathrm{T}−\mathrm{m}_{\mathrm{2}} \mathrm{g}=\mathrm{m}_{\mathrm{2}} \mathrm{a}_{\mathrm{2}} \\ $$$$\mathrm{a}_{\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{a}−\mathrm{a}_{\mathrm{2}} \right) \\ $$$$ \\ $$$$\mathrm{T}=\mathrm{F}−\mathrm{M} \\ $$$$\mathrm{2}\left(\mathrm{F}−\mathrm{Ma}\right)=\mathrm{m}_{\mathrm{1}} \left(\mathrm{g}+\mathrm{a}_{\mathrm{1}} \right) \\ $$$$\Rightarrow\mathrm{a}_{\mathrm{1}} =\frac{\mathrm{2}\left(\mathrm{F}−\mathrm{Ma}\right)}{\mathrm{m}_{\mathrm{1}} }−\mathrm{g} \\ $$$$\mathrm{F}−\mathrm{Ma}=\mathrm{m}_{\mathrm{2}} \left(\mathrm{g}+\mathrm{a}_{\mathrm{2}} \right) \\ $$$$\Rightarrow\mathrm{a}_{\mathrm{2}} =\frac{\mathrm{F}−\mathrm{Ma}}{\mathrm{m}_{\mathrm{2}} }−\mathrm{g} \\ $$$$\frac{\mathrm{4}\left(\mathrm{F}−\mathrm{Ma}\right)}{\mathrm{m}_{\mathrm{1}} }−\mathrm{2g}=\mathrm{a}−\frac{\mathrm{F}−\mathrm{Ma}}{\mathrm{m}_{\mathrm{2}} }+\mathrm{g} \\ $$$$\frac{\mathrm{4F}}{\mathrm{m}_{\mathrm{1}} }+\frac{\mathrm{F}}{\mathrm{m}_{\mathrm{2}} }−\mathrm{3g}=\left(\mathrm{1}+\frac{\mathrm{4M}}{\mathrm{m}_{\mathrm{1}} }+\frac{\mathrm{M}}{\mathrm{m}_{\mathrm{2}} }\right)\mathrm{a} \\ $$$$\Rightarrow\mathrm{a}=\frac{\left(\mathrm{m}_{\mathrm{1}} +\mathrm{4m}_{\mathrm{2}} \right)\mathrm{F}−\mathrm{3m}_{\mathrm{1}} \mathrm{m}_{\mathrm{2}} \mathrm{g}}{\left(\mathrm{m}_{\mathrm{1}} +\mathrm{4m}_{\mathrm{2}} \right)\mathrm{M}+\mathrm{m}_{\mathrm{1}} \mathrm{m}_{\mathrm{2}} } \\ $$$$=\frac{\left(\mathrm{1}+\mathrm{4}×\mathrm{2}\right)×\mathrm{20}−\mathrm{3}×\mathrm{1}×\mathrm{2}×\mathrm{9}.\mathrm{8}}{\left(\mathrm{1}+\mathrm{4}×\mathrm{2}\right)×\mathrm{2}+\mathrm{1}×\mathrm{2}} \\ $$$$=\frac{\mathrm{180}−\mathrm{6}×\mathrm{9}.\mathrm{8}}{\mathrm{20}} \\ $$$$=\mathrm{6}.\mathrm{06}\:\mathrm{m}/\mathrm{s}^{\mathrm{2}} \:\left(\leftarrow\right) \\ $$$$\Rightarrow\mathrm{a}_{\mathrm{1}} =\frac{\mathrm{2}\left(\mathrm{20}−\mathrm{2}×\mathrm{6}.\mathrm{06}\right)}{\mathrm{1}}−\mathrm{9}.\mathrm{8}=\mathrm{5}.\mathrm{96}\:\mathrm{m}/\mathrm{s}^{\mathrm{2}} \:\left(\uparrow\right) \\ $$$$\Rightarrow\mathrm{a}_{\mathrm{2}} =\frac{\mathrm{20}−\mathrm{2}×\mathrm{6}.\mathrm{06}}{\mathrm{2}}−\mathrm{9}.\mathrm{8}=−\mathrm{5}.\mathrm{86}\:\mathrm{m}/\mathrm{s}^{\mathrm{2}} \:\left(\downarrow\right) \\ $$
Commented by Tinkutara last updated on 01/Oct/17
I don′t think all blocks have same accelerations. If you take g = 9.8 m/s^2 , then I can check your answer from book.
$$\mathrm{I}\:\mathrm{don}'\mathrm{t}\:\mathrm{think}\:\mathrm{all}\:\mathrm{blocks}\:\mathrm{have}\:\mathrm{same} \\ $$$$\mathrm{accelerations}.\:\mathrm{If}\:\mathrm{you}\:\mathrm{take}\:{g}\:=\:\mathrm{9}.\mathrm{8}\:\mathrm{m}/\mathrm{s}^{\mathrm{2}} , \\ $$$$\mathrm{then}\:\mathrm{I}\:\mathrm{can}\:\mathrm{check}\:\mathrm{your}\:\mathrm{answer}\:\mathrm{from}\:\mathrm{book}. \\ $$
Commented by mrW1 last updated on 02/Oct/17
they have the same value of acceleration just because of these specific parameters.
$$\mathrm{they}\:\mathrm{have}\:\mathrm{the}\:\mathrm{same}\:\mathrm{value}\:\mathrm{of}\:\mathrm{acceleration} \\ $$$$\mathrm{just}\:\mathrm{because}\:\mathrm{of}\:\mathrm{these}\:\mathrm{specific}\:\mathrm{parameters}. \\ $$
Commented by Tinkutara last updated on 02/Oct/17
Can you explain why a_1 =(1/2)(a−a_2 )?
$$\mathrm{Can}\:\mathrm{you}\:\mathrm{explain}\:\mathrm{why}\:\mathrm{a}_{\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{a}−\mathrm{a}_{\mathrm{2}} \right)? \\ $$
Commented by mrW1 last updated on 02/Oct/17
since the rope is inextensible, the motion of mass m_1 is dependent from the motion of the other two masses. position of mass M =x (+ve if ←) position of mass m_2 =x_2 (+ve if ↑) position of mass m_1 =x_1 (+ve if ↑) we get x_1 =(x/2)−(x_2 /2) a=(d^2 x/dt^2 ) a_1 =(d^2 x_1 /dt^2 ) a_2 =(d^2 x_2 /dt^2 ) ⇒a_1 =(1/2)(a−a_2 )
$$\mathrm{since}\:\mathrm{the}\:\mathrm{rope}\:\mathrm{is}\:\mathrm{inextensible},\:\mathrm{the} \\ $$$$\mathrm{motion}\:\mathrm{of}\:\mathrm{mass}\:\mathrm{m}_{\mathrm{1}} \:\mathrm{is}\:\mathrm{dependent}\:\mathrm{from} \\ $$$$\mathrm{the}\:\mathrm{motion}\:\mathrm{of}\:\mathrm{the}\:\mathrm{other}\:\mathrm{two}\:\mathrm{masses}. \\ $$$$\mathrm{position}\:\mathrm{of}\:\mathrm{mass}\:\mathrm{M}\:=\mathrm{x}\:\left(+\mathrm{ve}\:\mathrm{if}\:\leftarrow\right) \\ $$$$\mathrm{position}\:\mathrm{of}\:\mathrm{mass}\:\mathrm{m}_{\mathrm{2}} \:=\mathrm{x}_{\mathrm{2}} \:\left(+\mathrm{ve}\:\mathrm{if}\:\uparrow\right) \\ $$$$\mathrm{position}\:\mathrm{of}\:\mathrm{mass}\:\mathrm{m}_{\mathrm{1}} \:=\mathrm{x}_{\mathrm{1}} \:\left(+\mathrm{ve}\:\mathrm{if}\:\uparrow\right) \\ $$$$\mathrm{we}\:\mathrm{get}\:\mathrm{x}_{\mathrm{1}} =\frac{\mathrm{x}}{\mathrm{2}}−\frac{\mathrm{x}_{\mathrm{2}} }{\mathrm{2}} \\ $$$$\mathrm{a}=\frac{\mathrm{d}^{\mathrm{2}} \mathrm{x}}{\mathrm{dt}^{\mathrm{2}} } \\ $$$$\mathrm{a}_{\mathrm{1}} =\frac{\mathrm{d}^{\mathrm{2}} \mathrm{x}_{\mathrm{1}} }{\mathrm{dt}^{\mathrm{2}} } \\ $$$$\mathrm{a}_{\mathrm{2}} =\frac{\mathrm{d}^{\mathrm{2}} \mathrm{x}_{\mathrm{2}} }{\mathrm{dt}^{\mathrm{2}} } \\ $$$$\Rightarrow\mathrm{a}_{\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{a}−\mathrm{a}_{\mathrm{2}} \right) \\ $$
Commented by Tinkutara last updated on 02/Oct/17
Thank you very much Sir!
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{Sir}! \\ $$

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