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Question Number 21793 by Tinkutara last updated on 04/Oct/17
A plank of mass 10 kg rests on a smooth horizontal surface. Two blocks A and B of masses m_A = 2 kg and m_B = 1 kg lies at a distance of 3 m on the plank. The friction coefficient between the blocks and plank are μ_A = 0.3 and μ_B = 0.1. Now a force F = 15 N is applied to the plank in horizontal direction. Find the times (in sec) after which block A collides with B.
$$\mathrm{A}\:\mathrm{plank}\:\mathrm{of}\:\mathrm{mass}\:\mathrm{10}\:\mathrm{kg}\:\mathrm{rests}\:\mathrm{on}\:\mathrm{a}\:\mathrm{smooth} \\ $$$$\mathrm{horizontal}\:\mathrm{surface}.\:\mathrm{Two}\:\mathrm{blocks}\:\mathrm{A}\:\mathrm{and} \\ $$$$\mathrm{B}\:\mathrm{of}\:\mathrm{masses}\:\mathrm{m}_{\mathrm{A}} \:=\:\mathrm{2}\:\mathrm{kg}\:\mathrm{and}\:\mathrm{m}_{\mathrm{B}} \:=\:\mathrm{1}\:\mathrm{kg} \\ $$$$\mathrm{lies}\:\mathrm{at}\:\mathrm{a}\:\mathrm{distance}\:\mathrm{of}\:\mathrm{3}\:\mathrm{m}\:\mathrm{on}\:\mathrm{the}\:\mathrm{plank}. \\ $$$$\mathrm{The}\:\mathrm{friction}\:\mathrm{coefficient}\:\mathrm{between}\:\mathrm{the} \\ $$$$\mathrm{blocks}\:\mathrm{and}\:\mathrm{plank}\:\mathrm{are}\:\mu_{\mathrm{A}} \:=\:\mathrm{0}.\mathrm{3}\:\mathrm{and}\:\mu_{\mathrm{B}} \:= \\ $$$$\mathrm{0}.\mathrm{1}.\:\mathrm{Now}\:\mathrm{a}\:\mathrm{force}\:\mathrm{F}\:=\:\mathrm{15}\:\mathrm{N}\:\mathrm{is}\:\mathrm{applied}\:\mathrm{to} \\ $$$$\mathrm{the}\:\mathrm{plank}\:\mathrm{in}\:\mathrm{horizontal}\:\mathrm{direction}.\:\mathrm{Find} \\ $$$$\mathrm{the}\:\mathrm{times}\:\left(\mathrm{in}\:\mathrm{sec}\right)\:\mathrm{after}\:\mathrm{which}\:\mathrm{block}\:\mathrm{A} \\ $$$$\mathrm{collides}\:\mathrm{with}\:\mathrm{B}. \\ $$
Commented by Tinkutara last updated on 04/Oct/17
Commented by mrW1 last updated on 04/Oct/17
a_A =((μ_A m_A g)/m_A )=μ_A g a_B =((μ_B m_B g)/m_B )=μ_B g < a_A Δa=a_A −a_B =(μ_A −μ_B )g Δs=(1/2)Δat^2 ⇒t=(√((2Δs)/(Δa)))=(√((2Δs)/((μ_A −μ_B )g))) =(√((2×3)/((0.3−0.1)×10))) =(√3) ≈1.73 s F−μ_A m_A g−μ_B m_B g=Ma ⇒a=((F−(μ_A m_A +μ_B m_B )g)/M) =((15−(0.3×2+0.1×1)×10)/(10)) =0.8 m/s^2
$$\mathrm{a}_{\mathrm{A}} =\frac{\mu_{\mathrm{A}} \mathrm{m}_{\mathrm{A}} \mathrm{g}}{\mathrm{m}_{\mathrm{A}} }=\mu_{\mathrm{A}} \mathrm{g} \\ $$$$\mathrm{a}_{\mathrm{B}} =\frac{\mu_{\mathrm{B}} \mathrm{m}_{\mathrm{B}} \mathrm{g}}{\mathrm{m}_{\mathrm{B}} }=\mu_{\mathrm{B}} \mathrm{g}\:<\:\mathrm{a}_{\mathrm{A}} \\ $$$$\Delta\mathrm{a}=\mathrm{a}_{\mathrm{A}} −\mathrm{a}_{\mathrm{B}} =\left(\mu_{\mathrm{A}} −\mu_{\mathrm{B}} \right)\mathrm{g} \\ $$$$\Delta\mathrm{s}=\frac{\mathrm{1}}{\mathrm{2}}\Delta\mathrm{at}^{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{t}=\sqrt{\frac{\mathrm{2}\Delta\mathrm{s}}{\Delta\mathrm{a}}}=\sqrt{\frac{\mathrm{2}\Delta\mathrm{s}}{\left(\mu_{\mathrm{A}} −\mu_{\mathrm{B}} \right)\mathrm{g}}} \\ $$$$=\sqrt{\frac{\mathrm{2}×\mathrm{3}}{\left(\mathrm{0}.\mathrm{3}−\mathrm{0}.\mathrm{1}\right)×\mathrm{10}}} \\ $$$$=\sqrt{\mathrm{3}} \\ $$$$\approx\mathrm{1}.\mathrm{73}\:\mathrm{s} \\ $$$$ \\ $$$$\mathrm{F}−\mu_{\mathrm{A}} \mathrm{m}_{\mathrm{A}} \mathrm{g}−\mu_{\mathrm{B}} \mathrm{m}_{\mathrm{B}} \mathrm{g}=\mathrm{Ma} \\ $$$$\Rightarrow\mathrm{a}=\frac{\mathrm{F}−\left(\mu_{\mathrm{A}} \mathrm{m}_{\mathrm{A}} +\mu_{\mathrm{B}} \mathrm{m}_{\mathrm{B}} \right)\mathrm{g}}{\mathrm{M}} \\ $$$$=\frac{\mathrm{15}−\left(\mathrm{0}.\mathrm{3}×\mathrm{2}+\mathrm{0}.\mathrm{1}×\mathrm{1}\right)×\mathrm{10}}{\mathrm{10}} \\ $$$$=\mathrm{0}.\mathrm{8}\:\mathrm{m}/\mathrm{s}^{\mathrm{2}} \\ $$
Commented by mrW1 last updated on 05/Oct/17
I know the answer was wrong, since a=0.8 m/s^2 which is less than a_A and a_B . that means the plank moves more slowly than A and B, but this is contradiction. therefore I made following consideration:
$$\mathrm{I}\:\mathrm{know}\:\mathrm{the}\:\mathrm{answer}\:\mathrm{was}\:\mathrm{wrong},\:\mathrm{since} \\ $$$$\mathrm{a}=\mathrm{0}.\mathrm{8}\:\mathrm{m}/\mathrm{s}^{\mathrm{2}} \:\mathrm{which}\:\mathrm{is}\:\mathrm{less}\:\mathrm{than}\:\mathrm{a}_{\mathrm{A}} \:\mathrm{and} \\ $$$$\mathrm{a}_{\mathrm{B}} .\:\mathrm{that}\:\mathrm{means}\:\mathrm{the}\:\mathrm{plank}\:\mathrm{moves}\:\mathrm{more} \\ $$$$\mathrm{slowly}\:\mathrm{than}\:\mathrm{A}\:\mathrm{and}\:\mathrm{B},\:\mathrm{but}\:\mathrm{this}\:\mathrm{is} \\ $$$$\mathrm{contradiction}. \\ $$$$ \\ $$$$\mathrm{therefore}\:\mathrm{I}\:\mathrm{made}\:\mathrm{following}\:\mathrm{consideration}: \\ $$
Commented by mrW1 last updated on 05/Oct/17
case 1: only block B slides on plank R_B =μ_B m_B g=m_B a_B ⇒a_B =μ_B g=0.1×10=1 a_A =a F−R_B =(M+m_A )a ⇒a=((F−μ_B m_B g)/(M+m_A ))=((15−0.1×1×10)/(10+2))=(7/6)>a_B =1 R_A =m_A a=2×(7/6)=(7/3)≈2.33 N μ_A m_A g=0.3×2×10=6 N > R_A ⇒block A doesn′t slide, ok! Δa=a_A −a_B =(7/6)−1=(1/6) m/s^2 t=(√((2Δs)/(Δa)))=(√((2×3)/(1/6)))=6 s ⇒after 6 seconds A meets B. case 2: both blocks slide on plank impossible, as shown above case 3: none of the blocks slides on plank a_A =a_B =a F=(M+m_A +m_B )a ⇒a=(F/(M+m_A +m_B ))=((15)/(10+2+1))=((15)/(13)) m/s^2 R_B =m_B a=1×((15)/(13))≈1.15 N μ_B m_B g=0.1×1×10=1 N < R_B ⇒block B muss slide! ⇒the answer is that only block B slides on plank. block A keeps in rest on plank. after 6 sec block A meets B.
$$\mathrm{case}\:\mathrm{1}:\:\mathrm{only}\:\mathrm{block}\:\mathrm{B}\:\mathrm{slides}\:\mathrm{on}\:\mathrm{plank} \\ $$$$\mathrm{R}_{\mathrm{B}} =\mu_{\mathrm{B}} \mathrm{m}_{\mathrm{B}} \mathrm{g}=\mathrm{m}_{\mathrm{B}} \mathrm{a}_{\mathrm{B}} \\ $$$$\Rightarrow\mathrm{a}_{\mathrm{B}} =\mu_{\mathrm{B}} \mathrm{g}=\mathrm{0}.\mathrm{1}×\mathrm{10}=\mathrm{1} \\ $$$$\mathrm{a}_{\mathrm{A}} =\mathrm{a} \\ $$$$\mathrm{F}−\mathrm{R}_{\mathrm{B}} =\left(\mathrm{M}+\mathrm{m}_{\mathrm{A}} \right)\mathrm{a} \\ $$$$\Rightarrow\mathrm{a}=\frac{\mathrm{F}−\mu_{\mathrm{B}} \mathrm{m}_{\mathrm{B}} \mathrm{g}}{\mathrm{M}+\mathrm{m}_{\mathrm{A}} }=\frac{\mathrm{15}−\mathrm{0}.\mathrm{1}×\mathrm{1}×\mathrm{10}}{\mathrm{10}+\mathrm{2}}=\frac{\mathrm{7}}{\mathrm{6}}>\mathrm{a}_{\mathrm{B}} =\mathrm{1} \\ $$$$\mathrm{R}_{\mathrm{A}} =\mathrm{m}_{\mathrm{A}} \mathrm{a}=\mathrm{2}×\frac{\mathrm{7}}{\mathrm{6}}=\frac{\mathrm{7}}{\mathrm{3}}\approx\mathrm{2}.\mathrm{33}\:\mathrm{N} \\ $$$$\mu_{\mathrm{A}} \mathrm{m}_{\mathrm{A}} \mathrm{g}=\mathrm{0}.\mathrm{3}×\mathrm{2}×\mathrm{10}=\mathrm{6}\:\mathrm{N}\:>\:\mathrm{R}_{\mathrm{A}} \\ $$$$\Rightarrow\mathrm{block}\:\mathrm{A}\:\mathrm{doesn}'\mathrm{t}\:\mathrm{slide},\:\mathrm{ok}! \\ $$$$ \\ $$$$\Delta\mathrm{a}=\mathrm{a}_{\mathrm{A}} −\mathrm{a}_{\mathrm{B}} =\frac{\mathrm{7}}{\mathrm{6}}−\mathrm{1}=\frac{\mathrm{1}}{\mathrm{6}}\:\mathrm{m}/\mathrm{s}^{\mathrm{2}} \\ $$$$\mathrm{t}=\sqrt{\frac{\mathrm{2}\Delta\mathrm{s}}{\Delta\mathrm{a}}}=\sqrt{\frac{\mathrm{2}×\mathrm{3}}{\frac{\mathrm{1}}{\mathrm{6}}}}=\mathrm{6}\:\mathrm{s} \\ $$$$\Rightarrow\mathrm{after}\:\mathrm{6}\:\mathrm{seconds}\:\mathrm{A}\:\mathrm{meets}\:\mathrm{B}. \\ $$$$ \\ $$$$\mathrm{case}\:\mathrm{2}:\:\mathrm{both}\:\mathrm{blocks}\:\mathrm{slide}\:\mathrm{on}\:\mathrm{plank} \\ $$$$\mathrm{impossible},\:\mathrm{as}\:\mathrm{shown}\:\mathrm{above} \\ $$$$ \\ $$$$\mathrm{case}\:\mathrm{3}:\:\mathrm{none}\:\mathrm{of}\:\mathrm{the}\:\mathrm{blocks}\:\mathrm{slides}\:\mathrm{on}\:\mathrm{plank} \\ $$$$\mathrm{a}_{\mathrm{A}} =\mathrm{a}_{\mathrm{B}} =\mathrm{a} \\ $$$$\mathrm{F}=\left(\mathrm{M}+\mathrm{m}_{\mathrm{A}} +\mathrm{m}_{\mathrm{B}} \right)\mathrm{a} \\ $$$$\Rightarrow\mathrm{a}=\frac{\mathrm{F}}{\mathrm{M}+\mathrm{m}_{\mathrm{A}} +\mathrm{m}_{\mathrm{B}} }=\frac{\mathrm{15}}{\mathrm{10}+\mathrm{2}+\mathrm{1}}=\frac{\mathrm{15}}{\mathrm{13}}\:\mathrm{m}/\mathrm{s}^{\mathrm{2}} \\ $$$$\mathrm{R}_{\mathrm{B}} =\mathrm{m}_{\mathrm{B}} \mathrm{a}=\mathrm{1}×\frac{\mathrm{15}}{\mathrm{13}}\approx\mathrm{1}.\mathrm{15}\:\mathrm{N} \\ $$$$\mu_{\mathrm{B}} \mathrm{m}_{\mathrm{B}} \mathrm{g}=\mathrm{0}.\mathrm{1}×\mathrm{1}×\mathrm{10}=\mathrm{1}\:\mathrm{N}\:<\:\mathrm{R}_{\mathrm{B}} \\ $$$$\Rightarrow\mathrm{block}\:\mathrm{B}\:\mathrm{muss}\:\mathrm{slide}! \\ $$$$ \\ $$$$\Rightarrow\mathrm{the}\:\mathrm{answer}\:\mathrm{is}\:\mathrm{that}\:\mathrm{only}\:\mathrm{block}\:\mathrm{B}\:\mathrm{slides} \\ $$$$\mathrm{on}\:\mathrm{plank}.\:\mathrm{block}\:\mathrm{A}\:\mathrm{keeps}\:\mathrm{in}\:\mathrm{rest}\:\mathrm{on} \\ $$$$\mathrm{plank}.\:\mathrm{after}\:\mathrm{6}\:\mathrm{sec}\:\mathrm{block}\:\mathrm{A}\:\mathrm{meets}\:\mathrm{B}. \\ $$
Commented by mrW1 last updated on 05/Oct/17
a=(F/(M+m_A +m_B ))≥μ_B g F≥μ_B (M+m_A +m_B )g=0.1×(10+2+1)×10=13 N that means if F≥13 N, block B slides on plank. a=((F−μ_B m_B g)/(M+m_A ))≥μ_A g F≥[μ_A (M+m_A )+μ_B m_B ]g=[0.3×(10+2)+0.1×1]×10=37 N that means if F≥37 N, also block A slides on the plank. F<13 N ⇒ case 3: no block slides 13 N≤F<37 N ⇒ case 1: only block B slides F≥37 N ⇒ case 2: both blocks slide in case 3: t→∞ in case 2: t=(√3) sec (see above) in case 1: Δa=((F−μ_B m_B g)/(M+m_A ))−μ_B g=((F−μ_B (M+m_A +m_B )g)/(M+m_A )) t=(√((2Δs)/(Δa)))=(√((2s(M+m_A ))/(F−μ_B (M+m_A +m_B )g))) =(√((2×3×(10+2))/(F−0.1×(10+2+1)×10))) =(√((72)/(F−13)))
$$\mathrm{a}=\frac{\mathrm{F}}{\mathrm{M}+\mathrm{m}_{\mathrm{A}} +\mathrm{m}_{\mathrm{B}} }\geqslant\mu_{\mathrm{B}} \mathrm{g} \\ $$$$\mathrm{F}\geqslant\mu_{\mathrm{B}} \left(\mathrm{M}+\mathrm{m}_{\mathrm{A}} +\mathrm{m}_{\mathrm{B}} \right)\mathrm{g}=\mathrm{0}.\mathrm{1}×\left(\mathrm{10}+\mathrm{2}+\mathrm{1}\right)×\mathrm{10}=\mathrm{13}\:\mathrm{N} \\ $$$$\mathrm{that}\:\mathrm{means}\:\mathrm{if}\:\mathrm{F}\geqslant\mathrm{13}\:\mathrm{N},\:\mathrm{block}\:\mathrm{B}\:\mathrm{slides} \\ $$$$\mathrm{on}\:\mathrm{plank}. \\ $$$$ \\ $$$$\mathrm{a}=\frac{\mathrm{F}−\mu_{\mathrm{B}} \mathrm{m}_{\mathrm{B}} \mathrm{g}}{\mathrm{M}+\mathrm{m}_{\mathrm{A}} }\geqslant\mu_{\mathrm{A}} \mathrm{g} \\ $$$$\mathrm{F}\geqslant\left[\mu_{\mathrm{A}} \left(\mathrm{M}+\mathrm{m}_{\mathrm{A}} \right)+\mu_{\mathrm{B}} \mathrm{m}_{\mathrm{B}} \right]\mathrm{g}=\left[\mathrm{0}.\mathrm{3}×\left(\mathrm{10}+\mathrm{2}\right)+\mathrm{0}.\mathrm{1}×\mathrm{1}\right]×\mathrm{10}=\mathrm{37}\:\mathrm{N} \\ $$$$\mathrm{that}\:\mathrm{means}\:\mathrm{if}\:\mathrm{F}\geqslant\mathrm{37}\:\mathrm{N},\:\mathrm{also}\:\mathrm{block}\:\mathrm{A} \\ $$$$\mathrm{slides}\:\mathrm{on}\:\mathrm{the}\:\mathrm{plank}. \\ $$$$ \\ $$$$\mathrm{F}<\mathrm{13}\:\mathrm{N}\:\Rightarrow\:\mathrm{case}\:\mathrm{3}:\:\mathrm{no}\:\mathrm{block}\:\mathrm{slides} \\ $$$$\mathrm{13}\:\mathrm{N}\leqslant\mathrm{F}<\mathrm{37}\:\mathrm{N}\:\Rightarrow\:\mathrm{case}\:\mathrm{1}:\:\mathrm{only}\:\mathrm{block}\:\mathrm{B}\:\mathrm{slides} \\ $$$$\mathrm{F}\geqslant\mathrm{37}\:\mathrm{N}\:\Rightarrow\:\mathrm{case}\:\mathrm{2}:\:\mathrm{both}\:\mathrm{blocks}\:\mathrm{slide} \\ $$$$ \\ $$$$\mathrm{in}\:\mathrm{case}\:\mathrm{3}: \\ $$$$\mathrm{t}\rightarrow\infty \\ $$$$ \\ $$$$\mathrm{in}\:\mathrm{case}\:\mathrm{2}: \\ $$$$\mathrm{t}=\sqrt{\mathrm{3}}\:\mathrm{sec}\:\left(\mathrm{see}\:\mathrm{above}\right) \\ $$$$ \\ $$$$\mathrm{in}\:\mathrm{case}\:\mathrm{1}: \\ $$$$\Delta\mathrm{a}=\frac{\mathrm{F}−\mu_{\mathrm{B}} \mathrm{m}_{\mathrm{B}} \mathrm{g}}{\mathrm{M}+\mathrm{m}_{\mathrm{A}} }−\mu_{\mathrm{B}} \mathrm{g}=\frac{\mathrm{F}−\mu_{\mathrm{B}} \left(\mathrm{M}+\mathrm{m}_{\mathrm{A}} +\mathrm{m}_{\mathrm{B}} \right)\mathrm{g}}{\mathrm{M}+\mathrm{m}_{\mathrm{A}} } \\ $$$$\mathrm{t}=\sqrt{\frac{\mathrm{2}\Delta\mathrm{s}}{\Delta\mathrm{a}}}=\sqrt{\frac{\mathrm{2s}\left(\mathrm{M}+\mathrm{m}_{\mathrm{A}} \right)}{\mathrm{F}−\mu_{\mathrm{B}} \left(\mathrm{M}+\mathrm{m}_{\mathrm{A}} +\mathrm{m}_{\mathrm{B}} \right)\mathrm{g}}} \\ $$$$=\sqrt{\frac{\mathrm{2}×\mathrm{3}×\left(\mathrm{10}+\mathrm{2}\right)}{\mathrm{F}−\mathrm{0}.\mathrm{1}×\left(\mathrm{10}+\mathrm{2}+\mathrm{1}\right)×\mathrm{10}}} \\ $$$$=\sqrt{\frac{\mathrm{72}}{\mathrm{F}−\mathrm{13}}} \\ $$
Commented by Tinkutara last updated on 05/Oct/17
Thank you very much Sir!
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{Sir}! \\ $$

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