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Question-59378

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Question Number 59378 by tanmay last updated on 09/May/19
Answered by tanmay last updated on 09/May/19
Total force=2T common accelaration a_c =((2T)/(m_(man) +m_(plank) ))=((200)/(150)) a_c =(4/3) F_ =(4/3)×50=((200)/3) (F_(friction) )_(max) =μmg=(1/6)×50×10=((250)/3) ((so F_(friction) =((200)/3)N→towards right )/) i think so
$${Total}\:{force}=\mathrm{2}{T} \\ $$$${common}\:{accelaration}\:{a}_{{c}} =\frac{\mathrm{2}{T}}{{m}_{{man}} +{m}_{{plank}} }=\frac{\mathrm{200}}{\mathrm{150}} \\ $$$${a}_{{c}} =\frac{\mathrm{4}}{\mathrm{3}} \\ $$$${F}_{} =\frac{\mathrm{4}}{\mathrm{3}}×\mathrm{50}=\frac{\mathrm{200}}{\mathrm{3}} \\ $$$$\left({F}_{{friction}} \right)_{{max}} =\mu{mg}=\frac{\mathrm{1}}{\mathrm{6}}×\mathrm{50}×\mathrm{10}=\frac{\mathrm{250}}{\mathrm{3}} \\ $$$$\frac{{so}\:{F}_{{friction}} =\frac{\mathrm{200}}{\mathrm{3}}{N}\rightarrow{towards}\:{right}\:}{} \\ $$$${i}\:{think}\:{so} \\ $$
Answered by mr W last updated on 10/May/19
assume man doesn′t slide on the plank, i.e. friction force F between man and plank is less than μmg=((50×10)/6)=((250)/3) N. man and plank move towards right with acceleration a, a=((2T)/(m+M))=((2×100)/(50+100))=(4/3) m/s^2 plank: Ma=T+F ⇒F=Ma−T=100×(4/3)−100=((100)/3) N i.e. the friction force acting on the plank is ((100)/3) N, towards right. that means the friction force acting on the man is also ((100)/3)N, but towards left. since friction force is ((100)/3) which is less than ((250)/3), the assumption is true that man and plank make the same motion. or look at the man: ma=T+F ⇒F=ma−T=50×(4/3)−100=−((100)/3) N i.e. F is towards left. ⇒answer (A) is correct.
$${assume}\:{man}\:{doesn}'{t}\:{slide}\:{on}\:{the}\:{plank}, \\ $$$${i}.{e}.\:{friction}\:{force}\:{F}\:{between}\:{man}\:{and} \\ $$$${plank}\:{is}\:{less}\:{than}\:\mu{mg}=\frac{\mathrm{50}×\mathrm{10}}{\mathrm{6}}=\frac{\mathrm{250}}{\mathrm{3}}\:{N}. \\ $$$${man}\:{and}\:{plank}\:{move}\:{towards}\:{right} \\ $$$${with}\:{acceleration}\:{a}, \\ $$$${a}=\frac{\mathrm{2}{T}}{{m}+{M}}=\frac{\mathrm{2}×\mathrm{100}}{\mathrm{50}+\mathrm{100}}=\frac{\mathrm{4}}{\mathrm{3}}\:{m}/{s}^{\mathrm{2}} \\ $$$$ \\ $$$${plank}: \\ $$$${Ma}={T}+{F} \\ $$$$\Rightarrow{F}={Ma}−{T}=\mathrm{100}×\frac{\mathrm{4}}{\mathrm{3}}−\mathrm{100}=\frac{\mathrm{100}}{\mathrm{3}}\:{N} \\ $$$${i}.{e}.\:{the}\:{friction}\:{force}\:{acting}\:{on}\:{the}\:{plank} \\ $$$${is}\:\frac{\mathrm{100}}{\mathrm{3}}\:{N},\:{towards}\:{right}.\:{that}\:{means} \\ $$$${the}\:{friction}\:{force}\:{acting}\:{on}\:{the}\:{man} \\ $$$${is}\:{also}\:\frac{\mathrm{100}}{\mathrm{3}}{N},\:{but}\:{towards}\:{left}. \\ $$$${since}\:{friction}\:{force}\:{is}\:\frac{\mathrm{100}}{\mathrm{3}}\:{which}\:{is} \\ $$$${less}\:{than}\:\frac{\mathrm{250}}{\mathrm{3}},\:{the}\:{assumption}\:{is} \\ $$$${true}\:{that}\:{man}\:{and}\:{plank}\:{make}\:{the} \\ $$$${same}\:{motion}. \\ $$$$ \\ $$$${or}\:{look}\:{at}\:{the}\:{man}: \\ $$$${ma}={T}+{F} \\ $$$$\Rightarrow{F}={ma}−{T}=\mathrm{50}×\frac{\mathrm{4}}{\mathrm{3}}−\mathrm{100}=−\frac{\mathrm{100}}{\mathrm{3}}\:{N} \\ $$$${i}.{e}.\:{F}\:{is}\:{towards}\:{left}. \\ $$$$ \\ $$$$\Rightarrow{answer}\:\left({A}\right)\:{is}\:{correct}. \\ $$

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