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Question-102880




Question Number 102880 by Dwaipayan Shikari last updated on 11/Jul/20
Commented by Dwaipayan Shikari last updated on 11/Jul/20
Sorry sir . I did not include it . The weight recorder is mass  less.The man stands on it.if Recorder records 72kg as the  weight of the man the. Here the weight recorder is a trianguler  platform. And it slides. Friction is along the slope
$${Sorry}\:{sir}\:.\:{I}\:{did}\:{not}\:{include}\:{it}\:.\:{The}\:{weight}\:{recorder}\:{is}\:{mass} \\ $$$${less}.{The}\:{man}\:{stands}\:{on}\:{it}.{if}\:{Recorder}\:{records}\:\mathrm{72}{kg}\:{as}\:{the} \\ $$$${weight}\:{of}\:{the}\:{man}\:{the}.\:{Here}\:{the}\:{weight}\:{recorder}\:{is}\:{a}\:{trianguler} \\ $$$${platform}.\:{And}\:{it}\:{slides}.\:{Friction}\:{is}\:{along}\:{the}\:{slope} \\ $$$$ \\ $$
Commented by mr W last updated on 11/Jul/20
do you understand based on the text  how this recorder works and what  the system does? does the man  stand on the recorder which slideo  along the wedge? can the wedge move  on the ground? it talks about  friction, which friction is meant?  between the recorder and the wedge  or elsewhere?
$${do}\:{you}\:{understand}\:{based}\:{on}\:{the}\:{text} \\ $$$${how}\:{this}\:{recorder}\:{works}\:{and}\:{what} \\ $$$${the}\:{system}\:{does}?\:{does}\:{the}\:{man} \\ $$$${stand}\:{on}\:{the}\:{recorder}\:{which}\:{slideo} \\ $$$${along}\:{the}\:{wedge}?\:{can}\:{the}\:{wedge}\:{move} \\ $$$${on}\:{the}\:{ground}?\:{it}\:{talks}\:{about} \\ $$$${friction},\:{which}\:{friction}\:{is}\:{meant}? \\ $$$${between}\:{the}\:{recorder}\:{and}\:{the}\:{wedge} \\ $$$${or}\:{elsewhere}? \\ $$
Commented by Dwaipayan Shikari last updated on 11/Jul/20
Commented by prakash jain last updated on 11/Jul/20
Can you use smaller lines. You may  be using a tablet but most of us use  phones.
$$\mathrm{Can}\:\mathrm{you}\:\mathrm{use}\:\mathrm{smaller}\:\mathrm{lines}.\:\mathrm{You}\:\mathrm{may} \\ $$$$\mathrm{be}\:\mathrm{using}\:\mathrm{a}\:\mathrm{tablet}\:\mathrm{but}\:\mathrm{most}\:\mathrm{of}\:\mathrm{us}\:\mathrm{use} \\ $$$$\mathrm{phones}. \\ $$
Commented by Dwaipayan Shikari last updated on 11/Jul/20
I will maintain it from the next time
$${I}\:{will}\:{maintain}\:{it}\:{from}\:{the}\:{next}\:{time} \\ $$
Answered by mr W last updated on 11/Jul/20
Commented by mr W last updated on 11/Jul/20
N=mg cos θ  f=μN=μmg cos θ  ma=mg sin θ−f=mg sin θ−μmg cos θ  a=g(sin θ−μ cos θ)  ⇒μ=(sin θ−(a/g))(1/(cos θ))  mg−N_1 =ma sin θ  ⇒a=((mg−N_1 )/(m sin θ))=((80×10−72×10)/(80×0.6))=(5/3) m/s^2   ⇒μ=((3/5)−(5/(3×10)))(1/(4/5))=((13)/(24))=0.54
$${N}={mg}\:\mathrm{cos}\:\theta \\ $$$${f}=\mu{N}=\mu{mg}\:\mathrm{cos}\:\theta \\ $$$${ma}={mg}\:\mathrm{sin}\:\theta−{f}={mg}\:\mathrm{sin}\:\theta−\mu{mg}\:\mathrm{cos}\:\theta \\ $$$${a}={g}\left(\mathrm{sin}\:\theta−\mu\:\mathrm{cos}\:\theta\right) \\ $$$$\Rightarrow\mu=\left(\mathrm{sin}\:\theta−\frac{{a}}{{g}}\right)\frac{\mathrm{1}}{\mathrm{cos}\:\theta} \\ $$$${mg}−{N}_{\mathrm{1}} ={ma}\:\mathrm{sin}\:\theta \\ $$$$\Rightarrow{a}=\frac{{mg}−{N}_{\mathrm{1}} }{{m}\:\mathrm{sin}\:\theta}=\frac{\mathrm{80}×\mathrm{10}−\mathrm{72}×\mathrm{10}}{\mathrm{80}×\mathrm{0}.\mathrm{6}}=\frac{\mathrm{5}}{\mathrm{3}}\:{m}/{s}^{\mathrm{2}} \\ $$$$\Rightarrow\mu=\left(\frac{\mathrm{3}}{\mathrm{5}}−\frac{\mathrm{5}}{\mathrm{3}×\mathrm{10}}\right)\frac{\mathrm{1}}{\frac{\mathrm{4}}{\mathrm{5}}}=\frac{\mathrm{13}}{\mathrm{24}}=\mathrm{0}.\mathrm{54} \\ $$
Commented by mr W last updated on 11/Jul/20
Commented by Dwaipayan Shikari last updated on 11/Jul/20
Thanking you. Great!
$${Thanking}\:{you}.\:{Great}! \\ $$

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