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Question Number 27332 by ajfour last updated on 05/Jan/18

Commented by ajfour last updated on 05/Jan/18

Find acceleration of blue and brown blocks. Friction coefficient is 𝛍 everywhere (sufficiently less, and permits motion).

Findaccelerationofblueandbrownblocks.Frictioncoefficientisμeverywhere(sufficientlyless,andpermitsmotion).

Answered by mrW1 last updated on 05/Jan/18

Commented by mrW1 last updated on 05/Jan/18

Commented by mrW1 last updated on 06/Jan/18

(a_2 /a_1 )=tan θ N_2 =N sin θ−f cos θ=N(sin θ−μ cos θ) mg−f_2 −N cos θ−f sin θ=ma_2 mg−μN_2 −N (cos θ+μ sin θ)=ma_2 mg−μN(sin θ−μ cos θ)−N (cos θ+μ sin θ)=ma_2 mg−N(2μsin θ−μ^2 cos θ+cos θ)=ma_2 N_1 =mg+N cos θ+f sin θ=mg+N(cos θ+μ sin θ) N sin θ−f cos θ−f_1 =ma_1 N(sin θ−μ cos θ)−μN_1 =ma_1 N(sin θ−μ cos θ)−μmg−μN(cos θ+μ sin θ)=ma_1 N(sin θ−μ^2 sin θ−2μ cos θ)−μmg=ma_1 ((mg−N(2μsin θ−μ^2 cos θ+cos θ))/(N(sin θ−μ^2 sin θ−2μ cos θ)−μmg))=((ma_2 )/(ma_1 ))=tan θ=((sin θ)/(cos θ)) mg cos θ−N(2μsin θ cos θ−μ^2 cos^2 θ+cos^2 θ)=N(sin^2 θ−μ^2 sin^2 θ−2μ sin θ cos θ)−μmg sin θ N(2μsin θ cos θ−μ^2 cos^2 θ+cos^2 θ+sin^2 θ−μ^2 sin^2 θ−2μ sin θ cos θ)=mg(cos θ+μsin θ) N(1−μ^2 )=mg(cos θ+μsin θ) ⇒N=(((cos θ+μ sin θ)mg)/(1−μ^2 )) a_1 =((N(sin θ−μ^2 sin θ−2μ cos θ))/m)−μg a_1 =(((cos θ+μ sin θ)(sin θ−μ^2 sin θ−2μ cos θ)g)/(1−μ^2 ))−μg ⇒a_1 =[(cos θ+μ sin θ)(sin θ−((2μ)/(1−μ^2 )) cos θ)−μ]g a_2 =g−((N(2μ sin θ+cos θ−μ^2 cos θ))/m) a_2 =g−(((cos θ+μ sin θ)(2μ sin θ+cos θ−μ^2 cos θ)g)/(1−μ^2 )) ⇒a_2 =[1−(cos θ+μ sin θ)(((2μ)/(1−μ^2 )) sin θ+cos θ)]g

a2a1=tanθN2=Nsinθ−fcosθ=N(sinθ−μcosθ)mg−f2−Ncosθ−fsinθ=ma2mg−μN2−N(cosθ+μsinθ)=ma2mg−μN(sinθ−μcosθ)−N(cosθ+μsinθ)=ma2mg−N(2μsinθ−μ2cosθ+cosθ)=ma2N1=mg+Ncosθ+fsinθ=mg+N(cosθ+μsinθ)Nsinθ−fcosθ−f1=ma1N(sinθ−μcosθ)−μN1=ma1N(sinθ−μcosθ)−μmg−μN(cosθ+μsinθ)=ma1N(sinθ−μ2sinθ−2μcosθ)−μmg=ma1mg−N(2μsinθ−μ2cosθ+cosθ)N(sinθ−μ2sinθ−2μcosθ)−μmg=ma2ma1=tanθ=sinθcosθmgcosθ−N(2μsinθcosθ−μ2cos2θ+cos2θ)=N(sin2θ−μ2sin2θ−2μsinθcosθ)−μmgsinθN(2μsinθcosθ−μ2cos2θ+cos2θ+sin2θ−μ2sin2θ−2μsinθcosθ)=mg(cosθ+μsinθ)N(1−μ2)=mg(cosθ+μsinθ)⇒N=(cosθ+μsinθ)mg1−μ2a1=N(sinθ−μ2sinθ−2μcosθ)m−μga1=(cosθ+μsinθ)(sinθ−μ2sinθ−2μcosθ)g1−μ2−μg⇒a1=[(cosθ+μsinθ)(sinθ−2μ1−μ2cosθ)−μ]ga2=g−N(2μsinθ+cosθ−μ2cosθ)ma2=g−(cosθ+μsinθ)(2μsinθ+cosθ−μ2cosθ)g1−μ2⇒a2=[1−(cosθ+μsinθ)(2μ1−μ2sinθ+cosθ)]g

Commented by ajfour last updated on 05/Jan/18

Excellently presented and handled Sir!

ExcellentlypresentedandhandledSir!

Commented by mrW1 last updated on 06/Jan/18

Thank you sir! I have a problem with the result, to be exact, with the term (1/(1−μ^2 )). It means μ≠1. But why is it so? What happens if μ=1?

Thankyousir!Ihaveaproblemwiththeresult,tobeexact,withtheterm11−μ2.Itmeansμ≠1.Butwhyisitso?Whathappensifμ=1?

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