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Question Number 53684 by ajfour last updated on 25/Jan/19

Commented by ajfour last updated on 25/Jan/19

ceiling to floor height is h.

ceilingtofloorheightish.

Commented by mr W last updated on 25/Jan/19

if no friction, then N=((Mg)/2)

ifnofriction,thenN=Mg2

Commented by mr W last updated on 25/Jan/19

Commented by ajfour last updated on 25/Jan/19

thanks for the tolerance, Sir!

thanksforthetolerance,Sir!

Answered by mr W last updated on 25/Jan/19

Commented by mr W last updated on 25/Jan/19

assume μ=friction on ground a sin α+b sin β=h ⇒sin α=((h−b sin β)/a) ⇒cos α=((√(a^2 −(h−b sin β)^2 ))/a) f=μN N b cos β=Mg ((b cos β)/2)+f b sin β ⇒(1−μ tan β)N=((Mg)/2) N (b cos β−a cos α)+mg((a cos α)/2)=Mg (((b cos β)/2)−a cos α)+fh 2N(b cos β−a cos α−μh)=Mgb cos β−(2M+m)ga cos α ((Mg)/(1−μ tan β))(b cos β−a cos α−μh)=Mgb cos β−(2M+m)ga cos α (1/(1−μ tan β))(b cos β−a cos α−μh)=b cos β−(2+(m/M))a cos α ((b cos β−μh)/(1−μ tan β))=b cos β−(2+(m/M)−(1/(1−μ tan β)))a cos α ⇒((μ(h−b sin β))/(1−μ tan β))=(2+(m/M)−(1/(1−μ tan β)))(√(a^2 −(h−b sin β)^2 )) ⇒β=.... ⇒N=((Mg)/(2(1−μ tan β)))

assumeμ=frictionongroundasinα+bsinβ=h⇒sinα=h−bsinβa⇒cosα=a2−(h−bsinβ)2af=μNNbcosβ=Mgbcosβ2+fbsinβ⇒(1−μtanβ)N=Mg2N(bcosβ−acosα)+mgacosα2=Mg(bcosβ2−acosα)+fh2N(bcosβ−acosα−μh)=Mgbcosβ−(2M+m)gacosαMg1−μtanβ(bcosβ−acosα−μh)=Mgbcosβ−(2M+m)gacosα11−μtanβ(bcosβ−acosα−μh)=bcosβ−(2+mM)acosαbcosβ−μh1−μtanβ=bcosβ−(2+mM−11−μtanβ)acosα⇒μ(h−bsinβ)1−μtanβ=(2+mM−11−μtanβ)a2−(h−bsinβ)2⇒β=....⇒N=Mg2(1−μtanβ)

Commented by ajfour last updated on 25/Jan/19

Very Nice Sir; thanks for attempting this way too.

VeryNiceSir;thanksforattemptingthiswaytoo.

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