Question Number 31792 by ajfour last updated on 14/Mar/18
Commented by ajfour last updated on 14/Mar/18
Commented by ajfour last updated on 15/Mar/18
![T(cos β−cos α)+N_1 sin α−N_2 sin β=MA ......(i) mgsin α+mAcos α−T=ma T+mAcos β−mgsin β=ma ⇒ 2T=mg(sin α+sin β)+mA(cos α−cos β) ......(ii) N_1 =mgcos α−mAsin α ...(iii) N_2 =mgcos β+mAsin β ...(iv) using (ii), (iii), and (iv) in (i) : [mg(sin α+sin β)+mA(cos α−cos β)][cos β−cos α] +2(mgcos α−mAsin α)sin α −2(mgcos β+mAsin β)sin β=2MA A=((mg[(sin α+sin β)(cos β−cos α)−2sin αcos α+2sin βcos β ])/(2M+m[(cos β−cos α)^2 +2(sin^2 α+sin^2 β)])) A=((mg sin(𝛂−𝛃)[1−3cos(𝛂+𝛃)])/(2M+m[4−(cos𝛂+cos𝛃)^2 ])) .](https://www.tinkutara.com/question/Q31801.png)
$${T}\left(\mathrm{cos}\:\beta−\mathrm{cos}\:\alpha\right)+{N}_{\mathrm{1}} \mathrm{sin}\:\alpha−{N}_{\mathrm{2}} \mathrm{sin}\:\beta={MA} \\ $$$$……\left({i}\right) \\ $$$${mg}\mathrm{sin}\:\alpha+{mA}\mathrm{cos}\:\alpha−{T}={ma} \\ $$$${T}+{mA}\mathrm{cos}\:\beta−{mg}\mathrm{sin}\:\beta={ma} \\ $$$$\Rightarrow\:\mathrm{2}{T}={mg}\left(\mathrm{sin}\:\alpha+\mathrm{sin}\:\beta\right)+{mA}\left(\mathrm{cos}\:\alpha−\mathrm{cos}\:\beta\right) \\ $$$$……\left({ii}\right) \\ $$$${N}_{\mathrm{1}} ={mg}\mathrm{cos}\:\alpha−{mA}\mathrm{sin}\:\alpha\:\:\:…\left({iii}\right) \\ $$$${N}_{\mathrm{2}} ={mg}\mathrm{cos}\:\beta+{mA}\mathrm{sin}\:\beta\:\:\:\:…\left({iv}\right) \\ $$$${using}\:\left({ii}\right),\:\left({iii}\right),\:{and}\:\left({iv}\right)\:{in}\:\left(\boldsymbol{{i}}\right)\:: \\ $$$$\left[{mg}\left(\mathrm{sin}\:\alpha+\mathrm{sin}\:\beta\right)+{mA}\left(\mathrm{cos}\:\alpha−\mathrm{cos}\:\beta\right)\right]\left[\mathrm{cos}\:\beta−\mathrm{cos}\:\alpha\right] \\ $$$$+\mathrm{2}\left({mg}\mathrm{cos}\:\alpha−{mA}\mathrm{sin}\:\alpha\right)\mathrm{sin}\:\alpha \\ $$$$−\mathrm{2}\left({mg}\mathrm{cos}\:\beta+{mA}\mathrm{sin}\:\beta\right)\mathrm{sin}\:\beta=\mathrm{2}{MA} \\ $$$${A}=\frac{{mg}\left[\left(\mathrm{sin}\:\alpha+\mathrm{sin}\:\beta\right)\left(\mathrm{cos}\:\beta−\mathrm{cos}\:\alpha\right)−\mathrm{2sin}\:\alpha\mathrm{cos}\:\alpha+\mathrm{2sin}\:\beta\mathrm{cos}\:\beta\:\right]}{\mathrm{2}{M}+{m}\left[\left(\mathrm{cos}\:\beta−\mathrm{cos}\:\alpha\right)^{\mathrm{2}} +\mathrm{2}\left(\mathrm{sin}\:^{\mathrm{2}} \alpha+\mathrm{sin}\:^{\mathrm{2}} \beta\right)\right]}\: \\ $$$$\boldsymbol{{A}}=\frac{\boldsymbol{{mg}}\:\boldsymbol{{sin}}\left(\boldsymbol{\alpha}−\boldsymbol{\beta}\right)\left[\mathrm{1}−\mathrm{3}\boldsymbol{{cos}}\left(\boldsymbol{\alpha}+\boldsymbol{\beta}\right)\right]}{\mathrm{2}\boldsymbol{{M}}+\boldsymbol{{m}}\left[\mathrm{4}−\left(\boldsymbol{{cos}\alpha}+\boldsymbol{{cos}\beta}\right)^{\mathrm{2}} \right]}\:. \\ $$
Commented by ajfour last updated on 15/Mar/18
$${Sir},\:{kindly}\:{solve}\:{and}\:{help}\:{checking} \\ $$$${my}\:{solution}. \\ $$
Commented by mrW2 last updated on 15/Mar/18
$${nice}\:{and}\:{right}\:{solution}\:{sir}! \\ $$