Question Number 24149 by ajfour last updated on 13/Nov/17

Answered by mrW1 last updated on 13/Nov/17
![x_1 =s_1 y_1 =0 x_2 =x_1 +s_2 cos α=s_1 +s_2 cos α y_2 =s_2 sin α x_3 =x_2 +s_3 cos 2α=s_1 +s_2 cos α+s_3 cos 2α y_3 =y_2 +s_3 sin 2α=s_2 sin α+s_3 sin 2α let a=(d^2 s_1 /dt^2 )=acceleration of M let b=(d^2 s_2 /dt^2 ) let c=(d^2 s_3 /dt^2 ) a_(1x) =a a_(1y) =0 a_(2x) =a+bcos α a_(2y) =bsin α a_(3x) =a+bcos α+ccos 2α a_(3y) =bsin α+csin 2α −N_2 sin 2α=m_0 a_(3x) =m_0 (a+bcos α+ccos 2α) N_2 cos 2α−m_0 g=m_0 a_(3y) =m_0 (bsin α+csin 2α) −N_1 sin α+N_2 sin 2α=ma_(2x) =m(a+bcos α) N_1 cos α−N_2 cos 2α−mg=ma_(2y) =m(bsin α) N_1 sin α=Ma_(1x) =Ma N_0 −N_1 cos α−Mg=Ma_(1y) =0 ⇒N_1 =((Ma)/(sin α)) ⇒N_2 sin 2α=m(a+bcos α)+Ma ⇒Matan α−[Ma+m(a+bcos α)]cot 2α−mg=m(bsin α) ⇒Mtan α a−[Ma+ma+mbcos α]cot 2α−mg=mbsin α ⇒[Mtan α −(M+m)cot 2α]a−m[sin α+cos α cot 2α]b=mg ...(i) ⇒−m(a+bcos α)−Ma=m_0 (a+bcos α+ccos 2α) ⇒[(m+m_0 )cos α]b+[m_0 cos 2α]c=−(M+m+m_0 )a ⇒[ma+mbcos α+Ma]cot 2α−m_0 g=m_0 (bsin α+csin 2α) ⇒[m_0 sin α+mcos αcot 2α]b−[m_0 sin 2α]c=m_0 g−[(M+m)cot 2α]a ⇒[(M+m+m_0 )sin 2α+(M+m)cot 2α cos 2α]a+[(m+m_0 )cos α sin 2α+(m_0 sin α+mcos αcot 2α)cos 2α]b=m_0 gcos 2α ((Mtan α −(M+m)cot 2α)/(m(sin α+cos α cot 2α)))×a −b=(g/(sin α+cos α cot 2α)) (((M+m+m_0 )sin 2α+(M+m)cot 2α cos 2α)/((m_0 sin α+mcos αcot 2α)cos 2α))×a+b=((m_0 gcos 2α)/((m_0 sin α+mcos αcot 2α)cos 2α)) ⇒a=(((1/(sin α+cos α cot 2α))+(m_0 /(m_0 sin α+mcos αcot 2α)))/(((Mtan α −(M+m)cot 2α)/(m(sin α+cos α cot 2α)))+(((M+m+m_0 )tan 2α+(M+m)cot 2α)/(m_0 sin α+mcos αcot 2α))))×g ⇒a=((mm_0 sin α+m(m+m_0 )cos αcot 2α)/([Mtan α −(M+m)cot 2α][m_0 sin α+mcos αcot 2α]+m(sin α+cos α cot 2α)[(M+m+m_0 )tan 2α+(M+m)cot 2α]))×g](https://www.tinkutara.com/question/Q24169.png)
$${x}_{\mathrm{1}} ={s}_{\mathrm{1}} \\ $$$${y}_{\mathrm{1}} =\mathrm{0} \\ $$$${x}_{\mathrm{2}} ={x}_{\mathrm{1}} +{s}_{\mathrm{2}} \mathrm{cos}\:\alpha={s}_{\mathrm{1}} +{s}_{\mathrm{2}} \mathrm{cos}\:\alpha \\ $$$${y}_{\mathrm{2}} ={s}_{\mathrm{2}} \mathrm{sin}\:\alpha \\ $$$${x}_{\mathrm{3}} ={x}_{\mathrm{2}} +{s}_{\mathrm{3}} \mathrm{cos}\:\mathrm{2}\alpha={s}_{\mathrm{1}} +{s}_{\mathrm{2}} \mathrm{cos}\:\alpha+{s}_{\mathrm{3}} \mathrm{cos}\:\mathrm{2}\alpha \\ $$$${y}_{\mathrm{3}} ={y}_{\mathrm{2}} +{s}_{\mathrm{3}} \mathrm{sin}\:\mathrm{2}\alpha={s}_{\mathrm{2}} \mathrm{sin}\:\alpha+{s}_{\mathrm{3}} \mathrm{sin}\:\mathrm{2}\alpha \\ $$$${let}\:{a}=\frac{{d}^{\mathrm{2}} {s}_{\mathrm{1}} }{{dt}^{\mathrm{2}} }={acceleration}\:{of}\:{M} \\ $$$${let}\:{b}=\frac{{d}^{\mathrm{2}} {s}_{\mathrm{2}} }{{dt}^{\mathrm{2}} } \\ $$$${let}\:{c}=\frac{{d}^{\mathrm{2}} {s}_{\mathrm{3}} }{{dt}^{\mathrm{2}} } \\ $$$${a}_{\mathrm{1}{x}} ={a} \\ $$$${a}_{\mathrm{1}{y}} =\mathrm{0} \\ $$$${a}_{\mathrm{2}{x}} ={a}+{b}\mathrm{cos}\:\alpha \\ $$$${a}_{\mathrm{2}{y}} ={b}\mathrm{sin}\:\alpha \\ $$$${a}_{\mathrm{3}{x}} ={a}+{b}\mathrm{cos}\:\alpha+{c}\mathrm{cos}\:\mathrm{2}\alpha \\ $$$${a}_{\mathrm{3}{y}} ={b}\mathrm{sin}\:\alpha+{c}\mathrm{sin}\:\mathrm{2}\alpha \\ $$$$ \\ $$$$−{N}_{\mathrm{2}} \mathrm{sin}\:\mathrm{2}\alpha={m}_{\mathrm{0}} {a}_{\mathrm{3}{x}} ={m}_{\mathrm{0}} \left({a}+{b}\mathrm{cos}\:\alpha+{c}\mathrm{cos}\:\mathrm{2}\alpha\right) \\ $$$${N}_{\mathrm{2}} \mathrm{cos}\:\mathrm{2}\alpha−{m}_{\mathrm{0}} {g}={m}_{\mathrm{0}} {a}_{\mathrm{3}{y}} ={m}_{\mathrm{0}} \left({b}\mathrm{sin}\:\alpha+{c}\mathrm{sin}\:\mathrm{2}\alpha\right) \\ $$$$ \\ $$$$−{N}_{\mathrm{1}} \mathrm{sin}\:\alpha+{N}_{\mathrm{2}} \mathrm{sin}\:\mathrm{2}\alpha={ma}_{\mathrm{2}{x}} ={m}\left({a}+{b}\mathrm{cos}\:\alpha\right) \\ $$$${N}_{\mathrm{1}} \mathrm{cos}\:\alpha−{N}_{\mathrm{2}} \mathrm{cos}\:\mathrm{2}\alpha−{mg}={ma}_{\mathrm{2}{y}} ={m}\left({b}\mathrm{sin}\:\alpha\right) \\ $$$$ \\ $$$${N}_{\mathrm{1}} \mathrm{sin}\:\alpha={Ma}_{\mathrm{1}{x}} ={Ma} \\ $$$${N}_{\mathrm{0}} −{N}_{\mathrm{1}} \mathrm{cos}\:\alpha−{Mg}={Ma}_{\mathrm{1}{y}} =\mathrm{0} \\ $$$$ \\ $$$$\Rightarrow{N}_{\mathrm{1}} =\frac{{Ma}}{\mathrm{sin}\:\alpha} \\ $$$$\Rightarrow{N}_{\mathrm{2}} \mathrm{sin}\:\mathrm{2}\alpha={m}\left({a}+{b}\mathrm{cos}\:\alpha\right)+{Ma} \\ $$$$ \\ $$$$\Rightarrow{Ma}\mathrm{tan}\:\alpha−\left[{Ma}+{m}\left({a}+{b}\mathrm{cos}\:\alpha\right)\right]\mathrm{cot}\:\mathrm{2}\alpha−{mg}={m}\left({b}\mathrm{sin}\:\alpha\right) \\ $$$$\Rightarrow{M}\mathrm{tan}\:\alpha\:{a}−\left[{Ma}+{ma}+{mb}\mathrm{cos}\:\alpha\right]\mathrm{cot}\:\mathrm{2}\alpha−{mg}={mb}\mathrm{sin}\:\alpha \\ $$$$\Rightarrow\left[{M}\mathrm{tan}\:\alpha\:−\left({M}+{m}\right)\mathrm{cot}\:\mathrm{2}\alpha\right]{a}−{m}\left[\mathrm{sin}\:\alpha+\mathrm{cos}\:\alpha\:\mathrm{cot}\:\mathrm{2}\alpha\right]{b}={mg}\:\:…\left({i}\right) \\ $$$$ \\ $$$$\Rightarrow−{m}\left({a}+{b}\mathrm{cos}\:\alpha\right)−{Ma}={m}_{\mathrm{0}} \left({a}+{b}\mathrm{cos}\:\alpha+{c}\mathrm{cos}\:\mathrm{2}\alpha\right) \\ $$$$\Rightarrow\left[\left({m}+{m}_{\mathrm{0}} \right)\mathrm{cos}\:\alpha\right]{b}+\left[{m}_{\mathrm{0}} \mathrm{cos}\:\mathrm{2}\alpha\right]{c}=−\left({M}+{m}+{m}_{\mathrm{0}} \right){a} \\ $$$$ \\ $$$$\Rightarrow\left[{ma}+{mb}\mathrm{cos}\:\alpha+{Ma}\right]\mathrm{cot}\:\mathrm{2}\alpha−{m}_{\mathrm{0}} {g}={m}_{\mathrm{0}} \left({b}\mathrm{sin}\:\alpha+{c}\mathrm{sin}\:\mathrm{2}\alpha\right) \\ $$$$\Rightarrow\left[{m}_{\mathrm{0}} \mathrm{sin}\:\alpha+{m}\mathrm{cos}\:\alpha\mathrm{cot}\:\mathrm{2}\alpha\right]{b}−\left[{m}_{\mathrm{0}} \mathrm{sin}\:\mathrm{2}\alpha\right]{c}={m}_{\mathrm{0}} {g}−\left[\left({M}+{m}\right)\mathrm{cot}\:\mathrm{2}\alpha\right]{a} \\ $$$$ \\ $$$$\Rightarrow\left[\left({M}+{m}+{m}_{\mathrm{0}} \right)\mathrm{sin}\:\mathrm{2}\alpha+\left({M}+{m}\right)\mathrm{cot}\:\mathrm{2}\alpha\:\mathrm{cos}\:\mathrm{2}\alpha\right]{a}+\left[\left({m}+{m}_{\mathrm{0}} \right)\mathrm{cos}\:\alpha\:\mathrm{sin}\:\mathrm{2}\alpha+\left({m}_{\mathrm{0}} \mathrm{sin}\:\alpha+{m}\mathrm{cos}\:\alpha\mathrm{cot}\:\mathrm{2}\alpha\right)\mathrm{cos}\:\mathrm{2}\alpha\right]{b}={m}_{\mathrm{0}} {g}\mathrm{cos}\:\mathrm{2}\alpha \\ $$$$ \\ $$$$\frac{{M}\mathrm{tan}\:\alpha\:−\left({M}+{m}\right)\mathrm{cot}\:\mathrm{2}\alpha}{{m}\left(\mathrm{sin}\:\alpha+\mathrm{cos}\:\alpha\:\mathrm{cot}\:\mathrm{2}\alpha\right)}×{a}\:−{b}=\frac{{g}}{\mathrm{sin}\:\alpha+\mathrm{cos}\:\alpha\:\mathrm{cot}\:\mathrm{2}\alpha} \\ $$$$\frac{\left({M}+{m}+{m}_{\mathrm{0}} \right)\mathrm{sin}\:\mathrm{2}\alpha+\left({M}+{m}\right)\mathrm{cot}\:\mathrm{2}\alpha\:\mathrm{cos}\:\mathrm{2}\alpha}{\left({m}_{\mathrm{0}} \mathrm{sin}\:\alpha+{m}\mathrm{cos}\:\alpha\mathrm{cot}\:\mathrm{2}\alpha\right)\mathrm{cos}\:\mathrm{2}\alpha}×{a}+{b}=\frac{{m}_{\mathrm{0}} {g}\mathrm{cos}\:\mathrm{2}\alpha}{\left({m}_{\mathrm{0}} \mathrm{sin}\:\alpha+{m}\mathrm{cos}\:\alpha\mathrm{cot}\:\mathrm{2}\alpha\right)\mathrm{cos}\:\mathrm{2}\alpha} \\ $$$$ \\ $$$$\Rightarrow{a}=\frac{\frac{\mathrm{1}}{\mathrm{sin}\:\alpha+\mathrm{cos}\:\alpha\:\mathrm{cot}\:\mathrm{2}\alpha}+\frac{{m}_{\mathrm{0}} }{{m}_{\mathrm{0}} \mathrm{sin}\:\alpha+{m}\mathrm{cos}\:\alpha\mathrm{cot}\:\mathrm{2}\alpha}}{\frac{{M}\mathrm{tan}\:\alpha\:−\left({M}+{m}\right)\mathrm{cot}\:\mathrm{2}\alpha}{{m}\left(\mathrm{sin}\:\alpha+\mathrm{cos}\:\alpha\:\mathrm{cot}\:\mathrm{2}\alpha\right)}+\frac{\left({M}+{m}+{m}_{\mathrm{0}} \right)\mathrm{tan}\:\mathrm{2}\alpha+\left({M}+{m}\right)\mathrm{cot}\:\mathrm{2}\alpha}{{m}_{\mathrm{0}} \mathrm{sin}\:\alpha+{m}\mathrm{cos}\:\alpha\mathrm{cot}\:\mathrm{2}\alpha}}×{g} \\ $$$$\Rightarrow{a}=\frac{{mm}_{\mathrm{0}} \mathrm{sin}\:\alpha+{m}\left({m}+{m}_{\mathrm{0}} \right)\mathrm{cos}\:\alpha\mathrm{cot}\:\mathrm{2}\alpha}{\left[{M}\mathrm{tan}\:\alpha\:−\left({M}+{m}\right)\mathrm{cot}\:\mathrm{2}\alpha\right]\left[{m}_{\mathrm{0}} \mathrm{sin}\:\alpha+{m}\mathrm{cos}\:\alpha\mathrm{cot}\:\mathrm{2}\alpha\right]+{m}\left(\mathrm{sin}\:\alpha+\mathrm{cos}\:\alpha\:\mathrm{cot}\:\mathrm{2}\alpha\right)\left[\left({M}+{m}+{m}_{\mathrm{0}} \right)\mathrm{tan}\:\mathrm{2}\alpha+\left({M}+{m}\right)\mathrm{cot}\:\mathrm{2}\alpha\right]}×{g} \\ $$
Commented by ajfour last updated on 13/Nov/17

$${Sir},\:{after}\:{lot}\:{of}\:{calculations}, \\ $$$${I}\:{however},\:{got}\:{a}\:{simpler}\: \\ $$$${expression}\:{for}\:{acceleration}\:{of}\:{M}. \\ $$$$\:\boldsymbol{{a}}=\frac{{m}\left({m}+{m}_{\mathrm{0}} \right){g}\mathrm{sin}\:\alpha\mathrm{cos}\:\alpha}{{M}\left({m}+{m}_{\mathrm{0}} \mathrm{sin}\:^{\mathrm{2}} \alpha\right)+{m}\mathrm{sin}\:^{\mathrm{2}} \alpha\left({m}−{m}_{\mathrm{0}} \mathrm{cos}\:\mathrm{2}\alpha\right)}\:. \\ $$
Commented by mrW1 last updated on 13/Nov/17

$${Great}! \\ $$