Question Number 40166 by ajfour last updated on 16/Jul/18
Answered by MrW3 last updated on 17/Jul/18
![a=acceleration of m (↑) A=acceleration of M_0 (→) a_1 =acceleration of M (↙) on M_0 with a_1 =a+A T=tension in string T−mg=ma ⇒T=m(g+a) Mg sin θ−T=M(a_1 −A cos θ)=M(a+A−A cos θ) Mg sin θ−m(g+a)=M(a+A−A cos θ) Mg sin θ−mg−ma=Ma+M(1−cos θ)A ⇒ (M+m)a+M(1−cos θ)A=(M sin θ−m)g ...(i) T−T cos θ+Mg cos θ sin θ=M_0 A m(1−cos θ)(g+a)+Mg cos θ sin θ=M_0 A ⇒m(1−cos θ)a+M_0 A=Mg cos θ sin θ+mg(1−cos θ) ...(ii) (i)×m(1−cos θ): ⇒(M+m)m(1−cos θ)a+Mm(1−cos θ)^2 A=(M sin θ−m)m(1−cos θ)g ...(iii) (ii)×(M+m): ⇒(M+m)m(1−cos θ)a+(M+m)M_0 A=M(M+m)g cos θ sin θ+(M+m)mg(1−cos θ) ...(iv) (iv)−(iii): [(M+m)M_0 −Mm(1−cos θ)^2 ]A={(M+m)[Mcos θsin θ+m(1−cos θ)]−(M sin θ−m)m(1−cos θ)}g ⇒A=(({(M+m)[Mcos θsin θ+m(1−cos θ)]−(M sin θ−m)m(1−cos θ)}g)/((M+m)M_0 −Mm(1−cos θ)^2 ))](Q40197.png)
$${a}={acceleration}\:{of}\:{m}\:\left(\uparrow\right) \\ $$$${A}={acceleration}\:{of}\:{M}_{\mathrm{0}} \:\left(\rightarrow\right) \\ $$$${a}_{\mathrm{1}} ={acceleration}\:{of}\:{M}\:\left(\swarrow\right)\:{on}\:{M}_{\mathrm{0}} \\ $$$${with}\:{a}_{\mathrm{1}} ={a}+{A} \\ $$$${T}={tension}\:{in}\:{string} \\ $$$${T}−{mg}={ma} \\ $$$$\Rightarrow{T}={m}\left({g}+{a}\right) \\ $$$${Mg}\:\mathrm{sin}\:\theta−{T}={M}\left({a}_{\mathrm{1}} −{A}\:\mathrm{cos}\:\theta\right)={M}\left({a}+{A}−{A}\:\mathrm{cos}\:\theta\right) \\ $$$${Mg}\:\mathrm{sin}\:\theta−{m}\left({g}+{a}\right)={M}\left({a}+{A}−{A}\:\mathrm{cos}\:\theta\right) \\ $$$${Mg}\:\mathrm{sin}\:\theta−{mg}−{ma}={Ma}+{M}\left(\mathrm{1}−\mathrm{cos}\:\theta\right){A} \\ $$$$\Rightarrow\:\left({M}+{m}\right){a}+{M}\left(\mathrm{1}−\mathrm{cos}\:\theta\right){A}=\left({M}\:\mathrm{sin}\:\theta−{m}\right){g}\:\:...\left({i}\right) \\ $$$${T}−{T}\:\mathrm{cos}\:\theta+{Mg}\:\mathrm{cos}\:\theta\:\mathrm{sin}\:\theta={M}_{\mathrm{0}} {A} \\ $$$${m}\left(\mathrm{1}−\mathrm{cos}\:\theta\right)\left({g}+{a}\right)+{Mg}\:\mathrm{cos}\:\theta\:\mathrm{sin}\:\theta={M}_{\mathrm{0}} {A} \\ $$$$\Rightarrow{m}\left(\mathrm{1}−\mathrm{cos}\:\theta\right){a}+{M}_{\mathrm{0}} {A}={Mg}\:\mathrm{cos}\:\theta\:\mathrm{sin}\:\theta+{mg}\left(\mathrm{1}−\mathrm{cos}\:\theta\right)\:\:...\left({ii}\right) \\ $$$$\left({i}\right)×{m}\left(\mathrm{1}−\mathrm{cos}\:\theta\right): \\ $$$$\Rightarrow\left({M}+{m}\right){m}\left(\mathrm{1}−\mathrm{cos}\:\theta\right){a}+{Mm}\left(\mathrm{1}−\mathrm{cos}\:\theta\right)^{\mathrm{2}} {A}=\left({M}\:\mathrm{sin}\:\theta−{m}\right){m}\left(\mathrm{1}−\mathrm{cos}\:\theta\right){g}\:\:...\left({iii}\right) \\ $$$$\left({ii}\right)×\left({M}+{m}\right): \\ $$$$\Rightarrow\left({M}+{m}\right){m}\left(\mathrm{1}−\mathrm{cos}\:\theta\right){a}+\left({M}+{m}\right){M}_{\mathrm{0}} {A}={M}\left({M}+{m}\right){g}\:\mathrm{cos}\:\theta\:\mathrm{sin}\:\theta+\left({M}+{m}\right){mg}\left(\mathrm{1}−\mathrm{cos}\:\theta\right)\:\:...\left({iv}\right) \\ $$$$\left({iv}\right)−\left({iii}\right): \\ $$$$\left[\left({M}+{m}\right){M}_{\mathrm{0}} −{Mm}\left(\mathrm{1}−\mathrm{cos}\:\theta\right)^{\mathrm{2}} \right]{A}=\left\{\left({M}+{m}\right)\left[{M}\mathrm{cos}\:\theta\mathrm{sin}\:\theta+{m}\left(\mathrm{1}−\mathrm{cos}\:\theta\right)\right]−\left({M}\:\mathrm{sin}\:\theta−{m}\right){m}\left(\mathrm{1}−\mathrm{cos}\:\theta\right)\right\}{g} \\ $$$$\Rightarrow{A}=\frac{\left\{\left({M}+{m}\right)\left[{M}\mathrm{cos}\:\theta\mathrm{sin}\:\theta+{m}\left(\mathrm{1}−\mathrm{cos}\:\theta\right)\right]−\left({M}\:\mathrm{sin}\:\theta−{m}\right){m}\left(\mathrm{1}−\mathrm{cos}\:\theta\right)\right\}{g}}{\left({M}+{m}\right){M}_{\mathrm{0}} −{Mm}\left(\mathrm{1}−\mathrm{cos}\:\theta\right)^{\mathrm{2}} } \\ $$
Commented by ajfour last updated on 17/Jul/18
$${Thank}\:{you}\:{Sir},\:{seems}\:{true}\:; \\ $$$${and}\:{excellent}\:{working}! \\ $$