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Question Number 40166 by ajfour last updated on 16/Jul/18

	Answered by MrW3 last updated on 17/Jul/18
	$a=acceleration of m (↑) A=acceleration of M_0 (→) a_1 =acceleration of M (↙) on M_0 with a_1 =a+A T=tension in string T−mg=ma ⇒T=m(g+a) Mg sin θ−T=M(a_1 −A cos θ)=M(a+A−A cos θ) Mg sin θ−m(g+a)=M(a+A−A cos θ) Mg sin θ−mg−ma=Ma+M(1−cos θ)A ⇒ (M+m)a+M(1−cos θ)A=(M sin θ−m)g ...(i) T−T cos θ+Mg cos θ sin θ=M_0 A m(1−cos θ)(g+a)+Mg cos θ sin θ=M_0 A ⇒m(1−cos θ)a+M_0 A=Mg cos θ sin θ+mg(1−cos θ) ...(ii) (i)×m(1−cos θ): ⇒(M+m)m(1−cos θ)a+Mm(1−cos θ)^2 A=(M sin θ−m)m(1−cos θ)g ...(iii) (ii)×(M+m): ⇒(M+m)m(1−cos θ)a+(M+m)M_0 A=M(M+m)g cos θ sin θ+(M+m)mg(1−cos θ) ...(iv) (iv)−(iii): [(M+m)M_0 −Mm(1−cos θ)^2 ]A={(M+m)[Mcos θsin θ+m(1−cos θ)]−(M sin θ−m)m(1−cos θ)}g ⇒A=(({(M+m)[Mcos θsin θ+m(1−cos θ)]−(M sin θ−m)m(1−cos θ)}g)/((M+m)M_0 −Mm(1−cos θ)^2 ))$
	$a = a c c e l e r a t i o n o f m (↑)$ $A = a c c e l e r a t i o n o f M_{0} (\to)$ $a_{1} = a c c e l e r a t i o n o f M (↙) o n M_{0}$ $w i t h a_{1} = a + A$ $T = t e n s i o n i n s t r i n g$ $T - m g = m a$ $\Rightarrow T = m (g + a)$ $M g \sin θ - T = M (a_{1} - A \cos θ) = M (a + A - A \cos θ)$ $M g \sin θ - m (g + a) = M (a + A - A \cos θ)$ $M g \sin θ - m g - m a = M a + M (1 - \cos θ) A$ $\Rightarrow (M + m) a + M (1 - \cos θ) A = (M \sin θ - m) g . . . (i)$ $T - T \cos θ + M g \cos θ \sin θ = M_{0} A$ $m (1 - \cos θ) (g + a) + M g \cos θ \sin θ = M_{0} A$ $\Rightarrow m (1 - \cos θ) a + M_{0} A = M g \cos θ \sin θ + m g (1 - \cos θ) . . . (i i)$ $(i) \times m (1 - \cos θ) :$ $\Rightarrow (M + m) m (1 - \cos θ) a + M m {(1 - \cos θ)}^{2} A = (M \sin θ - m) m (1 - \cos θ) g . . . (i i i)$ $(i i) \times (M + m) :$ $\Rightarrow (M + m) m (1 - \cos θ) a + (M + m) M_{0} A = M (M + m) g \cos θ \sin θ + (M + m) m g (1 - \cos θ) . . . (i v)$ $(i v) - (i i i) :$ $[(M + m) M_{0} - M m {(1 - \cos θ)}^{2}] A = {(M + m) [M \cos θ \sin θ + m (1 - \cos θ)] - (M \sin θ - m) m (1 - \cos θ)} g$ $\Rightarrow A = \frac{{(M + m) [M \cos θ \sin θ + m (1 - \cos θ)] - (M \sin θ - m) m (1 - \cos θ)} g}{(M + m) M_{0} - M m {(1 - \cos θ)}^{2}}$
	Commented by ajfour last updated on 17/Jul/18

	$T h a n k y o u S i r, s e e m s t r u e;$ $a n d e x c e l l e n t w o r k i n g!$
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