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Question Number 217252 by mr W last updated on 07/Mar/25

Commented by mr W last updated on 07/Mar/25

$A s Q 217238, b u t t h e o t h e r e n d o f$ $t h e s t r i n g i s c o n n e c t e d w i t h a b l o c k$ $w i t h m a s s m w h i c h c a n s l i d e o n t h e$ $t a b l e f r i c t i o n l e s s l y .$ $f i n d t h e a c c e l e r a t i o n s o f t h e o b j e c t s$ $a n d t h e t e n s i o n i n s t r i n g .$
Commented by Tawa11 last updated on 07/Mar/25

$a = \frac{Mg}{M + m}$ $T = \frac{Mmg}{M + m}$ $Sir .$
Commented by mr W last updated on 07/Mar/25

$h o w ?$
Commented by Tawa11 last updated on 07/Mar/25

$Is it correct sir ?$
Commented by mr W last updated on 07/Mar/25

$n o!$
	Answered by mahdipoor last updated on 07/Mar/25
	$T=mA=mx^(..) { ((F=ma^− ⇒ −T+Mg=Ma=My^(..) )),((ΣM_(CM) =I^− α ⇒TR=Iθ^(..) )) :} y−θR=x ⇒ a−θ^(..) R=A ⇒⇒ T=m(a−θ^(..) R)=m((g−(T/M))−(((TR)/((MR^2 )/2)))R) ⇒T=((Mmg)/(M+3m))=cte ⇒A=(T/m) ⇒a=g−(T/M)=g−(m/M)A$
	$T = m A = m \overset{. .}{x}$ ${\begin{cases} F = m \bar{a} \Rightarrow - T + M g = M a = M \overset{. .}{y} \\ Σ M_{C M} = \bar{I} α \Rightarrow T R = I \overset{. .}{θ} \end{cases}$ $y - θ R = x \Rightarrow a - \overset{. .}{θ} R = A$ $\Rightarrow\Rightarrow$ $T = m (a - \overset{. .}{θ} R) = m ((g - \frac{T}{M}) - (\frac{T R}{\frac{{M R}^{2}}{2}}) R)$ $\Rightarrow T = \frac{M m g}{M + 3 m} = c t e$ $\Rightarrow A = \frac{T}{m}$ $\Rightarrow a = g - \frac{T}{M} = g - \frac{m}{M} A$
	Commented by mr W last updated on 07/Mar/25
	��
	Commented by Tawa11 last updated on 08/Mar/25

	$Thanks sir .$
	Answered by aleks041103 last updated on 07/Mar/25

	$m A = T$ $M a = M g - T$ $(\frac{1}{2} {M R}^{2}) ε = T R \Rightarrow T = \frac{1}{2} M ε R$ $ε R = a - A$ $\Rightarrow T = \frac{1}{2} M (a - A)$ $m A = \frac{1}{2} M (a - A) \Rightarrow M a = (2 m + M) A \Rightarrow a = (1 + \frac{2 m}{M}) A$ $M a + m A = M g$ $\Rightarrow (3 m + M) A = M g$ $\Rightarrow A = \frac{M}{3 m + M} g a n d a = \frac{2 m + M}{3 m + M} g$
	Commented by mr W last updated on 07/Mar/25
	��
	Commented by Tawa11 last updated on 08/Mar/25

	$Thanks sir$
	Answered by mr W last updated on 08/Mar/25

	$α = a n g u l a r a c c e l e r a t i o n o f c y l i n d e r (↶)$ $a = a c c e l e r a t i o n o f c y l i n d e r (↓)$ $A = a c c e l e r a t i o n o f b l o c k o n t a b l e (\leftarrow)$ $w e h a v e :$ $a = α R + A$ $I = \frac{{M R}^{2}}{2}$ $α I = T R$ $\frac{α {M R}^{2}}{2} = T R$ $\Rightarrow T = \frac{α M R}{2}$ $M a = M g - T$ $M (α R + A) = M g - T$ $\Rightarrow T = M (g - α R - A)$ $T = m A$ $\frac{α M R}{2} = M (g - α R - A)$ $\Rightarrow A = g - \frac{3 α R}{2}$ $m A = \frac{α M R}{2}$ $m (g - \frac{3 α R}{2}) = \frac{α M R}{2}$ $\Rightarrow α = \frac{2 g}{R (3 + \frac{M}{m})}$ $\Rightarrow A = g - \frac{3 R}{2} \times \frac{2 g}{R (3 + \frac{M}{m})} = \frac{g}{1 + \frac{3 m}{M}}$ $\Rightarrow a = \frac{2 g R}{R (3 + \frac{M}{m})} + g (1 - \frac{3}{3 + \frac{M}{m}}) = \frac{(1 + \frac{2 m}{M}) g}{1 + \frac{3 m}{M}}$ $\Rightarrow T = m g (1 - \frac{3}{3 + \frac{M}{m}}) = \frac{M g}{3 + \frac{M}{m}}$ $i f m \to 0 : (\to f r e e f a l l)$ $α \to 0$ $a \to g, A \to g$ $T \to 0$ $i f m \to \infty : (\to c a s e i n Q 217238)$ $α \to \frac{2 g}{3 R}$ $a \to \frac{2 g}{3}, A \to 0$ $T \to \frac{M g}{3}$
	Commented by Tawa11 last updated on 08/Mar/25

	$Thanks sir .$
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