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Question Number 217640 by Tawa11 last updated on 17/Mar/25

Commented by mr W last updated on 17/Mar/25

$F s h o u l d a l s o b e g i v e n t o g e t a$ $c o n c r e t e r e s u l t .$
	Answered by mahdipoor last updated on 17/Mar/25
	$for B : (ΣF=ma) (→ ≡ +) F−T_1 =M_1 a_1 =10a_1 (1) for A : T_2 =M_2 a_2 =5a_2 (2) for C : T_3 =M_3 a_3 =2a_3 (3) for pully : a^− =a_1 , { ((a^− +θ^(..) r=a_1 +θ^(..) r=a_2 (4))),((a^− −θ^(..) r=a_1 −θ^(..) r=a_3 (5))) :} (↻ ≡ +) ΣF=ma^− ⇒ T_1 −T_2 −T_3 =2a_1 (6) ΣM^− =I^− θ^(..) ⇒ (T_3 −T_2 )r=((mr^2 )/2)θ^(..) ⇒T_3 −T_2 =rθ^(..) (7) 7 unkown (3T+3a+θ^(..) r) and 7 eq ⇒ a_1 =... a_2 =... a_3 =...$
	$f o r B : (Σ F = m a) (\to \equiv +)$ $F - T_{1} = M_{1} a_{1} = 10 a_{1} (1)$ $f o r A :$ $T_{2} = M_{2} a_{2} = 5 a_{2} (2)$ $f o r C :$ $T_{3} = M_{3} a_{3} = 2 a_{3} (3)$ $f o r p u l l y :$ $\bar{a} = a_{1}, {\begin{cases} \bar{a} + \overset{. .}{θ} r = a_{1} + \overset{. .}{θ} r = a_{2} (4) \\ \bar{a} - \overset{. .}{θ} r = a_{1} - \overset{. .}{θ} r = a_{3} (5) \end{cases} (↻ \equiv +)$ $Σ F = m \bar{a} \Rightarrow T_{1} - T_{2} - T_{3} = 2 a_{1} (6)$ $Σ \bar{M} = \bar{I} \overset{. .}{θ} \Rightarrow (T_{3} - T_{2}) r = \frac{{m r}^{2}}{2} \overset{. .}{θ} \Rightarrow T_{3} - T_{2} = r \overset{. .}{θ} (7)$ $7 u n k o w n (3 T + 3 a + \overset{. .}{θ} r) a n d 7 e q \Rightarrow$ $a_{1} = . . . a_{2} = . . . a_{3} = . . .$
	Commented by Tawa11 last updated on 17/Mar/25

	$Thanks sir .$ $I appreciate .$
	Answered by mr W last updated on 17/Mar/25

	Commented by mr W last updated on 17/Mar/25

	$α = a n g u l a r a c c . o f p u l l e y (↶)$ $A_{1} = a c c . o f M_{1} a n d m$ $A_{2} = a c c . o f M_{2} = A_{1} + α r$ $A_{3} = a c c . o f M_{3} = A_{1} - α r$ $F_{2} = M_{2} A_{2} = M_{2} (A_{1} + α r)$ $F_{3} = M_{3} A_{3} = M_{3} (A_{1} - α r)$ $F - F_{2} - F_{3} = (M_{1} + m) A_{1}$ $F = (M_{1} + m) A_{1} + M_{2} (A_{1} + α r) + M_{3} (A_{1} - α r)$ $\Rightarrow (M_{1} + m + M_{2} + M_{3}) A_{1} + (M_{2} - M_{3}) α r = F . . . (i)$ $F_{3} r - F_{2} r = I α = \frac{{m r}^{2} α}{2}$ $M_{3} (A_{1} - α r) - M_{2} (A_{1} + α r) = \frac{m r α}{2}$ $\Rightarrow (M_{2} - M_{3}) A_{1} + (M_{2} + M_{3} + \frac{m}{2}) α r = 0 . . . (i i)$ $f r o m (i i) :$ $α r = - \frac{(M_{2} - M_{3}) A_{1}}{(M_{2} + M_{3} + \frac{m}{2})}$ $t h i s i n t o (i) :$ $[M_{1} + m + M_{2} + M_{3} - \frac{{(M_{2} - M_{3})}^{2}}{M_{2} + M_{3} + \frac{m}{2}}] A_{1} = F$ $\Rightarrow A_{1} = \frac{F}{M_{1} + m + M_{2} + M_{3} - \frac{{(M_{2} - M_{3})}^{2}}{M_{2} + M_{3} + \frac{m}{2}}}$ $\Rightarrow A_{2} = (1 - \frac{M_{2} - M_{3}}{M_{2} + M_{3} + \frac{m}{2}}) \frac{F}{M_{1} + m + M_{2} + M_{3} - \frac{{(M_{2} - M_{3})}^{2}}{M_{2} + M_{3} + \frac{m}{2}}}$ $\Rightarrow A_{3} = (1 + \frac{M_{2} - M_{3}}{M_{2} + M_{3} + \frac{m}{2}}) \frac{F}{M_{1} + m + M_{2} + M_{3} - \frac{{(M_{2} - M_{3})}^{2}}{M_{2} + M_{3} + \frac{m}{2}}}$
	Commented by Tawa11 last updated on 17/Mar/25

	$Great sir .$ $Weldone .$ $I appreciate sir .$
	Commented by Tawa11 last updated on 17/Mar/25

	Commented by Tawa11 last updated on 17/Mar/25

	$Sir, why is it (M + m) A_{1}$ $why not {mA}_{1} .$ $I am just asking to learn why sir .$
	Commented by mr W last updated on 17/Mar/25

	$m a n d M_{1} h a v e t h e s a m e a c c . A_{1},$ $i s t h i s c l e a r ?$ $w e h a v e$ $F - F_{1} = M_{1} A_{1} . . . (i)$ $F_{1} - F_{2} - F_{3} = {m A}_{1} . . (i i)$ $i s t h i s c l e a r ?$ $w e c a n a l s o t r e a t m a n d M_{1} a s o n e$ $o b j e c t w h i c h h a s a c c . A_{1}, t h e n w e$ $h a v e$ $F - F_{2} - F_{3} = (M_{1} + m) A_{1}$ $c e r t a i n l y y o u c a n g e t t h e s a m e b y$ $a d d i n g (i) + (i i) .$
	Commented by Tawa11 last updated on 17/Mar/25

	$I understand now .$ $Thank you sir .$
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