Question-217640

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

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} = \dots a_{2} = \dots a_{3} = \dots

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 \dots (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 \dots (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} \dots (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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