Neglecting friction and mass of pulleys, what is the acceleration of mass B?


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Question Number 24482 by Tinkutara last updated on 18/Nov/17

$Neglecting friction and mass of pulleys,$ $what is the acceleration of mass B ?$
Commented by ajfour last updated on 18/Nov/17

$L e t t e n s i o n i n s t r i n g o n w h i c h$ $B h a n g s b e T .$ $a_{B} = 2 a_{A} . . . . (i)$ $m g - T = m (2 a_{A}) . . . . . (i i)$ $2 T - m g = {m a}_{A} . . . . . (i i i)$ $\Rightarrow a d d i n g 2 \times (i i) a n d (i i i),$ $m g = 5 {m a}_{A}$ $\Rightarrow a_{B} = 2 a_{A} = \frac{2 g}{5} .$
Commented by Tinkutara last updated on 18/Nov/17

Commented by mrW1 last updated on 18/Nov/17

$a_{B} = 2 a_{A}$ $m_{B} g - T_{B} = m_{B} a_{B}$ $\Rightarrow T_{B} = m_{B} (g - a_{B})$ $T_{A} - m_{A} g = m_{A} a_{A}$ $\Rightarrow T_{A} = m_{A} (a_{A} + g) = m_{A} (\frac{a_{B}}{2} + g)$ $T_{A} = 2 T_{B}$ $m_{A} (\frac{a_{B}}{2} + g) = 2 m_{B} (g - a_{B})$ $(m_{A} + 4 m_{B}) a_{B} = 2 (2 m_{B} - m_{A}) g$ $\Rightarrow a_{B} = \frac{2 (2 m_{B} - m_{A})}{m_{A} + 4 m_{B}} \times g$ $w i t h m_{A} = m_{B} = m$ $\Rightarrow a_{B} = \frac{2 (2 - 1)}{1 + 4} \times g = 0.4 g$ $i f m_{A} = 2 m_{B}$ $\Rightarrow a_{B} = 0, i . e . n o m o t i o n .$ $f o r m_{A} > 2 m_{B}$ $\Rightarrow a_{B} < 0, i . e . b l o c k B m o v e s u p w a r d s .$
Commented by Tinkutara last updated on 19/Nov/17

$Thank you very much Sirs!$
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