Neglecting-friction-and-mass-of-pulleys-what-is-the-acceleration-of-mass-B-

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} \dots . (i)

m g - T = m (2 a_{A}) \dots . . (i i)

2 T - m g = {m a}_{A} \dots . . (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!