Question Number 24772 by Tinkutara last updated on 25/Nov/17
$$\mathrm{The}\:\mathrm{ratio}\:\mathrm{of}\:\mathrm{acceleration}\:\mathrm{of}\:\mathrm{points}\:{A}, \\ $$$${B}\:\mathrm{and}\:{C}\:\mathrm{is}\:\left[\mathrm{assume}\:\mathrm{all}\:\mathrm{surfaces}\:\mathrm{are}\right. \\ $$$$\left.\mathrm{smooth},\:\mathrm{pulley}\:\mathrm{and}\:\mathrm{strings}\:\mathrm{are}\:\mathrm{light}\right] \\ $$
Commented by Tinkutara last updated on 25/Nov/17
Commented by mrW1 last updated on 26/Nov/17
$${I}\:{think}\:{B}\:{and}\:{C}\:{can}\:{move}\:{independently} \\ $$$${from}\:{A},\:{therefore}\:{there}\:{is}\:{no}\:{certain} \\ $$$${ratio}\:{between}\:{a}_{{A}} \:{and}\:{a}_{{B}} . \\ $$
Commented by Tinkutara last updated on 26/Nov/17
$$\mathrm{Answer}\:\mathrm{is}\:\mathrm{wrong}. \\ $$$$\mathrm{Why}\:{a}_{{B}} ={a}_{{C}} =\frac{\mathrm{9}{F}}{{m}}? \\ $$
Commented by mrW1 last updated on 26/Nov/17
$${a}_{{A}} =\frac{\mathrm{3}{F}}{{m}} \\ $$$${a}_{{F}} =\mathrm{3}{a}_{{A}} =\frac{\mathrm{9}{F}}{{m}}\:\left({acceleration}\:{of}\:{the}\:{end}\:{of}\:{string}\right) \\ $$$${a}_{{B}} ={a}_{{C}} =\mathrm{0} \\ $$
Commented by Tinkutara last updated on 26/Nov/17
$${But}\:{answer}\:{is}\:{still}\:{wrong}\:{as}\:{ratio}\:{is} \\ $$$${integer}. \\ $$
Commented by mrW1 last updated on 26/Nov/17
$${This}\:{is}\:{I}\:{think}\:{the}\:{right}\:{answer}. \\ $$
Commented by ajfour last updated on 26/Nov/17
$${Is}\:{B}\:{a}\:{rope}\:{point}\:{or}\:{centre}\:{of} \\ $$$${pulley}\:? \\ $$
Commented by mrW1 last updated on 26/Nov/17
$${B}\:\left({as}\:{well}\:{as}\:{C}\right)\:{is}\:{center}\:{of}\:{pulley}. \\ $$
Commented by ajfour last updated on 26/Nov/17
$$\mathrm{3}{x}+\mathrm{2}{l}+\mathrm{3}{y}={constant} \\ $$$$\mathrm{3}\left({a}_{{C}} −{a}_{{A}} \right)−\mathrm{3}{a}_{{B}} +{a}_{{F}} =\mathrm{0}\:\:\:\:….\left({i}\right) \\ $$$${Fv}_{{F}} ={mv}_{{A}} {a}_{{A}} \\ $$$$\Rightarrow{Fa}_{{F}} ={ma}_{{A}} ^{\mathrm{2}} +\mathrm{0} \\ $$$${Further}\:\:\:\:\mathrm{3}{F}={ma}_{{A}} \\ $$$${a}_{{A}} =\frac{\mathrm{3}{F}}{{m}}\: \\ $$$$\Rightarrow\:\:\:{a}_{{F}} =\mathrm{3}{a}_{{A}} \:=\:\frac{\mathrm{9}{F}}{{m}}\:. \\ $$$${Now}\:{from}\:\left({i}\right): \\ $$$${a}_{{C}} ={a}_{{B}} \:=\:\left({whatever},\:{depends}\:{on}\right. \\ $$$$\left.{mass}\:{of}\:{small}\:{pulleys}\right). \\ $$
Commented by Tinkutara last updated on 26/Nov/17
$${Answer}\:{is}\:\mathrm{3}:\mathrm{1}:\mathrm{1} \\ $$
Commented by jota+ last updated on 26/Nov/17
$${La}\:{resultante}\:{sobre}\:{el}\:{sistema}\:{B} \\ $$$${y}\:{C}\:{is}\:{zero}.\:\mathrm{2}{F}−\mathrm{2}{F}=\mathrm{0} \\ $$$$\Rightarrow\:{a}_{{B}} ={a}_{{C}} =\mathrm{0} \\ $$
Commented by mrW1 last updated on 26/Nov/17
$${Answer}\:\mathrm{3}:\mathrm{1}:\mathrm{1}\:{is}\:{wrong}\:{in}\:{each}\:{case}, \\ $$$${that}\:{wurde}\:{mean}\:{a}_{{A}} =\:\mathrm{3}{a}_{{B}} =\mathrm{3}{a}_{{C}} . \\ $$$${But}\:{this}\:{is}\:{wrong}\:{even}\:{if}\:{B}\:{and}\:{C} \\ $$$${are}\:{the}\:{rope}\:{points}.\:{If}\:{B}\:{and}\:{C}\:{are} \\ $$$${the}\:{points}\:{on}\:{the}\:{rope},\:{then}\:{the} \\ $$$${answer}\:{should}\:{be}\:\mathrm{1}:\mathrm{3}:\mathrm{3}\:{with} \\ $$$${a}_{{A}} =\frac{\mathrm{3}{F}}{{m}} \\ $$$${a}_{{B}} ={a}_{{C}} =\mathrm{3}{a}_{{A}} =\frac{\mathrm{9}{F}}{{m}} \\ $$$${This}\:{was}\:{my}\:{original}\:{answer}. \\ $$$$ \\ $$$${If}\:{B}\:{and}\:{C}\:{are}\:{the}\:{center}\:{of}\:{pulley}, \\ $$$${then}\:{a}_{{A}} :{a}_{{B}} :{a}_{{C}} ={uncertain}. \\ $$
Commented by Tinkutara last updated on 26/Nov/17
$${Thanks}\:{Sir}!\:{I}\:{will}\:{confirm}. \\ $$