Question Number 20761 by Tinkutara last updated on 02/Sep/17
$${A}\:\mathrm{5}\:{kg}\:{block}\:{B}\:{is}\:{suspended}\:{from}\:{a} \\ $$$${cord}\:{attached}\:{to}\:{a}\:\mathrm{40}\:{kg}\:{cart}\:{A}.\:{Find} \\ $$$${the}\:{accelerations}\:{of}\:{both}\:{the}\:{block}\:{and} \\ $$$${cart}.\:\left({All}\:{surfaces}\:{are}\:{frictionless}\right) \\ $$$$\left({g}\:=\:\mathrm{10}\:{m}/{s}^{\mathrm{2}} \right) \\ $$
Commented by Tinkutara last updated on 02/Sep/17
Commented by ajfour last updated on 02/Sep/17
$${let}\:{vertical}\:{acceleration}\:{of}\: \\ $$$${block}\:{be}\:\boldsymbol{{a}},\:{and}\:{acceleeation}\:{of} \\ $$$${cart}\:{be}\:\boldsymbol{{A}}. \\ $$$${mg}â{T}={ma} \\ $$$${T}=\:\left({M}+{m}\right){A} \\ $$$${A}={a} \\ $$$$\Rightarrow\:\:\:\:\:{mg}=\left({M}+\mathrm{2}{m}\right){a} \\ $$$$\:\:\:\:\boldsymbol{{a}}=\boldsymbol{{A}}=\frac{{mg}}{{M}+\mathrm{2}{m}}\:=\frac{\left(\mathrm{5}{kg}\right)\left(\mathrm{10}{m}/{s}^{\mathrm{2}} \right)}{\mathrm{50}{kg}} \\ $$$$\:\:\:\:\boldsymbol{{a}}=\boldsymbol{{A}}\:=\mathrm{1}{m}/{s}^{\mathrm{2}} \: \\ $$$${net}\:{acceleration}\:{of}\:{block} \\ $$$$\:\:\:{is}\:=\:\sqrt{\mathrm{2}}\:{m}/{s}^{\mathrm{2}} \:. \\ $$
Commented by Tinkutara last updated on 02/Sep/17
$$\mathrm{2}\:\mathrm{doubts}: \\ $$$$\left(\mathrm{1}\right)\:\mathrm{Why}\:{T}=\left({M}+{m}\right){A}? \\ $$$$\left(\mathrm{2}\right)\:\mathrm{Why}\:{mg}=\left({M}+\mathrm{2}{m}\right){a}? \\ $$
Commented by ajfour last updated on 02/Sep/17
$$\left(\mathrm{1}\right)\:{for}\:{the}\:{horizontal}\:{motion}\:{both} \\ $$$${cart}\:{and}\:{block}\:{can}\:{be}\:{treated}\:{as} \\ $$$${one}\:{of}\:{mass}\:\left({M}+{m}\right). \\ $$$$\left(\mathrm{2}\right)\:{I}\:{added}\:{the}\:{first}\:{two}\:{equations} \\ $$$${and}\:{with}\:{a}={A}\:{we}\:{do}\:{get} \\ $$$$\:\:\:\:\:\:{mg}=\left({M}+\mathrm{2}{m}\right){a}\:. \\ $$
Commented by Tinkutara last updated on 02/Sep/17
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{Sir}! \\ $$