Question Number 35737 by rahul 19 last updated on 22/May/18
Commented by rahul 19 last updated on 24/May/18
Thank you sir . ��
Commented by ajfour last updated on 23/May/18
$$\left({c}\right)\:{Tension}=\mathrm{10}{N} \\ $$$${a}_{{A}} ={a}_{{B}} =\mathrm{1}{m}/{s}^{\mathrm{2}} \\ $$$${a}_{{C}} =\mathrm{0}\:{m}/{s}^{\mathrm{2}} \:. \\ $$
Commented by rahul 19 last updated on 23/May/18
$${pls}\:{explain}. \\ $$$$\left({confused}\:{in}\:{block}\:{C},\:{why}\:{it}\:{will}\:{be}\:{at}\:\right. \\ $$$$\left.{rest}.?\right) \\ $$
Commented by ajfour last updated on 23/May/18
$${pulleys}\:{are}\:{part}\:{of}\:{block}\:{C}. \\ $$$${Net}\:{horizontal}\:{force}\:{on}\:{C} \\ $$$${including}\:{the}\:{pulleys}\:{is}\:{zero}. \\ $$$$\left({T}\:\uparrow+{T}\rightarrow\right)+\left({T}\:\downarrow+{T}\:\leftarrow\right)\:=\mathrm{0} \\ $$