Question Number 138631 by physicstutes last updated on 15/Apr/21
Commented by physicstutes last updated on 15/Apr/21
$$\mathrm{From}\:\mathrm{the}\:\mathrm{figure}\:\mathrm{above}\:\mathrm{show}\:\mathrm{completely}\:\mathrm{how}\:\mathrm{the}\:\mathrm{force}\:\mathrm{on}\:\mathrm{the} \\ $$$$\mathrm{pulley}\:\mathrm{can}\:\mathrm{be}\:\mathrm{obtained}. \\ $$
Commented by Dwaipayan Shikari last updated on 15/Apr/21
$${For}\:\mathrm{3}\:{kg}\:{block} \\ $$$$\mathrm{30}−\mathrm{3}{gsin}\mathrm{30}°=\mathrm{3}{a}\Rightarrow\mathrm{10}−\frac{{g}}{\mathrm{2}}={a} \\ $$
Commented by physicstutes last updated on 15/Apr/21
$$\mathrm{true},\:\mathrm{but}\:\mathrm{how}\:\mathrm{do}\:\mathrm{we}\:\mathrm{calculate}\:\mathrm{the}\:\mathrm{force}\:\mathrm{exerted} \\ $$$$\mathrm{on}\:\mathrm{the}\:\mathrm{pulley} \\ $$
Answered by mr W last updated on 15/Apr/21
Commented by physicstutes last updated on 16/Apr/21
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{so}\:\mathrm{much}\:\mathrm{sir} \\ $$