Question Number 18922 by Tinkutara last updated on 01/Aug/17
$$\mathrm{Consider}\:\mathrm{a}\:\mathrm{block}\:\mathrm{sliding}\:\mathrm{over}\:\mathrm{a}\:\mathrm{smooth} \\ $$$$\mathrm{inclined}\:\mathrm{surface}\:\mathrm{of}\:\mathrm{inclination}\:\theta.\:\mathrm{Relating} \\ $$$$\mathrm{to}\:\mathrm{Newton}'\mathrm{s}\:\mathrm{second}\:\mathrm{law}\:\mathrm{applied}\:\mathrm{on}\:\mathrm{the} \\ $$$$\mathrm{block},\:\mathrm{select}\:\mathrm{the}\:\mathrm{incorrect}\:\mathrm{alternative}. \\ $$$$\left(\mathrm{1}\right)\:\Sigma{F}_{{y}'} \:\neq\:\mathrm{0} \\ $$$$\left(\mathrm{2}\right)\:\Sigma{F}_{{y}} \:=\:\mathrm{0} \\ $$$$\left(\mathrm{3}\right)\:\Sigma{F}_{{x}} \:=\:−{mg}\:\mathrm{sin}\:\theta \\ $$$$\left(\mathrm{4}\right)\:\Sigma{F}_{{x}'} \:<\:\mathrm{0} \\ $$
Commented by Tinkutara last updated on 01/Aug/17
Commented by Tinkutara last updated on 01/Aug/17
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{ajfour}\:\mathrm{Sir}! \\ $$
Commented by ajfour last updated on 01/Aug/17
$$\Sigma\mathrm{F}_{\mathrm{x}} =\mathrm{mgsin}\:\theta \\ $$