Menu Close

Question-12672




Question Number 12672 by tawa last updated on 28/Apr/17
Answered by sandy_suhendra last updated on 28/Apr/17
Commented by sandy_suhendra last updated on 28/Apr/17
ΣF_y =0  Fcos60°+Fcos60°−W=0  (1/2)F+(1/2)F−162 = 0  F = 162 N
$$\Sigma\mathrm{F}_{\mathrm{y}} =\mathrm{0} \\ $$$$\mathrm{Fcos60}°+\mathrm{Fcos60}°−\mathrm{W}=\mathrm{0} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\mathrm{F}+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{F}−\mathrm{162}\:=\:\mathrm{0} \\ $$$$\mathrm{F}\:=\:\mathrm{162}\:\mathrm{N} \\ $$
Commented by tawa last updated on 28/Apr/17
Wow, God bless you sir.
$$\mathrm{Wow},\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$

Leave a Reply

Your email address will not be published. Required fields are marked *