Question Number 107764 by hhryhrry2 last updated on 12/Aug/20
$$\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{tan}\theta\:=\:\:\frac{\Sigma\:\mathrm{F}_{\mathrm{y}} }{\Sigma\:\mathrm{F}_{\mathrm{x}} } \\ $$$$ \\ $$
Commented by ajfour last updated on 12/Aug/20
$$\frac{{d}\bar {{p}}}{{dt}}=\sqrt{\left(\Sigma{F}_{{x}} \right)^{\mathrm{2}} +\left(\Sigma{F}_{{y}} \right)^{\mathrm{2}} }\:\left(\mathrm{cos}\:\theta\hat {{i}}+\mathrm{sin}\:\theta\hat {{j}}\right) \\ $$