Vector Addition

Forces $\frac{?}{\text{b}\text{a}\text{r}}(F{)}_1$ and $\frac{?}{\text{b}\text{a}\text{r}}(F{)}_2$ act at point $O$ as shown to produce the horizontal

resultant $\frac{?}{\text{b}\text{a}\text{r}}(R)\frac{?}{b}=ar(F{)}_1\frac{+}{b}ar(F{)}_2=24N.$

Determine the magnitude of force $\frac{?}{\text{b}\text{a}\text{r}}(F{)}_1.$

$F\frac{4}{5}+F_1\frac{3}{5}$

A$.20\mathrm{.}2N$

$F_2=F_{2}sin29-F_{2}cos29$

C$.21\mathrm{.}2N$

$\sqrt[\phantom{2}]{2}*484-\sqrt[\phantom{2}]{2}*974$

$F_1\frac{4}{5}+F_{2}sin20$

$F_2=\frac{5}{4}(\mathrm{.}484)$

$F_2=\mathrm{.}6\mathrm{.}05$

A$.15\mathrm{.}6N$

B$.15\mathrm{.}1N$

C$.13\mathrm{.}9N$

D$.14\mathrm{.}5N$

$F_1=\mathrm{.}8F_1+\mathrm{.}6F_1$

$F_2=-484F_2-\mathrm{.}9746F_2$

${\displaystyle \underset{?}{\overset{?}{}}}{F}_x=\mathrm{.}8F_1+484F_2=0,\mathrm{.}605(24)=14,5N$