+ $(\vec{F_1}, \vec{F_2})=0°$

$F_{12}=F_1+F_2=16+12=28N$

+ $(\vec{F_1}, \vec{F_2})=60°$

$F_{12}=\sqrt{F_1^2+F_2^2+2F_1F_2cos\alpha}$

$=\sqrt{16^2+12^2+2.16.12.cos60}=4\sqrt{37}N$

+ $(\vec{F_1}, \vec{F_2})=120°$

$F_{12}=\sqrt{F_1^2+F_2^2+2F_1F_2cos\alpha}$

$=\sqrt{16^2+12^2+2.16.12.cos120}=4\sqrt{13}N$

+ $(\vec{F_1}, \vec{F_2})=180°$

$F_{12}=|F_1-F_2|=|16-12|=4N$

Góc hợp bởi hai hợp lực để F12=20N:

$F_{12}=\sqrt{F_1^2+F_2^2+2F_1F_2cos\alpha}$

$\Leftrightarrow20=\sqrt{16^2+12^2+2.16.12.cos\alpha}$

$\Rightarrow cos \alpha=0$

$\Rightarrow \alpha=90°$

Đáp án:

\(\begin{align}
  & F=28N \\ 
 & F=4\sqrt{37}N \\ 
 & F=20N \\ 
 & F=4\sqrt{13}N \\ 
\end{align}\)

Giải thích các bước giải:

ta có: 
\(\overrightarrow{F}={{\overrightarrow{F}}_{1}}+{{\overrightarrow{F}}_{2}}\)

TH1: 
\((\overrightarrow{{{F}_{1}}};{{\overrightarrow{F}}_{2}})={{0}^{0}}\)
\(\Rightarrow F={{F}_{1}}+{{F}_{2}}\Rightarrow F=16+12=28N\)

TH2: 
\((\overrightarrow{{{F}_{1}}};{{\overrightarrow{F}}_{2}})={{60}^{0}}\)

\(\Rightarrow {{F}^{2}}=F_{1}^{2}+F_{2}^{2}+2{{F}_{1}}{{F}_{2}}\cos \alpha \)

\(\begin{align}
  & \Rightarrow {{F}^{2}}={{16}^{2}}+{{12}^{2}}+2.16.12\cos {{60}^{0}} \\ 
 & \Rightarrow F=4\sqrt{37}N \\ 
\end{align}\)

TH3: 
\((\overrightarrow{{{F}_{1}}};{{\overrightarrow{F}}_{2}})={{90}^{0}}\)
\(\begin{align}
  & \Rightarrow {{F}^{2}}=F_{1}^{2}+F_{2}^{2} \\ 
 & \Rightarrow {{F}^{2}}={{16}^{2}}+{{12}^{2}} \\ 
 & \Rightarrow F=20N \\ 
\end{align}\)

TH4: 
\((\overrightarrow{{{F}_{1}}};{{\overrightarrow{F}}_{2}})={{120}^{0}}\)
\(\begin{align}
  & \Rightarrow {{F}^{2}}=F_{1}^{2}+F_{2}^{2}+2{{F}_{1}}{{F}_{2}}\cos \alpha \Leftrightarrow {{F}^{2}}={{16}^{2}}+{{12}^{2}}+2.16.12\cos {{120}^{0}} \\ 
 & \Rightarrow F=4\sqrt{13}N \\ 
\end{align}\)