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Đáp án:

 \(228,84\left( N \right)\)

Giải thích các bước giải:

Ta có:

\(\begin{array}{l}
{F_{15}} = \sqrt {F_1^2 + F_5^2}  = \sqrt {{{200}^2} + {{120}^2}}  = 40\sqrt {34} \left( N \right)\\
\cos \left( {\widehat {{F_1};{F_{15}}}} \right) = \dfrac{{{F_1}}}{{{F_{15}}}} = \dfrac{{200}}{{40\sqrt {34} }} \Rightarrow \widehat {{F_1};{F_{15}}} = 30,{96^o}\\
{F_{23}} = \sqrt {F_2^2 + F_3^2 + 2{F_2}{F_3}\cos 75}  = \sqrt {{{150}^2} + {{100}^2} + 2.150.100.\cos 75}  = 200,66\left( N \right)\\
\cos \left( {\widehat {{F_2};{F_{23}}}} \right) = \dfrac{{F_{23}^2 + F_2^2 - F_3^2}}{{2.{F_{23}}.{F_2}}} = \dfrac{{200,{{66}^2} + {{150}^2} - {{100}^2}}}{{2.200,66.150}} = 0,877\\
 \Rightarrow \widehat {{F_2};{F_{23}}} = 28,{78^o}\\
\cos \left( {\widehat {{F_{23}};{F_{15}}}} \right) = \widehat {{F_2};{F_{23}}} + {30^o} + \widehat {{F_1};{F_{15}}} = 28,{78^o} + {30^o} + 30,{96^o} = 89,{74^o}\\
{F_{1235}} = \sqrt {F_{15}^2 + F_{23}^2 + 2{F_{15}}.{F_{23}}.\cos \left( {\widehat {{F_{23}};{F_{15}}}} \right)} \\
 \Rightarrow {F_{1235}} = \sqrt {{{\left( {40\sqrt {34} } \right)}^2} + 200,{{66}^2} + 2.40\sqrt {34} .200,66.\cos 89,{{74}^o}} \\
 \Rightarrow {F_{1235}} = 308,37\left( N \right)\\
\cos \left( {\widehat {{F_{1235}};{F_{15}}}} \right) = \dfrac{{F_{1235}^2 + F_{15}^2 - F_{23}^2}}{{2.{F_{1235}}.{F_{15}}}} = \dfrac{{308,{{37}^2} + {{\left( {40\sqrt {34} } \right)}^2} - 200,{{66}^2}}}{{2.308,37.40\sqrt {34} }} = 0,759\\
 \Rightarrow \widehat {{F_{1235}};{F_{15}}} = 40,{59^o}\\
 \Rightarrow \widehat {{F_{1235}};{F_4}} = \widehat {{F_{1235}};{F_{15}}} + \left( {{{90}^o} - \widehat {{F_1};{F_{15}}}} \right) + {75^o}\\
 \Rightarrow \widehat {{F_{1235}};{F_4}} = 40,{59^o} + \left( {{{90}^o} - 30,{{96}^o}} \right) + {75^o}\\
 \Rightarrow \widehat {{F_{1235}};{F_4}} = 174,{63^o}\\
 \Rightarrow F = \sqrt {F_{1235}^2 + F_4^2 + 2{F_{1235}}.{F_4}\cos \left( {\widehat {{F_{1235}};{F_4}}} \right)} \\
 \Rightarrow F = 228,84\left( N \right)
\end{array}\)