Tham khảo hình vẽ.
Hệ cân bằng: \(\Leftrightarrow\overrightarrow{P}+\overrightarrow{T_A}+\overrightarrow{T_B}=\overrightarrow{0}\)
Theo quy tắc tỏng hợp lực: \(\overrightarrow{P}+\overrightarrow{T_A}=\overrightarrow{Q}\)
\(\Rightarrow\overrightarrow{Q}+\overrightarrow{T_B}=\overrightarrow{0}\)\(\Rightarrow\left|Q\right|=\left|T_B\right|\)
Có \(\widehat{AOB}=120^o\Rightarrow\alpha=\widehat{T_AOQ}=180^o-120^o=60^o\)
\(P=100N\)
Xét \(\Delta T_AOQ\) vuông tại \(T_A\) có:
\(\left\{{}\begin{matrix}tan\alpha=\dfrac{P}{T_A}\Rightarrow T_A=\dfrac{P}{tan\alpha}=\dfrac{100}{tan60^o}=57,735N\\sin\alpha=\dfrac{P}{Q}\Rightarrow Q=\dfrac{P}{sin\alpha}=\dfrac{100}{sin60^o}=115,47N\end{matrix}\right.\)
