Vì lực truyền qua dây được bảo toàn nên P2= T2= 20(N); P3= T3= 40(N)
Có \(\overrightarrow{T_2}+\overrightarrow{T_3}+\overrightarrow{P_1}+\overrightarrow{N}=\overrightarrow{0}\)
\(\Rightarrow\left\{{}\begin{matrix}Ox:T_2-T_3.\cos\alpha+m_1g\sin\alpha=0\\N_1=m_1g\cos\alpha\end{matrix}\right.\)
\(\Rightarrow20-40.\frac{\sqrt{3}}{2}+10m_1.\frac{1}{2}=0\Leftrightarrow m_1=4\sqrt{3}-4\left(kg\right)\)
\(\Rightarrow N_1=\left(4\sqrt{3}-4\right).10.\frac{\sqrt{3}}{2}=60-20\sqrt{3}\left(N\right)\)