Chcę żeby z tego wyszło mi "a"
\(\displaystyle{ \begin{cases} m_{1}a=m_{1}g - N_{1}
\\ m_{2}a=N_{1} - N_{2} \\ m_{3}a=N_{2}-m_{3}g \end{cases}}\)
Układ 3 równań
Układ 3 równań
\(\displaystyle{ N _{2} =m _{3} (a+g)}\)
\(\displaystyle{ N _{1}= m _{3} (a+g)+m _{2} a}\)
\(\displaystyle{ N _{1}=a(m _{3} +m _{2})+m _{3} g}\)
\(\displaystyle{ m _{1}a=m _{1}g-a(m _{3} +m _{2})-m _{3}g}\)
\(\displaystyle{ a(m _{3} +m _{2} +m _{1} )=g(m _{1} -m _{3})}\)
\(\displaystyle{ a= \frac{g(m _{1} -m _{3})}{(m _{3} +m _{2} +m _{1} )}}\)
\(\displaystyle{ N _{1}= m _{3} (a+g)+m _{2} a}\)
\(\displaystyle{ N _{1}=a(m _{3} +m _{2})+m _{3} g}\)
\(\displaystyle{ m _{1}a=m _{1}g-a(m _{3} +m _{2})-m _{3}g}\)
\(\displaystyle{ a(m _{3} +m _{2} +m _{1} )=g(m _{1} -m _{3})}\)
\(\displaystyle{ a= \frac{g(m _{1} -m _{3})}{(m _{3} +m _{2} +m _{1} )}}\)