Question

Pt a presa un material se foloseste o presa hidraulica care are raportul dintre diametrul pistonului mare si diametrul pistonului mic de 20.Sa se determine:
A)FORTA NECESARA DACA FORTA DE REZISTENTA ESTE DE 120 KN.
B)DEPLASARILE PISTOANELOR CUNOSCAND LUCRUL MECANIC AL FORTEI DE REZISENTA DE 0,6 KJ

Answer 1

   
[tex]\displaystyle\\ \texttt{Se da:} \\ \\ \frac{ D_2}{ D_1} = 20~KN \\ \\ F_{2} =120~KN\\ \\ L_{2} =0,6~KJ\\ \\ \texttt{Se cere:}\\ \\ a)~~ F_{1} = ?\\ \\ b)~~ d_{1}=?\\ \\ ~~~~ d_{2} =?\\ \\ Rezolvare:\\\\ a)\frac{ F_{1} }{ S_{1}}= \frac{ F_{2} }{ S_{2} } ~~\Longrightarrow~~ \frac{ F_{2} }{ F_{1} } = \frac{ S_{2} }{ S_{1} } = \Big(\frac{ D_{2} }{ D_{1} } \Big)^2= 20^{2} =\boxed{400}\\ \\ \Longrightarrow~~ F_{1}= \frac{ F_{2} }{ 400} = \frac{120.000}{400} = \boxed{300N} [/tex]
[tex]\displaystyle\\ \\ b)~ L_{2} = F_{2} \times d_{2}\\ \\ d_{2}= \frac{ L_{2} }{ F_{2} } = \frac{0.6~KJ}{120~KN} = \frac{600}{120.000} = \frac{1}{200} =0,005~m=\boxed{5~mm}\\ \\ V_{1}= S_{1}\times d_{1} \\ \\ V_{2} =S_{2}\times d_{2} \\ \\ \texttt{Dar }V_{1}= V_{2} ~~~~~(volumele) \\ \\ S_{1}\times d_{1}=S_{2}\times d_{2} \\ \\ \Longrightarrow~~~ d_{1}= \frac{ S_{2} }{ S_{1} }\times d_1 = 400 \times 5~mm = 2000~mm =\boxed {2~m} [/tex]