Respuesta :
Answer:
[tex]F_s=1075.9493\ N[/tex]
Explanation:
Given:
- area of piston on the smaller side of hydraulic lift, [tex]a_s=0.075\ m^2[/tex]
- area of piston on the larger side of hydraulic lift, [tex]a_l=0.237\ m^2[/tex]
- Weight of the engine on the larger side, [tex]W_l=3400\ N[/tex]
Now, using Pascal's law which state that the pressure change in at any point in a confined continuum of an incompressible fluid is transmitted throughout the fluid at its each point.
[tex]P_s=P_l[/tex]
[tex]\frac{F_s}{a_s}=\frac{W_l}{a_l}[/tex]
[tex]\frac{F_s}{0.075} =\frac{3400}{0.237}[/tex]
[tex]F_s=1075.9493\ N[/tex] is the required effort force.
Answer:
F = 1076 N
Explanation:
given,
small piston area, a = 0.075 m²
large piston area, A = 0.237 m²
weight on the large piston, W = 3400 N
force applied on the second piston, F = ?
using pascal law for the force calculation
[tex]\dfrac{F}{W}=\dfrac{a}{A}[/tex]
[tex]\dfrac{F}{3400}=\dfrac{0.075}{0.237}[/tex]
F = 0.3165 x 3400
F = 1076 N
The force applied to the small piston in order to lift the engine is equal to 1076 N.
