Answer:
a) F₁ = 1.48 x 10³ N
b) P = 1.88*10⁵ Pa
c) The work is equal in both pistons
Explanation:
Pascal´s Principle can be applied in the hydraulic press:
If we apply a small force (F₁) on a small area piston (A₁), then, a pressure (P) is generated that is transmitted equally to all the particles of the liquid until it reaches a larger area piston and therefore a force (F₂) can be exerted that is proportional to the area(A₂) of the piston.
Pressure is defined as the force per unit area:
[tex]P=\frac{F}{A}[/tex] Formula (1)
P₁=P₂
[tex]\frac{F_{1} }{A_{1} } =\frac{F_{2} }{A_{2} }[/tex] Formula (2)
Data
r₁= 5 cm = 0.05 m
r₂= 15 cm = 0.15 m
F₂= 13300N
Area of the pistons (A₁,A₂)
A=π*r² : Area of the circle
A₁ = π*(0.05)²=7.85*10⁻³ m²
A₂= π*(0.15)²= 70.69*10⁻³ m²
a) Force that compressed air must exert to lift a car weighing 13300 N
We replace data in the formula (2)
[tex]\frac{F_{1} }{A_{1} } =\frac{F_{2} }{A_{2} }[/tex]
[tex]F_{1} = \frac{13300*7.85*10^{-3} }{70.69*10^{-3} }[/tex]
F₁ = 1.48 x 10³ N
b) Air pressure produced by F₁
We replace data in the formula (1)
[tex]P=\frac{F}{A}[/tex]
F₁ = 1.48 x 10³ N , A₁ = 7.85*10⁻³ m²
[tex]P=\frac{1.48*10^{3} }{7.85*10^{-3} }[/tex]
P= 1.88*10⁵ Pa
c)The volume of liquid displaced by the small piston is distributed in a thin layer on the large piston, so that the product of the force by the displacement (the work) is equal in both pistons.
