Answer:
323.47 N
Explanation:
Applying pascals Principle,
F₁/A₁ = F₂/A₂...................... Equation 1
Where F₁ = Force exerted on the small piston, A₁ = Area of the small piston, F₂ = Force exerted on the second piston, A₂ = Area of the second piston
Note: Since the piston has a circular cross-section,
Then,
A₁ = πR₁².................... Equation 2
A₂ = πR₂².................... Equation 3
Where R₁ = Radius of the first piston, R₂ = Radius of the second piston
Substitute equation 2 and equation 3 into equation 1
F₁/πR₁² = F₂/πR₂²
F₁/R₁² = F₂/R₂²................. Equation 4
make F₁ the subject of the equation
F₁ = R₁²(F₂/R₂²)................. Equation 5
Given: F₂ = 11200 N, R₁ = 3.11 cm, R₂ = 18.3 cm
Substitute into equation 5
F₁ = 11200(3.11²/18.3²)
F₁ = (9.6721/334.89)11200
F₁ = 323.47 N.
Hence the force that must be exerted by the compressed air = 323.47 N
Answer:
The force that must the compressed air exert in orden to lift a car weighing 11200 N is 323.47 N
Explanation:
Given:
ri = radius of first piston = 3.11 cm
ro = radius of second piston = 18.3 cm
wc = weight of the car = 11200 N
The inward pressure is:
[tex]\frac{F_{i} }{A_{i} } =\frac{F_{o} }{A_{o} } \\F_{i} =\frac{A_{i} }{A_{o} } F_{o} \\F_{i}=\frac{\pi r_{i}^{2} }{\pi r_{o}^{2} } F_{o} \\F_{o}=w_{c} \\F_{i}=\frac{\pi r_{i}^{2} }{\pi r_{o}^{2} } w_{c}[/tex]
Replacing:
[tex]F_{i} =\frac{3.11^{2} }{18.3^{2} } *11200=323.47N[/tex]
