Answer :
The force that must be exerted on the master cylinder of the hydraulic lift to support the weight of car is 147.2 N .
The Pascal's Law states that the relation between the pressure , force and area is given as :
Pressure = Force/Area ;
given that the diameter of the master cylinder is = 1.93 cm ;
so radius of master cylinder = 1.93/2 = 0.965 cm = 0.00965 m ;
the diameter of slave cylinder is = 24.9 cm ;
the radius of the slave cylinder = 24.9/2 = 12.45 m = 0.1245 m ;
Area of slave cylinder (A₁)= π(0.1245)² = 0.0486 m²
Area of the master cylinder (A₂) = π(0.00965)² = 0.000292 m² ;
given mass of the car = 2500 Kg ,
the force exerted will be : F = m*a
F₁ = 2500*9.8 = 24500 N ;
F₁/A₁ = F₂/A₂
So , 24500/0.0486 = F₂/0.000292
F₂ = (24500/0.0486)×0.000292
F₂ = 147.2 N
Therefore , the force exerted on the master cylinder is 147.2 N .
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