A green sedan weighing 25,000 Newtons is put on a lift supported by a piston with
a cross sectional area of 0.1 square meters. What is the minimum force that must
be exerted by the air compressor that pushes down on the oil in the reservoir of the lift system in order to lift the sedan if the cross-sectional area of the reservoir
is 1.0 square meter? (Hint: use Pascal’s Law.)
To find the minimum force exerted by the air compressor, we can use Pascal's Law, which states that when pressure is applied to a fluid in a confined space, the pressure is transmitted equally in all directions.
First, let's calculate the pressure exerted on the oil in the reservoir. We can use the formula:
Pressure (P) = Force (F) / Area (A)
The weight of the sedan, 25,000 Newtons, is acting downwards. Since the sedan is in equilibrium (not moving up or down), the force exerted by the piston must balance the weight of the sedan. Therefore, the force exerted by the piston upwards is also 25,000 Newtons.
Next, we can calculate the pressure exerted on the oil in the reservoir using the formula:
Pressure (P) = Force (F) / Area (A)
Since the cross-sectional area of the reservoir is 1.0 square meter, the pressure exerted is:
P = 25,000 Newtons / 1.0 square meter
P = 25,000 Pascal
According to Pascal's Law, this pressure will be transmitted equally throughout the fluid in the confined space, including to the piston.
Now, let's calculate the minimum force required to lift the sedan. We know that the cross-sectional area of the piston is 0.1 square meters. Using the formula:
Force (F) = Pressure (P) x Area (A)
F = 25,000 Pascal x 0.1 square meters
F = 2,500 Newtons
Therefore, the minimum force that must be exerted by the air compressor is 2,500 Newtons.