The small piston of a hydraulic lift has a diameter of 8 cm and its large piston has a diameter of 40cm. The lift raises a load of 15,000 Newtons.

A. Determine the force that must be applied to the small piston.

B. Determine the pressure applied to the fluid in the lift.

a) 600N

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A. To determine the force that must be applied to the small piston, we can use Pascal's law, which states that the pressure applied to a fluid in a closed system is transmitted equally in all directions. This means that the pressure exerted on the small piston will be the same as the pressure exerted on the large piston.

We can calculate the pressure exerted on the large piston by using the formula:

Pressure = Force / Area

The area of the large piston can be calculated using the formula for the area of a circle:

Area = π * radius^2

Since the diameter of the large piston is given as 40 cm, the radius would be half of that, which is 20 cm. Converting the radius to meters, we get 0.2 meters.

Now, we can calculate the area of the large piston:

Area = π * (0.2m)^2

Next, we can calculate the pressure exerted on the large piston:

Pressure = 15,000 N / Area

B. To determine the pressure applied to the fluid in the lift, we will use the same pressure value calculated in Part A. The pressure is the same throughout the hydraulic system, so it is also the pressure applied to the fluid in the lift.

Now that we have calculated the pressure, we need to find the area of the small piston to determine the force that must be applied to it.

The area of the small piston can be calculated using the same formula:

Area = π * (radius)^2

Since the diameter of the small piston is given as 8 cm, the radius would be half of that, which is 4 cm. Converting the radius to meters, we get 0.04 meters.

Now, we can calculate the area of the small piston:

Area = π * (0.04m)^2

Finally, we can determine the force that must be applied to the small piston:

Force = Pressure * Area

By substituting the values we have calculated, we can find the force required to lift the load.