Question

the cross-section of two pistons in a hydraulic lift are 100 cm2 and 800 cm2 respectively. what is the minimum force required to lift 5000 kg-wt on a large piston surface? (in kn) -

Answer 1

The minimum force required to lift the 5000 kg-wt object on the large piston surface can be found using the equations force = pressure x area and pressure = weight/area. By calculating the pressure on the large piston and using it to calculate the force on the small piston, we can determine the minimum force in kilonewtons.

Step-by-step explanation:

The minimum force required to lift an object can be found using the equation force = pressure x area. In this case, we can calculate the pressure exerted on the large piston by dividing the weight of the object by the area of the large piston: pressure = weight/area. Then, we can calculate the force required on the small piston using the equation force = pressure x area and plugging in the pressure we calculated and the area of the small piston. The force will be in kilograms-force (kg-wt), so we can convert it to kilonewtons (kN) by dividing by 9.8 (since 1 kg-wt = 9.8 N).

Let's calculate:

1. The pressure on the large piston is pressure = 5000 kg-wt / 800 cm2.
2. Next, we calculate the force required on the small piston: force = pressure x area. Since we want the force in kN, we divide the force by 9.8.

By performing these calculations, we can find the minimum force required to lift the 5000 kg-wt object on the large piston surface.