Question

Let s=9.8 m/s² . A hydraulic system is used to lift a 94 kg motor in a garage. If the motor sits on a piston of area 0.23 m² , and a force is applied to a piston of area 82 cm² , what is the minimum force that must be applied to lift the motor in units of N ?

Answer 1

The minimum force that must be applied to lift the motor using a hydraulic system is 32.53 N.

Step-by-step explanation:

To calculate the minimum force that must be applied to lift the motor using a hydraulic system, we can use the equation of force ratio:

Force ratio = area of applied force / area of load

First, convert the area of the piston of the motor to square meters: 82 cm² = 82/10000 m² = 0.0082 m². Next, substitute the values into the equation:

Force ratio = 0.0082 / 0.23 = 0.03565

Finally, multiply the force ratio by the weight of the motor to find the minimum force required to lift it:

Minimum force = 0.03565 * 94 kg * 9.8 m/s² = 32.53 N