Question

Dentists' chairs are examples of hydraulic-lift systems. If a chair weighs 1400 N and rests on a piston with a cross-sectional area of 1220 cm2, what force must be applied to the smaller piston with a cross-sectional area of 72 cm2 to lift the chair?

Answer 1

Answer:

Force applied to smaller cross section is

= 82.63 N

Explanation:

As we know

$F_2 A_1 = F_1 A_2$

where $F1, F2$ signifies the weight of the two chair in a hydraulic-lift system

And $A_1, A_2$ signifies the area of the two respective chairs in a hydraulic-lift system

Given -

$F2=1400$ N

$A1 =1220$ Square centimeter

$A_2 = 72$ Square centimeter

Substituting the given values in above equation, we get -

$1400 * 72 = F1 * 1220\\F2 = 82.63$

Force applied to smaller cross section is

= 82.63 N