## Dentist's Chair

A dentist's chair of mass 250.0 kg is supported by a hydraulic lift having a large piston of cross-sectional area 1499.0 cm2. The dentist has a foot pedal attached to a small piston of cross-sectional area 75.0 cm2.

What force must be applied to the small piston to raise the chair?

What force must be applied to the small piston to raise the chair?

**What distance must the small piston be moved to raise the chair 0.15 m?****122.7 N**

**3.00 m**## Discussion

**NEW**

**part 1**

set up the ratio as follows and solve for the unknown force

(mass1 * 9.8)/(area1 in m^2)=(X) / (area2 in m^2)

so....

(mass1*9.8)/(area1 in m^2) ] / (area2 in m^2)= force required in Newtons

hope this works

**NEW**

im knid of confused on how to solve the second part, I've

been tryin to use the equation from pg323 WL=(FL)(YL)

W= force * distance

FL= mass * gravity

YL= is the distance moved

I plugged in my numbers:

Force from first part: 120N

Distance is .1m

(120*.1m) = ((89cm^2/100)*9.8)(YL)

ans

but obviously Im doing something wrong... I have one more

try, i'd appreciate the help

**NEW**

**Re: Part 2**

Take the cross-sectional area of the large piston divided by

the cross-sectional area of the small piston and times that

answer by the number they give you in part 2. Units= m

[1511/81]=18.65

18.65*.10= 1.87 m

Good luck!

**NEW**

**Re: part 1**

For some reason Im getting the answer for part 1 wrong even

though I followed the steps you show in your post. Do I

have to divide (mass*9.8) by the the area of the large

piston or multiply it?

**NEW**

*Anonymous 5*

Part 1:

[(mass in kilograms * 9.8)/(Large piston area in meters

squared)]*(Small Piston area in meters squared)= Force

**NEW**

**part 2**

*Anonymous 6*

Can someone break down the steps for part 2?

**NEW**

**Re: Re: Part 2**

awesome thanks Delilah