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