Question

the output piston of a hydraulic garage lift has a cross-sectional area of 0.20m2.

a) how much pressure on the input piston is required to support a car with a mass of 1.4 metric tons?

b)what force is applied to the input piston if it has a diameter of 5.0cm?

Answer 1

\[Pressure = \frac{Force}{Area}\]

Can you use this to find the pressure needed?

Answer 2

yeah.

Answer 3

I had an explanation typed out, but when I re-read it, it was kind of confusing. I can try to explain what to do if you are unsure, though.

Answer 4

sure..your explanation will be a big help for me :)

Answer 5

First, a picture:


To support the car, the force from the pressure needs to equal the force from the car (weight = force). The area in the equation for pressure is the cross-sectional area of the platform.

If you plug those numbers into the formula, you'll know the pressure needed to support the car.

Make sense?

Answer 6

ahhh.:))

Answer 7

For the second part, we have a different area, but the force and equation is the same.

You're given the DIAMETER of a circle to be used as the cross-sectional area. Do you remember how to find the area of a circle?

Answer 8

Pi r^2

Answer 9

Yep.
Pi * RADIUS^2

Remember, you are given a DIAMETER.

Find the new area, plug that in with the force, and find the pressure needed. And you're done!

Answer 10

w=mg , so (1.4 metric tons)(9.81), you said that it is equal to the force also...the the answer here will be divided by .20m2....

Answer 11

pressure/ pi*(2.5)^2.....am i right?

Answer 12

The force of the car is mg, correct (and very good!)

For part A, put mg in for force, and the area given for area. Force/Area. The result is your answer.

For part B, use the same force (mg), but a new area of pi*(.025)^2

We use .025 because the diameter you are given was in cm.

Answer 13

ahhh....yeah..yeah...i forgot to convert it..

Answer 14

thank you !

Answer 15

My pleasure :)

Answer 16

wait ..F=ma...in letter b....their are asking for force... so the anwer in A mulitplied to Pi * (0.025)^2......
right?

Answer 17

Oh, I misread. You are correct. Use the new area, and the pressure from part A to find the force.

Answer 18

okay .. :)) so,,am i right? :)

Answer 19

Yeah, you're right :)

Answer 20

okay,,thank you again :)