Why does it get EASIER the

Higher it goes?

1

10 threads/inch Thread Pitch = 0.10” = 0.254 cm

2

Radius = 19 cm1 Revolution = 2π x radius = 119.4 cm

3

Initial Position of Car Jack

Hypotenuse8.0” = 20.32 cm

Gap = 3.2 cm

Base = 14.0” = 35.56 cm

4

Final Position of Car Jack

Base = 9.0” = 22.86 cm

5

Work done = Force x Distance

My work = Fp on handle x 2π R

Work on car = Fp of Jack on car x Δ H

Since My work = Work on Car (If there’s no Friction)

Then the ratio of 2π R / Δ H tells me how much the jack multiplies my force

6

We know 2π R = 119.4cm, but what is Δ H when the crank is rotated once?

For the Initial position of the jack,

Triangle Base = (35.56 cm - 3.2 cm)/2

Triangle Base

For the Final position of the jack,

Triangle Base = (22.86 cm - 3.2 cm)/2Triangle Base

7

Now we have to find the height of the triangles before and after the crank is turned once (base is one thread shorter)

The Hypotenuse stays the same (20.32 cm)

Triangle BaseBEFORE = (35.56 cm - 3.20 cm)/2 = 16.18 cmPythagoras says: H = SQ RT (20.322 -16.182) = 12.29 cm

Triangle BaseAFTER ONE CRANK = (35.56 cm - 3.20 cm - .254 cm)/2 = 16.05 cm

Pythagoras says: H = SQ RT (20.322 -16.052) = 12.46 cm

So Δ HINITIAL POSITION is only 16.18 -16.05 = .17cm

8

The height of the Final Position triangle before and after the crank is turned once (base is one thread shorter) is

The Hypotenuse stays the same (20.32 cm)

Triangle BaseBEFORE = (22.86 cm - 3.20 cm)/2 = 9.83 cmPythagoras says: H = SQ RT (20.322 -9.832) = 17.78 cm

Triangle BaseAFTER ONE CRANK = (22.86 cm - 3.20 cm - .254 cm)/2 = 9.70 cm

Pythagoras says: H = SQ RT (20.322 - 8.562) = 17.85cm

So Δ HFINAL POSITION is only17.85 - 17.78 = 0.07cm 9

Now we can calculate the difference in force magnification between the initial position (low jack) and final position

(high jack)

My Work / Work on Car = 1 = (Fp on handle x 2π R) / (Fp of Jack on car x Δ H) = 1

(Fp of Jack on car) / (Fp on handle) = (2π R) / (Δ H) =119.4 cm / 0.17 cm = 700 for initial position of jack.

For the final position of the Jack(Fp of Jack on car) / (Fp on handle) = (2π R) / (Δ H) =

119.4 cm / 0.07 cm = 1700 for final position of jack.which means, that at the end, it was almost 2.5 times

easier to turn the crank.

10

Δ H is actually twice as big because there is a triangle above and below the threaded rod. Therefor, the ratio

2π R / Δ H is twice as big. This means that the input force is doubled. so the magnified force of 700 is really 350, and

1700 is 850.

But wait!!!

second triangle

Fp on handle x 2π R

=Fp of Jack on car x Δ H

11