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## Homework Statement

The cables of length 70m supporting a lift consist of two steel ropes each of 100 strangs giving a total cross sectional area of [tex]1.0 x 10^-4 m^2.[/tex] Consider a full lift to carry 8 passengers of average mass 75kg. Calculate by how much an empty lift moves down when it is entered by 8 passengers.

## Homework Equations

Young modulus of steel in this example is taken to be 2.06 x 10

^{11}.

[tex]Young modulus = \frac{Stress}{Strain} = \frac{(Force)(Length)}{(Area)(Extension)}[/tex]

Therefore,

[tex]

Extension = \frac{(Force)(Length)}{(Area)(Young)}[/tex]

Load = 8 people, mass 75kg

Force = 8(75)(9.81)(0.5) (?)

Young mod = 2.06x10

^{11}

Extension = unknown

Length = 70m

Area = 0.01m (?)

## The Attempt at a Solution

Slightly confused as to what the cross sectional area of this rope would be, I assume that it can't possibly be 0.1mm so I think that it must be 100 times that which is 0.01m or 1cm which sounds OK.

For the attempt, take x = extension, f=force, y = young mod., l=natural length, a = cross-sectional area.

[tex]x = \frac{fl}{ay}[/tex]

[tex]x = \frac{(75x9.81x0.5x8)(70)}{2.06x10^11(0.01)}[/tex]

[tex]x = 1.00 x 10^{-4} m[/tex]

This is wrong, when I use the force as 8(75g) I get 2.00 x 10

^{-4}m and when I use the area as 1.0 x 10

^{-4}m with that, I get the right answer of 0.2m or 20mm. But I thought that the load for 2 cables would be spead evenly and also that the cables could not possibly be 0.0001m

^{2}?