- #1

Littlemonkey

- 4

- 0

**A steel wire diameter 0.50mm and length 1.84m is suspended vertically from a fixed point and used to support a 60N weight hung from the lower end of the wire. The Young modulus for steel is 2.0 x 10^11 N m-2**

**a) The extension of the wire**

**b) The elastic energy stored in a wire given the elastic limit for steel is at a strain of 0.0025**

**Formula:**

Δl=(lo F)/EA

Δl=(lo F)/EA

**a)Δl=(1.84m×60N)/((2.0×10^11 Nm-2) )×(Π×0.00025^2 ) ) = 2.81×10-3m or 2.81mm**

I know the answer to a) is correct I found the book that the original question was taken from I also know the answer to b) is 84mJ but I have no clue where to even start any help would be greatly appreciated.

Also could someone check my calculations on the problem below as again I have the answer from the book but I cannot get the equation to work:

**A crane used to lift iron girders on a building site has a steel lifting cable diameter 40 mm. When the crane lifts an iron girder of weight 10 kN, the cable length with the girder just off the ground is 75.0 m. Calculate**

a) The extension of the cable for this length and load

a) The extension of the cable for this length and load

**Formula:**

Δl=(lo F)/EA

Δl=(lo F)/EA

**Answer:**

Δl=(75m×10kN)/((2.0×10^11 Nm-2) )×(Π×0.02^2 ) ) = 2.98×10-4 m or .298 mm

Δl=(75m×10kN)/((2.0×10^11 Nm-2) )×(Π×0.02^2 ) ) = 2.98×10-4 m or .298 mm

The answer in the book is 2.98×10-3 m or 2.98 mm I just can't work out where I am going wrong.