

2. (a) Twelve identical wooden greenheart columns of crosssection 25.0 cm X 25.0 cm and 3.00 m above ground support a structure of mass 60000 kg. Assuming that the load is distributed uniformly, calculate the (i) strain of each column and (ii) the compression of each column above ground.
Shear Modulus of Greenheart Wood = 2.00 X
10^{9} Nm^{2}
(b) Each column is buried firmly in a concrete cuboid 1m X 1m X 1m. having mass 2500kg. If this structure rests on a layer of sand 2.0 am deep, with Young's modulus 5.00 X
10^{7} Nm^{2}, by how much has the building sink after construction? Assume the sand layer rests on an incompressible rock and ignore friction with the sides of the concrete base. 
Fig 2.1 (Greenheart House on the Essequibo River in Guyana) 


What do we know?
Mass of structure = 6.00 X 10^{4} kg
Crosssectional Area of column = [2.50 X 10^{1} m]^{2} = 6.25 X 10^{2} m^{2}
Length of column = 3.00 m
Young's Modulus of Greenheart Wood = 2.00 X
10^{10} Nm^{2}
Crosssectional Area of concrete base = [1^{} m]^{2} = 1 ^{} m^{2}
Length of sand column = 2.00 m
Young's Modulus of Sand = 5.00 X
10^{7} Nm^{2}
What can we infer?
W = mg, where am is the mass of
the object.
Weight of structure W_{s}= 6.00 X 10^{4} N
Weight on each column W_{c}= 5.00 X 10^{5} N
Weight on each column + concrete base W_{c+b}= 5.00 X 10^{5} N
Plan to solve:
We know Stress = [^{[Force]} / _{[Cross Sectional Area]}]
Young's Modulus = [^{[Stress]} / _{[Strain]}]
Strain = [^{[ΔL]} / _{[L]}]
So we can get ΔX if we calculate Strain

(a) Thus, the shear stress on one column is:
Stress = [^{[Force]} / _{[Cross Sectional Area]}]
Stress on each column = [(5.00 X 10^{5} N) / (6.25 X 10^{2} m^{2})]
Stress on each column = 8.00 X 10^{6} Nm^{2}
Young's Modulus = [^{[Stress]} / _{[Strain]}]
2.00 X
10^{10} Nm^{2} = [(8.00 X 10^{6 }Nm^{2})] / Strain
Strain = [(8.00 X 10^{6 }Nm^{2})] / [(2.00 X
10^{10} Nm^{2})]
Strain on each column = 4.00 X 10^{4 }
Strain = [^{[ΔL]} / _{[L]}]
4.00 X 10^{4 }= [^{[ΔL]} / _{[3m]}]
[ΔL] = (3m) X (4.00 X 10^{4})
[ΔL] = (1.20 X 10^{3} m)
[ΔL] = 1.20 mm
(b) On the base of each column:
Crosssectional Area of concrete base = [1 m]^{2} = 1 m ^{2}
Length of sand column = 2.0 m
Young's Modulus of Sand = 5.00 X
10^{7} Nm^{2}
Weight on sand from each column and base = Weight from structure + Weight of Concrete base
Weight on sand from each column and base = 5.00 X 10^{5} N + 2.5 X 10^{4} N
Weight on sand from each column and base = 5.25 X 10^{5} N
Stress on sand base= [^{[Force]} / _{[Cross Sectional Area]}]
Stress on each sand column = [(5.25 X 10^{5} N) / (1 m^{2})]
Stress on each sand column = 5.25 X 10^{5} Nm^{2}
Young's Modulus = [^{[Stress]} / _{[Strain]}]
5.00 X
10^{7} Nm^{2} = [(5.24 X 10^{5} Nm^{2})] / Strain
Strain = [(5.25 X 10^{5} Nm^{2})] / [(5.00 X
10^{7} Nm^{2})]
Strain on each sand column = 1.05 X 10^{2 }
Strain = [^{[ΔL]} / _{[L]}]
1.05 X 10^{2 }= [^{[ΔL]} / _{[3 m]}]
[ΔL] = (2m) X (1.05 X 10^{2})
[ΔL] = (2.10 X 10^{2 }m)
[ΔL] of each sand column = 2.10 cm 


3. (a) The steel jack used to raise a pickup truck has four identical struts 3.00 cm X 3.00 cm X 30.0cm long and 2.00 mm thick.
(a)
What is the stress on the struts when it raises a mass of 500kg and the separation of the struts is 90°.
(b)
Calculate the strain each strut.
(c) Calculate the compression of each strut.
Young's Modulus for steel is 2.0 X 10^{11} Nm2


What do we know?
W = mg, where am is the mass of
the object.
Weight on Jack = 5.00 X 10^{3} N
Dimesnsions of Strut = 3.00 cm X 3.00 cm X 30.0cm long X 2.00 mm thick
Young's Modulus for steel is 2.0 X 10^{11} Nm^{2}
What can we infer?
Crosssectional Area of strut = {[2 X (3.00 cm X 2.00 mm)] + [(3 cm  4 mm) X 2mm]}
CrosCrosssectional Area of strut = {[0.060 m X 0.002 m)] + (0.0296 cm X 0.002 m)}
Crosssectional Area of strut = 1.79 X 10^{4} m^{2}
Force on Strut = ½ [(5000N) / Cos 45°]
Force on Strut = 3535 N
Plan to solve:
We know Stress = [^{[Force]} / _{[Cross Sectional Area]}]
Young's Modulus = [^{[Stress]} / _{[Strain]}]
Strain = [^{[ΔL]} / _{[L]}]
So we can get ΔX if we calculate Strain 
Weight on Jack = 5.00 X 10^{3} N
Force on Strut = ½ [(5000N) / (Cos 45°)]
Force on Strut = 3.535 X 10^{3} N
Crosssectional Area of strut = {[2 X (3.00 cm X 2.00 mm)] + [(3 cm  4 mm) X 2mm]}
CrosCrosssectional Area of strut = {[0.060 m X 0.002 m)] + (0.0296 cm X 0.002 m)}
Crosssectional Area of strut = 1.72 X 10^{4} m^{2}
We know Stress = [^{[Force]} / _{[Cross Sectional Area]}]
Stress = [(3.535 X 10^{3} N) / (1.72 X 10^{4} m^{2)}]
Stress = 2.06 X 10^{7} Nm^{2}
Young's Modulus = [^{(Stress)} / _{(Strain)}]
2.0 X 10^{11} Nm^{2} = (1.98 X 10^{7} Nm^{2})/ Strain
Strain =
(2.06 X 10^{7}) Nm^{2} / (2.0 X 10^{11} Nm^{2})
Strain = 1.03 X 10^{4}
Strain = [^{[ΔL]} / _{[L]}]
(1.03 X 10^{4}) = [^{[ΔL]} / _{[0.3m]}]
[ΔL] = (0.300 m) X (1.03 X 10^{4})
[ΔL] = (3.09 X 10^{5} m)
[ΔL] = 30.9 µm 

4. A steel cable 30 m long has a diameter of 3.0 cm and supports an elevator of mass 1200 kg.
(a)
What is the largest load that not permanent stretch the elevator cable if the Yield strength of Steel is 2.00 X 10^{8} Nm^{2}?
(b) If the Young's Modulus of steel is 2.0 X 10^{11} Nm^{2}, how much will the cable stretch before it deforms? 
What do we know?
W = mg, where am is the mass of
the object.
Weight of Elevator = 1.20 X 10^{3} N
Radius of Cable = 1.5 cm
Length of Cable = 30 m
Crosssectional Area of Cable = πr^{2}
Young's Modulus for steel is 2.0 X 10^{11} Nm^{2}
Yield strength of Steel = 2.00 X 10^{8} Nm^{2}
What can we infer?
Crosssectional Area of cable = π (2.25 X 10^{4} m^{2})
Crosssectional Area of Cable = 7.07 X 10^{4} m^{2}
Plan to solve:
We know Stress = [^{[Force]} / _{[Cross Sectional Area]}]
Young's Modulus = [^{[Stress]} / _{[Strain]}]
Strain = [^{[ΔL]} / _{[L]}]
Yield strength is the maximum stress that can be applied before permanent deformation So we can get
1. Force if we
calculate stress
ΔX if we calculate Strain 
Maximum Stress = [^{[Maximum Force]} / _{[Cross Sectional Area]}]
(2.00 X 10^{8} Nm^{2}) = Maximum Force / ( 7.07 X 10^{4}m^{2})
Maximum Force = ( 7.07 X 10^{4} m^{2}) X (2.00 X 10^{8} Nm^{2})
Maximum Force = (1.41 X 10^{5} N)
Load that will break the elevator cable = Maximum Force  Weight of Elevator
Load that will break the elevator cable = (1.41 X 10^{5}N)  (1.20 X 10^{3} N)
Load that will break the elevator cable = 1.40 X 10^{5} N
Young's Modulus = [^{[Stress]} / _{[Strain]}]
(2.0 X 10^{11} Nm^{2}) = ((2.00 X 10^{8} Nm^{2}) / Strain
Strain = ((2.00 X 10^{8} Nm^{2}) / (2.0 X 10^{11} Nm^{2})
Strain = 1.00 X 10^{3}
Strain = [^{[ΔL]} / _{[L]}]
1.00 X 10^{3} = [^{[ΔL]} / _{[30m]}] [ΔL] = (3.02 X 10^{2} m)
[ΔL] = 3.02 cm 


5. A steel wire of length 5.0 m and crosssectional area 3.0 X 10^{5} m^{2} stretches by the same amount as a copper wire of length 4.0 m and crosssectional area of 4.0 × 10^{−5} m^{2} under a the same load. What is the ratio of the Young’s modulus of steel to that of copper? 

What do we know?
Length of Steel Wire = 5.0 m
Crosssectional Area of Steel Wire = 3.0 X 10^{−5} m^{2}
Length of Copper Wire = 4.0 m
Crosssectional Area of Copper Wire = 4.0 × 10^{−5} m^{2}
Load on Steel Wire = W
Load on Copper Wire = W
Extension of
Steel Wire = ΔL
Extension of
Copper Wire = ΔL
What can we infer?
Plan to solve:
We know Stress = [^{[Force]} / _{[Cross Sectional Area]}]
Strain = [^{[ΔL]} / _{[L]}]
Young's Modulus = [^{[Stress]} / _{[Strain]}]
So we can find ratio 

Stress = [^{[Force]} / _{[Cross Sectional Area]}]
Stress of Steel Wire = [(W) / (3.0 X 10^{−5} m^{2})]
Stress of Copper Wire = [(W / (4.0 X 10^{−5} m^{2})
Strain = [^{[ΔL]} / _{[L]}]
Strain of Steel Wire = [(ΔL ) / (5.0 m)]
Strain of Copper Wire = [(ΔL ) / (4.0 m)
Young's Modulus = [^{[Stress]} / _{[Strain]}]
Young's Modulus of Steel Wire = {[(W) / (5.0 m)] / [(ΔL ) / (3.0 X 10^{−5} m^{2})]}  Eq1
Young's Modulus of Copper Wire = {[(W) / (4.0 m)] / [(ΔL ) / (4.0 X 10^{−5} m^{2})]}  Eq2
Dividing Eq1 by Eq2
{[(W) / (3.0 X 10^{−5} m^{2})] / [(ΔL ) / (5.0 m)]} / {[(W) / (4.0 X 10^{−5} m^{2})] / [(ΔL ) / (4.0 m)]}
{[(W) X (5.0 m) / (3.0 X 10^{−5} m^{2})] X [(ΔL )]} / {[(W) X (4.0 m) / (4.0 X 10^{−5} m^{2})] X [(ΔL )]}
{[(W) X (4.0 X 10^{−5} m^{2}) X (5.0 m)] X [(ΔL ) / (4.0 m)] X [(ΔL ) X (W) X (3.0 X 10^{−5} m^{2})]}
(4.0 X 10^{−5} m^{2}) X (5.0 m) / (4.0 m) X (3.0 X 10^{−5} m^{2})
=1.67 


