COMPLETION PROBLEMS--PHYSICS
Completion Problems (Elasticity)
1. A steel wire of length 4.7m and cross-sectional area 3×10-5m2 stretches by the same amount as a copper wire of length 3.5m and cross-sectional area of 4×10–5m2 under a given load. What is the ratio of the Young’s modulus of steel to that of copper?
For Steel l1= 4.7 m, a1 = 3.0 x 10-5 m2 Y1= Fl1/a1∆l Y1= (F x 4.7m)/ (3.0 x 10-5 x ∆l)
For Copper
L2=3.5 m, a2 = 4.0 x 10-5 m2 Y2= (F x 3.5m)/ (4.0 x 10-5 x ∆l)
Dividing Steel by Copper Y1/ Y2 = [(4.7m)/ (3.0 x 10-5 x ∆l)] x [(4.0 x 10-5 x ∆l) / (3.5m)] = (4.7 x 4.0)/ (3.0 x 3.5 ) = 1.79
2. Two wires of diameter 0.25cm, one made of steel and the other made of brass are loaded as shown. The unloaded length of steel wire is 1.5m and that of brass wire is 1.0m. Compute the elongations of the steel and the brass wires.
For steel wire:
For brass wire:
3. The edge of an aluminium cube is 10cm long. One face of the cube is firmly fixed to a vertical wall. A mass of 100 kg is then attached to the opposite face of the cube. The shear modulus of aluminium is 25 GPa. What is the vertical deflection of this face?
Edge of the aluminium cube, L = 10 cm = 0.1 m
The mass attached to the cube, m = 100 kg
Shear modulus (η) of aluminium = 25 GPa = 25 × 109 Pa
Shear modulus, η
Where,
F = Applied force = mg = 100 × 9.8 = 980 N
A = Area of one of the faces of the cube = 0.1 × 0.1 = 0.01 m2
ΔL = Vertical deflection of the cube
= 3.92 × 10–7 m
4. Four identical hollow cylindrical columns of mild steel support a big structure of mass 50,000 kg. The inner and outer radii of each column are 30 and 60 cm respectively. Assuming the load distribution to be uniform, calculate the compressional strain of each column.
Mass of the big structure, M = 50,000 kg
Inner radius of the column, r = 30 cm = 0.3 m
Outer radius of the column, R = 60 cm = 0.6 m
Young’s modulus of steel, Y = 2 × 1011 Pa
Total force exerted, F = Mg = 50000 × 9.8 N
Stress =
1. A steel wire of length 4.7m and cross-sectional area 3×10-5m2 stretches by the same amount as a copper wire of length 3.5m and cross-sectional area of 4×10–5m2 under a given load. What is the ratio of the Young’s modulus of steel to that of copper?
For Steel l1= 4.7 m, a1 = 3.0 x 10-5 m2 Y1= Fl1/a1∆l Y1= (F x 4.7m)/ (3.0 x 10-5 x ∆l)
For Copper
L2=3.5 m, a2 = 4.0 x 10-5 m2 Y2= (F x 3.5m)/ (4.0 x 10-5 x ∆l)
Dividing Steel by Copper Y1/ Y2 = [(4.7m)/ (3.0 x 10-5 x ∆l)] x [(4.0 x 10-5 x ∆l) / (3.5m)] = (4.7 x 4.0)/ (3.0 x 3.5 ) = 1.79
2. Two wires of diameter 0.25cm, one made of steel and the other made of brass are loaded as shown. The unloaded length of steel wire is 1.5m and that of brass wire is 1.0m. Compute the elongations of the steel and the brass wires.
For steel wire:
For brass wire:
3. The edge of an aluminium cube is 10cm long. One face of the cube is firmly fixed to a vertical wall. A mass of 100 kg is then attached to the opposite face of the cube. The shear modulus of aluminium is 25 GPa. What is the vertical deflection of this face?
Edge of the aluminium cube, L = 10 cm = 0.1 m
The mass attached to the cube, m = 100 kg
Shear modulus (η) of aluminium = 25 GPa = 25 × 109 Pa
Shear modulus, η
Where,
F = Applied force = mg = 100 × 9.8 = 980 N
A = Area of one of the faces of the cube = 0.1 × 0.1 = 0.01 m2
ΔL = Vertical deflection of the cube
= 3.92 × 10–7 m
4. Four identical hollow cylindrical columns of mild steel support a big structure of mass 50,000 kg. The inner and outer radii of each column are 30 and 60 cm respectively. Assuming the load distribution to be uniform, calculate the compressional strain of each column.
Mass of the big structure, M = 50,000 kg
Inner radius of the column, r = 30 cm = 0.3 m
Outer radius of the column, R = 60 cm = 0.6 m
Young’s modulus of steel, Y = 2 × 1011 Pa
Total force exerted, F = Mg = 50000 × 9.8 N
Stress =