Question

Two circular rods, one steel and the other copper, are joined end to end. Each rod is 0.750 m long and 1.50 cm in diameter. The combination is subjected to a tensile force with magnitude 4000 N. For each rod, what are (a) the strain and (b) the elongation?

Answer 1

Answer:

(a) Steel rod: $1.1 * 10^{-4}$

    Copper rod: $1.88 * 10^{-4}$

(b) Steel rod: $8.3 * 10^{-5} m$

Copper rod: $1.41 * 10^{-4} m$

Explanation:

Length of each rod = 0.75 m

Diameter of each rod = 1.50 cm = 0.015 m

Tensile force exerted = 4000 N

(a) Strain is given as the ratio of change in length to the original length of a body. Mathematically, it is given as

Strain = $\frac{1}{Y} * \frac{F}{A}$

where Y = Young modulus

F = Fore applied

A = Cross sectional area

For the steel rod:

Y =  200 000 000 000 $N/m^{2}$

F = 4000N

A = $\pi r^{2}$      (r = d/2 = 0.015/2 = 0.0075 m)

=> A = $\pi * (0.0075)^{2}$

=> A = 0.000177 $m^{2}$

∴ $Strain = \frac{4000}{200000000000 * 0.000177} \\\\Strain = \frac{4000}{35400000}\\ \\Strain = 0.000113 = 1.13 * 10^{-4}$

For the copper rod:

Y =  120 000 000 000 N/m²

F = 4000N

A = $\pi r^{2}$      (r = d/2 = 0.015/2 = 0.0075 m)

=> A = $\pi * (0.0075)^{2}$

=> A = 0.000177 $m^{2}$

$Strain = \frac{4000}{120 000 000 000 * 0.000177} \\\\Strain = \frac{4000}{21240000}\\ \\Strain = = 1.88 * 10^{-4}$

(b) We can find the elongation by multiplying the Strain by the original length of the rods:

Elongation = Strain * Length

For the steel rod:

Elongation = $1.1 * 10^{-4} * 0.75 = 8.3 * 10^{-5} m$

For the copper rod:

Elongation = $1.88 * 10^{-4} * 0.75 = 1.41 * 10^{-4} m$