Answer to Question #162360 in Material Science Engineering for Abirami K A

Question

A 2 m copper wire with square cross-section of 1 mm by 1 mm stretched 1.51 mm when it  supported a 10 kg mass. Suppose you cut a length of this wire 0.2 m long and hang a 10 kg mass  from it. How much will this shore wire stretch?

Answer 1

First, wed have a wire with long L₁ = 2 m and the cross-sectional area is A = 1 mm x 1 mm = 1 x 10^-6 m². This copper wire stretched by s₁ = 1.51 mm = 0.00151 m when it supported a weight with mass m = 10 kg. Second, we have the same copper wire with same cross-sectional area, but with long L₂ = 0.2 m.

We want to determine the stretch s₂. There is a stiffness of the wire itself k_s and there is a stiffness of the single interatomic spring k_i. The stiffness of the single interatomic spring k_i will be the same for the two states at long 2 m and long 0.2 m. So, the strategy is to find k_i from k_s1 and then use this value to get k_s2 and finally use the value of k_s2 to find the second stretch s₂.

The system here is the mass 10 kg and the surroundings are the Earth (gravitational force) and the wire (tension force). By using the momentum principle we could get the stiffness of this spring when the mass hangs motionless, its momentum is not changing at any time ∆t we find

∆P_y = F_net∆t = 0

As the force in the y-direction, and the tension force is in opposite direction of the gravitational force, therefore, the net force is given by F_net = F - mg and we could get the next

F_net∆t = 0

(F - mg)∆t = 0

F - mg = 0

F = mg

k_ss = mg

k_s = (mg) / s

k_s1 = (mg) / s₁ = (10 kg x 9.8 m/s²) / 0.00151 m = 649 x 10² N/m

The stiffness of the single interatomic bond is very much smaller than the stiffness of the entire wire so we could find the stiffness of the single interatomic bond k_i as next

k_i = N_bondk_s / N_chain

Where N_bond is a number of interatomic bonds in one chain and its calculated by

N_bond = Wire's length / Atomic diameter of copper = L₁ / d = 2 m / 2.28 x 10^-10 m = 8.77 x 10⁹ bonds

And N_chain is the number of chains in the wire and it calculated by

N_chain = Wire's Area / Area of copper atom = A_w / d² = 1 x 10⁶ m² / (2.28 x 10^-10 m)² = 1.92 x 10¹³ chains

Now let us plug our values for N_chain, N_bond and k_s1 to get k_i

k_i = N_bondk_s1 / N_chain = (8.77 x 10⁹)(649 x 10² N/m) / (1.92 x 10¹³) = 29.64 N/m

Now we will find our calculations for long L₂ = 0.2 m. The number of chains N_chain is the same as cross-sectional area is the same but the number of interatomic bonds N_bond will change as the length is changed

N_bond = Wire's length / Atomic diameter of copper = L₂ / d = 0.2 m / 2.28 x 10^-10 m = 0.877 x 10⁹ bonds

k_s2 = N_chaink_i / N_bond = (1.92 x 10¹³)(29.64 N/m) / (0.877 x 10⁹) = 649 x 10³ N/m

s₂ = (mg) / k_s2 = (10 kg x 9.8 m/s²) / (649 x 10³ N/m) = 0.151 x 10^-3 m = 0.151 mm