Question

7. A force stretches a wire by 1 mm. a. A second wire of the same material has the same cross section and twice the length. How far will it be stretched by the same force? Explain. b. A third wire of the same material has the same length and twice the diameter as the first. How far will it be stretched by the same force? Explain.

Answer 1

Answer:

(a) The second wire will be stretched by 2 mm

(b) The third wire will be stretched by 0.25 mm

Explanation:

Tensile stress on every engineering material is given as the ratio of applied force to unit area of the material.

σ = F / A

Tensile strain on every engineering material is given as the ratio of extension of the material to the original length

δ = e / L

The ratio of tensile stress to tensile strain is known as Young's modulus of the material.

$Y = \frac{FL}{Ae}$

Part A

cross sectional area and applied force are the same as the original but the length is doubled

$\frac{FL_1}{A_1e_1} = \frac{FL_o}{A_oe_o} \\\\\frac{L_1}{e_1} = \frac{L_o}{e_o}\\\\e_1 = \frac{L_1e_o}{L_o} \\\\But, L_1 =2L_o\\\\e_1 = \frac{2L_oe_o}{L_o}\\e_1 = 2e_o$

The second wire will be stretched by 2 mm

Part B

a third wire with the same length but twice the diameter of the first

$\frac{FL}{A_1e_1} = \frac{FL}{A_oe_o} \\\\\frac{1}{A_1e_1} = \frac{1}{A_oe_o}\\\\\frac{4}{\pi d_1^2e_1} = \frac{4}{\pi d^2_oe_o}\\\\\frac{1}{d_1^2e_1} = \frac{1}{d^2_oe_o}\\\\d_1^2e_1 = d^2_oe_o\\\\e_1 = \frac{d^2_oe_o}{d_1^2} \\\\e_1 =(\frac{d_o}{d_1})^2e_o\\\\But, d_1 = 2d_o\\\\e_1 =(\frac{d_o}{2d_o})^2e_o\\\\e_1 =(\frac{1}{2})^2e_o\\\\e_1 =(\frac{1}{4})e_o$

e₁ = ¹/₄ x 1 mm = 0.25 mm

The third wire will be stretched by 0.25 mm