Question

Four rods that obey Hooke's law are each put under tension. (a) A rod 50.0 cm50.0 cm long with cross-sectional area 1.00 mm21.00 mm2 and with a 200 N200 N force applied on each end. (b) A rod 25.0 cm25.0 cm long with cross-sectional area 1.00 mm21.00 mm2 and with a 200 N200 N force applied on each end. (c) A rod 20.0 cm20.0 cm long with cross-sectional area 2.00 mm22.00 mm2 and with a 100 N100 N force applied on each end. Order the rods according to the tensile stress on each rod, from smallest to largest.

Answer 1

Answer:

c < a = b

Explanation:

The tensile stress = Force applied/(Cross sectional area)

(a) The applied force = 200 N

The cross sectional area = 1.00 mm² = 1 × 10⁻⁶ m²

The tensile stress = 200 N/(1 × 10⁻⁶ m²) = 200,000,000 Pa = 200 MPa

(b) The applied force = 200 N

The cross sectional area = 1.00 mm² = 1 × 10⁻⁶ m²

The tensile stress = 200 N/(1 × 10⁻⁶ m²) = 200,000,000 Pa = 200 MPa

(c) The applied force = 100 N

The cross sectional area = 2.00 mm² = 2 × 10⁻⁶ m²

The tensile stress = 100 N/(2 × 10⁻⁶ m²) = 50,000,000 Pa = 50 MPa

Therefore, the tensile stress from smallest to largest are;

(a) 50 MPa, < (b) 200 MPa = (a) 200 MPa

Therefore, we have;

c < a = b.