∙A 5.0 kg mass is hung by a vertical steel wire 0.500 m long and 6.0 ×10^-3 cm^2 in cross-sectio

∙A 5.0 kg mass is hung by a vertical steel wire 0.500 m long...

$\bullet$ A 5.0 kg mass is hung by a vertical steel wire 0.500 m long and $6.0 \times 10^{-3} \mathrm{cm}^{2}$ in cross-sectional area. Hanging from the bottom of this mass is a similar steel wire, from which in turn hangs a 10.0 kg mass. For each wire, compute (a) the tensile strain and (b) the elongation.

Key Concepts

Young's Modulus

Young's modulus, also known as the modulus of elasticity, is a material property that relates tensile stress to tensile strain within the elastic limit of the material. It provides a measure of the stiffness of a material, with a higher modulus indicating a stiffer behavior under tension.

Hooke's Law

Hooke's law states that, within the elastic limit of a material, the extension or compression produced is directly proportional to the applied load. This principle underlies the linear relationship between tensile stress and tensile strain, which is characterized by Young's modulus in elastic materials.

Tensile Stress

Tensile stress is the force exerted per unit cross-sectional area of a material. It quantifies the internal forces acting within a material when it is subject to a pulling force and is calculated by dividing the applied force by the material's cross-sectional area.

Tensile Strain

Tensile strain is a measure of deformation representing the fractional change in length of a material under an applied tensile load. It is defined as the ratio of the change in length (elongation) to the original length, indicating how much a material stretches relative to its initial dimensions.

Transcript

00:01 For steel, we know that the young's modulus, we know that the young's modulus is equal to 2 .0 times 10 to the 11 pesco.

00:16 And so since both wires are steel in this problem, we're going to be using this.

00:21 Now, the force on the end of the lower wire is the mass hanging at the end of the lower wire, which is 10 kilograms times gravity, which is equal to 98 newtons.

00:32 This is the force on the lower wire.

00:36 The force on the upper wire is equal to this mass, 10, plus the additional mass, 5.

00:45 These masses are at the bottom end of the upper wire, and so we have to add them together times by g, and this is equal to 147 newtons.

00:56 And so now we can use the formula for the tensile strain.

01:01 The tensile strain is equal to delta l over l0, and this is equal to the perpendicular force over the cross -sectional area times young's modulus.

01:18 So we just figured out what the perpendicular forces are for both wires.

01:22 They're right here.

01:23 Young's modulus is the same for both wires as right here, and we're given the cross -sectional areas, but we're given them in centimeter squared.

01:31 So we have to first convert them into meter squared, and then once we convert them, we can stick everything in this formula to find the tensile strain.

01:39 I'll go ahead and skip the conversion since it's fairly straightforward.

01:42 But once you plug in those values here, you get the tensile strain for the upper wire is equal to 1 .2 times 10 to negative 3.

01:54 And there's no units on that since it's a length divided by a length...

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