Question

A circular steel wire 3.00 mm long must stretch no more than 0.25 cmcm when a tensile force of 780 NN is applied to each end of the wire.

Answer 1

Minimum diameter is required for the wire is 1.2X10⁻³m.

Tensile force, which consists of tensile stress and tensile strain, is the stretching force exerted on the material. This indicates that the forces are attempting to stretch the material that is subject to the force since it is under tension.

Given;

Tensile force (ft) = 780 N

length (l) = 3.00 mm

change in length (Δl) = 0.25 cm

determine the strain ε = Δl/l

determine the stress σ = Eε = E(Δl/l)

determine the cross-sectional area A = f/σ = ft/EΔl

determine the diameter d = 2$\sqrt{} \frac{A}{\pi }$ = $2\sqrt{} \frac{ft}{EXΔlX\pi }$

take Young's modulus for steel E=200 GPa, Substitute the values,

d = $\sqrt[2]{\frac{x780X2}{y200X10^{9} X 0.0025\pi } }$ = 1.2X10⁻³

minimum diameter is required for the wire is 1.2X10⁻³m.

Learn more about the Tensile force with the help of the given link:

brainly.com/question/4443489

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