A steel rod of radius R = 15 cm and length ℓ0 stands upright on a firm surface. A 65-kg man clim

step 1 Once again, welcome to a new problem. This time we have a steel rod. Obviously, steel has iron i

A steel rod of radius R = 15 cm and length ℓ0 stands upright...

A steel rod of radius $R = 15 cm$ and length $\ell _0$ stands upright on a firm surface. A 65-kg man climbs atop the rod. (a) Determine the percent decrease in the rod's length. (b)When a metal is compressed, each atom moves closer to its neighboring atom by exactly the same fractional amount. If iron atoms in steel are normally $2.0 \times 10^{-10}m$ apart, by what distance did this interatomic spacing have to change in order to produce the normal force required to support the man? [$Note$: Neighboring atoms repel each other, and this repulsion accounts for the observed normal force.]

Key Concepts

Stress

Stress is the measure of force applied per unit area within materials. It quantifies the internal forces that develop in a material in response to external loads and is central to understanding how materials deform under compression, tension, or bending.

Strain

Strain is the fractional or percentage change in a material’s dimensions as a result of applied stress. It represents the relative deformation (such as shortening or elongation) and is calculated by dividing the change in length by the original length. This concept is crucial for linking microscopic movements to the overall deformation observed in a structure.

Young's Modulus

Young's Modulus is a material property that defines the relationship between stress and strain in the elastic regime for a solid. It provides a quantitative measure of the stiffness of a material and is used to calculate how much a material will deform under a given load. This constant is fundamental in engineering and physics for predicting deformation behavior under compressive or tensile forces.

Atomic-scale Deformation

Atomic-scale deformation refers to the concept that the relative positions of atoms in a material change by a uniform fraction when the material is deformed elastically. This idea bridges macroscopic properties like compressibility with the microscopic interatomic forces, explaining how slight changes in atomic spacing can result in significant macroscopic forces such as the normal force experienced in contact scenarios.

Transcript

00:01 Once again, welcome to a new problem.

00:04 This time we have a steel rod.

00:08 Obviously, steel has iron in it.

00:11 But this time we have a steel rod like that with a specific length, which we're going to call l -0.

00:25 And then also the other thing about the steel rod is that, you know, it has a cross -sectional diameter.

00:34 The radius itself for the rod is r equals to 15 centimeters we can always change that right away to meters by multiplying by one meter of 100 centimeters and that gives us 0 .15 meters.

00:53 It's standing upright and there's a man that's going to step on top of it.

01:01 Okay there's a man that's going to step on top of it and this man happens to have a mass of a mass of 65 kilograms remember with the mass we get so this is 65 kilograms and obviously his wage is acting downwards but um g where g is the acceleration due to gravity that's the information we're given in part a.

01:44 In part a, we want to find, well, this is going to be part a right here.

01:53 In part a, we want to find the decrease in the rod's length when demand steps on it.

02:14 So there's a compression going on because there's a side.

02:17 There that's stopping the rod from going all the way inside the ground.

02:23 So that compression is going to be to a decrease in the rod's length and that decrease, we're calling it delta l.

02:31 Also in part b, the second part of the problem where, you know, if there's compression, the atoms are going to move closer to each other.

02:46 The, and we're saying, you know what distance is the what distance is the inter -atomic spacing such that the the atoms have a normal normal 2 .0 times 10 to the negative 10 newtons between them or yeah, between the, or rather instead of between, we're going to say the force, a normal 2 .0 times 10 to the negative 10 normal force to support the man.

04:01 So that's kind of what's happening in this problem.

04:04 You know, we have a steel road that's supporting a man and we want to find the decrease in length and also the inter, the distance, that distance.

04:14 We're going to call it delta, delta d.

04:20 You know, this distance we're going to call it delta d.

04:23 So we have delta l and delta e.

04:26 So the initial part of the problem, again, we use equilibrium such that, you know, the force that's being applied upwards that's supporting the weight f equals to mg and the mass of the man is 65 kilograms, multiply that by the acceleration due to gravity 9 .81 meters per second squared.

04:51 And that gives us 637 .65 newtons.

04:59 That's this force right here.

05:03 And the question is we want to calculate the percent decrease in length.

05:08 The relationship between the change in length and the original length is equivalent to the force of a pi -r -scent.

05:17 Squared e, where this e is called the proportionality constant.

05:30 The e is the proportionality constant.

05:34 So what do we think about this? i mean, this r, we can make it small r.

05:41 It's the same thing.

05:41 You know, it's just going to be plugged into that formula...