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

Question

A steel rod of radius R = 15 cm and length ℓ₀ stands upright on a firm surface. A 78-kg man climbs atop the rod.

(b) When a metal is compressed, each atom throughout its bulk moves closer to its neighboring atom by exactly the same fractional amount. If iron atoms in steel are normally 2.0 x 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.]

Accepted Answer

Hi, everyone. Let's take a look at this practice problem dealing with Young's modulus. In this problem, a 90 kg woman that's one to an aluminum rod with a radius of R is equal to 20 centimeters and a length L which stands firmly on a concrete floor. The aluminum atoms in the rod are normally 2.5 multiplied by 10 to the negative 10 m apart. By what distance must this inner atomic spacing change to producing no force required to support the woman? Young's modulus E or aluminum is 70 multiplied by 10 to the nine newtons per meter squared. We give four possible choices as our answers to A is 1.4 multiplied by 10 to the negative 18 m. Choice B is 4.3 multiplied by 10 to the negative 17 m. Choice C is 2.5 multiplied by 10 to the negative 17 m and choice D is 1.9 multiplied by 10 to the negative 16 m. Now, we ultimately want to calculate the change in the spacing of our atomic a of our aluminum atoms. And one way we're going to do that is to relate to the change in the length of the aluminum rod. And one way we can calculate that change in length is by using our definition of Young's modulus. To recall our definition for Young's modulus, we have E equal to FL divided by A delta L. So E here is Young's modulus. F here is the force that's being applied. A is the area L is the original length and delta L is the change in length. What we're actually interested in is the ratio of delta L divided by L. We're gonna solve that this equation for that quantity, we'll have delta L divided by L gonna be equal to F divided by A E. Now, here F needs to be equal to the weight of the woman because that's the force that's being applied. So we'll have F is equal to MG and that's also gonna be um equal to the normal force that's going to be pushing back up on the woman due to the aluminum rod. That's because the woman is gonna be stationary once the rod has been compressed to the appropriate level. And so the weight is going to be equal to the normal force. That's how we relate it back to the normal force. We also have the area and the cross sectional area of a rod is just a circle. So A is going to be equal by R squared. And I can plug those two expressions into our equation. We will have delta L divided by L is going to be equal to mg divided by ir squared E. So everything on the right hand side of that quantity is given in the problem. But now we need to relate this delta L to the change in the atomic spacing. And the way we're gonna do that is by looking at this ratio have delta L divided by L and this could be for any length all the way down to possibly the atomic scale. And so this is also going to be have to be equal to our change in our atomic spacing delta D divided by D. So this ratio must hold true at any level. So what we can do now is solve for our delta D. So delta D is going to be equal to delta L divided by L multiplied by D. However, I have an expression for the delta L divided by L. In terms of our given quantity, I can substitute that as N, they'll have D as equal to delta D is equal to the quantity of MG divided by IR squared E. That fraction gets multiplied by D. So I can plug in values now at this point. So I have delta D is equal to, with the mass, I'll have 90 kg for the acceleration due to gravity. I'll have 9.80 meters per second squared that gets divided by pie. And then this gets multiplied by the radius squared. The radius is 20 centimeters. But I need to convert that meters by dividing by 100. So that will be 0.2 m thats squared. And then the young modulus I was given as 70 multiplied by 10 to the nine Newton's per meter squared. And then that fraction gets multiplied by the original atomic spacing, which is the 2.5 multiplied by 10 to the negative 10 m. So everything on the right hand side of that equation is just a number. So plugging those values into my calculator, I get delta D is equal to 2.5 multiplied by 10 to the negative 17 m here have just kept two significant figures. And if I look at what answer choice that corresponds to that corresponds to answer C. So just a quick little recap, we started off by using our definition of the young's modulus, define the ratio of the delta L divided by L in terms of our given quantities. And once we did that, we related that ratio to the um atomic spacing and its change in the atomic spacing. And that allowed us to calculate our change in spacing. So I hope that this has been useful and I'll see you in the next video.

Key Concepts

Normal Force

Interatomic Spacing

Elastic Deformation

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