A copper wire sags 54.0 cm between two utility poles 30.0 m ap...

Question

A copper wire sags 54.0 cm between two utility poles 30.0 m apart when the temperature is -15° C. Estimate the amount of sag when the temperature is + 35° C. [Hint: An estimate can be made by assuming the shape of the wire is approximately an arc of a circle; hard equations can sometimes be solved by guessing values.]

Accepted Answer

Hi everyone. Let's take a look at this practice problem dealing with thermal expansion. So in this problem, a clothes line is made of steel is installed between two pillars separated by a distance of 40 m in the winter at temperatures of negative five °C. The clothes line sags 4.5 centimeters, make an approximation of the amount of which the clothes line sags in the summer at a temperature of 15 °C assume that the coefficient of thermal expansion of steel is 1.2 multiplied by 10 to the negative five inverse degrees Celsius. And we're given a hint and it says that the shape of the clothes line in the summer can be approximated by an arc of a circle. And also the equations um that are not solvable analytically may be solved by trying out different values um by trial and error. We're given four possible choices as our answers. Choice A is 17.9 centimeters. Choice B is 25.1 centimeters. Choice C is 29.1 centimeters and choice D is 37.9 centimeters. So we're gonna start off by drawing a diagram of our problem. So we're gonna have the two end points of the close line and we're gonna draw a small arc and here this is gonna be exaggerated a bit, just make the um sag noticeable in the line. And we're gonna actually bisect that clothes line and we're only gonna be working with half the clothes line. So half of that arc length, we're gonna label as L uh L knot. We're gonna label the sag distance snot and we're going to label half of the distance between the two poles as X. Now, we're told that the clothes line sags 4.5 centimeters, but the distance between the two pillars is 40 m, that distance is much greater than the sag length. That means that we can approximate our initial length of the close line as that of the distance between the two pillars, which would be 40 m. But here we're only working with half of that. We're gonna have L knot is equal to X. So that means L knot gonna be approximately equal to 20 m. So that's gonna be my initial length of wire that I'm gonna be looking at. And here I need to use my thermal expansion equation to see how much that length changes by going from the winter temperature to the summer temperature. To recall our formula for thermal expansion that is delta L is equal to alpha multiplied by L not multiplied by delta T where alpha here is the coefficient of thermal expansion L not, it is gonna be my original length and delta T is my change in temperature. So that means my delta L is gonna be equal to where alpha will go ahead and plug in our value that we were given that is the 1.2 multiplied by 10 to the negative five inverse degrees Celsius. We have our initial length which is gonna be our 20 m and then we'll have our change in temperature. Now, our final temperature is going to be the 15 °C. And we're sub subtract from that our initial temperature, which is the negative five °C. So I plug those values into my calculator. I get delta L gonna be equal to 4.8 multiplied by 10 to the negative three meters, but not really gonna want my change in length. What I actually want is my new length because we're gonna need that to figure out how much the clothes line has sagged. So that means that my new length L is gonna be equal to my original length. L not plus delta L. And so we'll get L is equal to the 20 m plus the 4.8 multiplied by 10 to the negative 3 m. And so L going to be equal to 20.00 or 8 m. So now that we have our new length, we're going to basically redraw our diagram that we had for the first part, again, we'll have our two end points and again, I'll draw an arc with the close line and now we're gonna label things a little bit differently. We're still going to look at just half of the closed line. We'll still have the distance X as being the halfway point to one of the poles, our psych distance, we're gonna label as S and we're gonna have the length of the arc going from the pole to the midpoint as L NL that we just calculated. Now, I want to use L to actually find S but I'm not sure that S is small enough to approximate uh approximate L as a straight line. So what we're going to need to do is actually use some trick. So the arc here, it's gonna have a radius R associated with it. So we're told that be approximated as the arc of a circle, that circle has a radius. We'll label that as R and that gives me a right triangle where one leg is gonna be X, the hypotenuse is R and the other leg is going to be R minus S since R is going to go all the way down to the line, and then I'll subtract off S and that gives me that um length for that leg of the triangle. And we also have an angle that we can label and then label that as data and it's gonna be in the upper corner of that right triangle. So if we could figure out what the radius of the circle is, are, we could use it along with Pythons theorem to figure out what S is. And we actually can use some trig to do this. First thing we'll notice is that um we have the arc length L, we have the radius R and the angle theta. So we just basically use our definition for a radiant. So we'll have data is equal to L divided by R. So that's just used in our definition of arc length and the angle and radiance. The other thing we can use is a trick function. And for this, we'll use sign and the sine of theta is going to be equal to the opposite divided by the hypotenuse which is just X divided by R. The reason why we chose these two equations because I don't know what X uh R is and I don't know what theta is. But now I have two equations with two unknowns. So I can plug in for theta and I'll have sign of L divided by R is equal to X divided by R. And this is really as far as I can go with this equation, this is what's called non analytical equation. That means I can't solve for R just by rearranging the variables. And so I would have to solve for R in other ways. One way is to just by um plugging different values through trial and error to see if I can figure out what the value of R is. But another way is to actually use an equation solver. And that's what we'll actually do. Let's go ahead and plug in values for the L and X just so we have everything written out. We'll have the sign of for L will have the 20.0048 m divided by R that's gonna be equal to 20 m divided by R. So if we plug this into an equation solver, like those that are found in most graphing calculators, we'll get R gonna be equal to 527.2 m. So now that I have R, I can basically use Python's theorem to figure out what S is. So if I look at the right triangle from Python's theorem, we'll have R minus the quantity of R minus S quantity squared plus X squared. It's gonna be equal to R squared. So we'll multiply out that binomial and I'll get R squared minus two RS plus S squared, then we'll have plus X squared and that's gonna be equal to R squared. The R squared will cancel out and here I'll have R multiplied by S and then I'll have another, another term that is S squared. No, I'm not sure how big S is, but R is extremely large in the hundreds of meters. So R is gonna be much greater than S and so I can ignore the S square term. So I can just basically set it to zero. And so what I'm left with is gonna be X squared minus two RS is equal to zero. So I need to solve this for S so I'll just move the uh two RS to the other side and divide three by two R. So I'll get S is equal to X squared divided by two R. And here I can plug in values. That's all I have S is equal to X. That's gonna be the 20 m. And that quantity is squared and this gets divided by two. And then we'll have our value for R which is the 527.2 m. And this gives me an S value. I wanna plug those into my calculator of 0.379 m. However, I look at my answer choices, they're all in centimeters. So I need to multiply this by the conversion factor of 100 centimeters divided by 1 m. So that means that s gonna be equal to 37.9 centimeters. And so when I look at my answer choices, this actually corresponds to answer choice D. So just to recap of what we've done, we start off by making the approximation that our length of the close line is going to be equal to the separation distance between the two poles. We then use our thermal expansion equation to find a new length and then we use some geometry to figure out how much the line will sag at this new length and the new temperature. So I hope that this has been useful and I'll see you in the next video.

Key Concepts

Thermal Expansion

Catenary Curve

Mechanical Properties of Materials

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