A brass rod is 185 cm long and 1.60 cm in diameter. What force must be applied to each end of the rod to prevent it from contracting when it is cooled from 120.0°C to 10.0°C?

Video transcript

Hey everyone welcome back in this problem. We are asked what force magnitude should be applied to the ends of a 50 m long copper rod of radius five millimeters to prevent it from elongation when its temperature increases by 25 degrees Celsius. Alright, so we're talking about force in its relation to elongation of the wire. So let's recall. We have the following equation. The force F divided by the cross sectional area of the rod. A It's going to be equal to negative. Why the Young's module asse? Alpha? The coefficient of linear expansion times, delta T. The change in temperature. Alright, so we're looking for the force F. Let's go ahead and figure out what all of these other variables are. And if we have enough information to use this equation. Ok, well, the area a cross sectional area. We're talking about a rod. So the cross sectional area is going to be a circle. So the area is going to be pie r squared, which is gonna be pie. The radius is five. So we have 5 mm instead of millimeters. We want to write this in terms of meters. Okay, so we're going to divide by 1000 and get 0.5 m squared. Okay, so millimeters to meters. We divide by 1000. Alright, and this is going to give us a cross sectional area of 2.5 times 10 To the -5 pi meters squared. The young's module asse. Okay, this is gonna be the Young's module is for copper because we have a copper rod. Okay, you can look this up in a table in your textbook or that your professor provided. It's going to be 1.1 times 10 to the newton per meter squared Alpha coefficient of linear expansion. Again is something we can look up. It's going to be 1.7 Times 10 to the -5. The unit is per degree C. And our change in temperature delta T. We're told that the temperature increases by 25 degrees Celsius. So our delta T. Is going to be positive and it's going to be positive 25 degrees Celsius. And you'll see that both alpha and delta T have degrees Celsius in them. Okay, so we want those to be consistent. Alright, so we have a Y, alpha and delta T. We can go ahead and plug into our equation and solve for F. So we have F divided by and again, F. Is what we're looking for. Divided by 2.5 times 10 to the negative five pi meters squared Is equal to negative 1.1 times 10 to the 11 newton per meter squared Times 1.7 times 10 to the -5 per degree C Times 25°C. So here, the unit of degrees Celsius will divide out. Okay, let's give ourselves some more room to work. If we multiply 2.5 times 10 to the negative five pi up to the right hand side. We're left with the force F. Is going to be equal to negative 3671. newtons. Okay, We multiply by meter squared, then we have divided by meters squared to the unit of meter squared, cancels. We're left with just a unit of newton, which is a unit we want for four. So our units check out. Okay. All right, So we have our force F. What we wanted to find was the magnitude of the force. Okay, So the magnitude of the force is just gonna be the positive of this. 3671. newtons. Okay, We're gonna write this in scientific notation and round To approximate the magnitude of our force to 3.7 times 10 to the three newtons. Okay. Alright. So if we go back up to our answer choices, we see that the force magnitude that should be applied to that copper rod to prevent it from elongation is going to be D 3.7 times 10 to the three newtons. Thanks everyone for watching. I hope this video helped see you in the next one