Two circular rods, one steel and the other copper, are joined end...

Question

Two circular rods, one steel and the other copper, are joined end to end. Each rod is 0.750 m long and 1.50 cm in diameter. The combination is subjected to a tensile force with mag-nitude 4000 N. For each rod, what are (a) the strain and (b) the elongation?

Accepted Answer

Hey, everyone in this problem, we're told that most bike frames are made of either carbon fiber or aluminum. OK. And tests are conducted to determine the structural performance properties of each of those. OK. So we have identical bars of carbon and aluminum. Each of length is 0.5 m and radius five millimeters are subjected to a 2000 Newton tensile force. We're given the young's modulus for carbon and aluminum. And we're gonna come back to those when we need them. And we're asked to calculate the strain S and the elongation delta L of each bar. OK. So we have four answer choices A through D each of them contain a different value for the strain and the elongation for both carbon and aluminum. OK. So we're gonna come back to those as we work through this problem. So for part one, we're asked to calculate the strain S OK. We're giving the young's modulus. OK. So let's recall the relationship. OK. We can write that the young's modulus, why it's going to be equal to the stress divided by the strength? All right, we wanna find the strain. So let's go ahead and isolate that in our equation, we get that the strain is going to be equal to the stress divided by the young's modulus. Why? Um But we aren't given the stress in our problem. OK. So we need a way to calculate the stress using the values we have. Well, it turns out that the stress and I recall can be written as force divided by the cross sectional area. A OK. OK. So for our problem, we have that the string that we're looking for is going to be equal to the force divided by the cross sectional area. A multiplied by the young's modulus. Y. All right. So we now have what we need to calculate this problem. OK. And let's start with this area. Y both of these bars have the same radius. So they're gonna have the same cross sectional area. So we can actually calculate A for both of them. Now, this is going to be a bar broad. And so the cross sectional area is gonna be the area of this oh pr squared substituting in our radius. A five millimeters, we wanna convert to our standard unit of meters. So we divide by 1000 0.005 m and that's squared, we multiply that by pi we get our cross sectional area to be pi multiplied by 2.5 multiplied by 10 to the exponent negative 5 m squared. And we're gonna leave it like that in that exact form until we substitute it in our equation. So we can avoid some round off error. All right. So the first thing we're gonna do is calculate the strain for the carbon. OK. So we're gonna call that grain C and we're gonna use subscript C for the carbon. We're gonna just substitute in these values. So we have, the four is going to be 2000 noons divided by the area we just calculated. So pi multiplied by 2.5 multiplied by 10 to the exponent negative 5 m squared. And we're gonna multiply by the young's modulus of carbon, which we're told is 1.81 multiplied by 10 to the exponent 11 FSA. All right. So we work all this out and we get that the strain or Hartman, it is going to be about 1.407 multiplied by 10 to the exponent negative or hm. All right. Now we're gonna do the same for the aluminum and we're gonna do this in green. So everything related to carbon in blue, everything related to aluminum in green. So we are looking for the strain again this time of our aluminum. So we're gonna use subscript A L. It's gonna be the force, it's the same force that they're both subjected to. So 2000 noons divided by the area which we said is the same because these bars are the same length in um the same radius. So pi multiplied by 2.5 multiplied by 10 to the exponent negative 5 m squared. And the difference in this equation is gonna be that young's modulus. And for aluminum, we're told that it is 6.9 multiplied by 10 to the exponent 10 pascals. And when we work all of this out on our calculators, we get that the strain or aluminum is gonna be about 3.69 multiplied by 10 to the exponent negative four. All right. So we're done part one for this problem. And let's take a look at our answer choices. So we found that the strain s our carbon is about 1.41 multiplied by 10 to the exponent negative four and four aluminum. It's about 3.69 multiplied by 10 to the exponent negative four. So it looks like option D is going to be the only correct option. OK. Option A has that the strain for carbon is 0.011 B has that as well. That's not correct. Um Option C 0.8 times 10 to the exponent negative two. Also not correct. OK. So D looks like it only has the only correct answers for the string. But let's double check, let's calculate the elongation so that we know how to do that. And so that we know we have not made a mistake. Ok. So now for part two, we are going to be calculating the elongation. Now, we know the strains, we've just calculated them. So let's recall the relationship between strain and elongation. So the strain is going to be equal to the elongation delta L divided by the initial length L not that tells us that the elongation delta L is gonna be the strain multiplied by that initial length element. OK. So we know the strain, we know the length of our bars. We were told that in the problem. So we can calculate the elongation. So four carbon delta LC, that's gonna be the strain. 1.407 multiplied by 10 to the exponent, negative four multiplied by that initial length. We're told that it is 0.5 m. And when we work this out, we get that the elongation for carbon is about 7.035 multiplied by 10 to the exponent negative 5 m doing the same for aluminum delta lal. This is going to be the string which we found to be 3.69 multiplied by 10 to the exponent negative four multiplied by that length of 0.5 m. And so we get an elongation for aluminum of about 1.845 multiplied by 10 to the exponent negative 4 m. All right. So going back to our answer choices, we think it's going to be D let's double check that that has the correct elongations as well. So option D has the elongation for the carbon 7.04 times 10 to the exponent negative 5 m, which is what we found. And for the aluminum 1.85 multiplied by 10 to the exponent negative 4 m. And so option D is the correct answer for this problem. Thanks everyone for watching. I hope this video helped see you in the next one.

Struggling with Physics?

Two circular rods, one steel and the other copper, are joined end to end. Each rod is 0.750 m long and 1.50 cm in diameter. The combination is subjected to a tensile force with mag-nitude 4000 N. For each rod, what are (a) the strain and (b) the elongation?

Watch next

Density practice set