A medium-sized jet has a 3.8-m-diameter fuselage and a loaded mass of 85,000 kg. The drag on an airplane is primarily due to the cylindrical fuselage, and aerodynamic shaping gives it a drag coefficient of 0.37. How much thrust must the jet's engines provide to cruise at 230 m/s at an altitude where the air density is 1.0 kg/m^3

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Accepted Answer

Hey everyone. So this problem is working with lift and drag forces. Let's see what we're working with. We have a cylindrical shaped rocket of radius 1.25 m traveling horizontally through the earth's atmosphere. The rockets drag coefficient which was determined experimentally in a wind tunnel is 0.28. We are asked to calculate the thrust produced by the rocket engine in order to travel at a constant speed of 175 m per second in a region of the atmosphere where the air density is 0.825 kg per meter cubed. Our multiple choice answers here are a 9.92 times 10 to the third newtons. B 1.74 times 10 to the four newtons C 2.11 times 10 to the four newtons or D 6.21 times 10 to the four newtons. So the first thing we're going to do here is draw our free body diagram and see what forces are acting on this rocket. So we know we have a thrust force that is moving the rocket horizontally and then we have the drag force that is acting opposite of that direction of motion. We also have our weights in the uh negative Y direction and then our lift force in the positive. And recall here that our lift force is perpendicular to uh the flight direction. The next thing we're going to do is recall Newton's second law which states that F equals ma so the sum of our forces is equal to our mass multiplied by our acceleration because our rocket is traveling at a constant speed that tells us that our acceleration is zero. So we can rewrite this and we can look at our forces just in the X direction as some of the forces in the X direction equals zero. And so we have our thrust force minus our drag force equals zero. And then we move or drive force to the other side of the equation or add our drag force to both sides. And we are left with our thrust force is equal to our drag force. Next, we can recall that our drag force equation is given as 1/2 multiplied by row C A V squared. And so while, while they are asking us to solve for the thrust force, we know that the thrust force is equal to the drag force. And so that would also be the equation in this problem for our thrust force. And from here, we can plug in chug because we have all of these values or can find all of them from the problem. So let's take each term 1 x 1. So our density was given to us as 0.8, 2, kilograms per meter cubed. Our drag coefficient C was given to us as 0.28. Our area, well, they didn't give us area. They told us that we have a cylindrical shaped rocket and the radius. So we can recall that because the cross section of a cylinder is a circle and the area of a circle is given as pi R squared. We can then plug in that radius. So we have pi multiplied by 1.25 m squared. And then our speed was given to us in the problem as 175 m/s. OK. So let's plug and chug from here one half multiplied by 0.8, 25 kilograms per meter cubed multiplied by 0.2, 8, multiplied by pi multiplied by 1.25 m squared. And then our last term Speed 175 meters per second squared. We can plug all of that into our calculator and we are left with 1.74 times 10 to the four S. And so that is the answer to this problem. And that aligns with answer choice B. So that's all we have for this one. We'll see you in the next video.

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Key Concepts

Drag Force

Thrust

Aerodynamic Coefficient