The forces acting on an 85,000-kg aircraft flying at constant ...

Question

The forces acting on an 85,000-kg aircraft flying at constant velocity are shown in Fig. 12–108. The engine thrust, F_T = 5.0 x 10⁵ N, acts on a line 1.6 m below the cm. Determine the drag force F_D and the distance above the cm that it acts. Assume ${\vec{F}}_{D}$ and ${\vec{F}}_{T}$ are horizontal. ( ${\vec{F}}_{L}$ is the “lift” force on the wing.)

Accepted Answer

Hey, everyone in this problem in eight multiplied by 10 to the exponent 4 kg aircraft is orbiting earth at a constant velocity. The forces acting on it are depicted in the figure below the thrust generated by the onboard propulsion system is given by FW is equal to 7.2 multiplied by 10 to the exponent five newtons acting on a line 1.8 m below the aircraft center of mass. We're asked to calculate the drag force FD exerted by the atmosphere and the distance above the center of mass where it acts. We're told to assume that both FD and FW are horizontal and we have F lift given as well virtual that that represents the lift force generated by the aircraft's wing if any. So we're given a diagram of this aircraft. We're gonna come back to in just a minute. But first, let's look at our answer choices. There are four of them. Option A 7.2 multiplied by 10 to the exponent five noons at 2 m. Option B, we have the same magnitude of the force, but it's acting at 1.8 m. Option C 7.9 multiplied by 10 to the exponent five newtons at 2 m. In option D, the same magnitude of force is option C but it's acting at 1.8 m. So let's go ahead and take a look at our diagram. So we have our aircraft, we have this lift force acting straight upwards and it's acting to the left of the center of mass by 3.5 m. We have the force of gravity acting at the center of mass. We have this FW. So the thrust force acting to the right and we know that the distance vertically from the center of mass to this thrust force is 1.8 m. OK? And then we have our drag force FD. OK. We're looking for that force FD. We're told it points to the left and we also wanna find the distance above the center of mass work. Now, what we're told is that we are orbiting at a constant velocity because we have a constant velocity. That means our acceleration is going to be equal to zero. If the acceleration is equal to zero, then Newton's second law tells us that the sum of the forces must also be equal to zero. Now, we can look at those in each direction. So let's start with the sum of the forces in the x direction. And we know that this is gonna be equal to zero. We're gonna take up into the right to be positive in that case in a positive direction, we have the thrust force FW in the negative direction. We have this drag force FD. So we get FW minus FV is gonna be equal to zero. That tells us that the drag force FD will be equal. That thrust force told that that's 7.2 multiplied by 10 to the exponent five noons. OK. So we already have the magnitude of that force that we were looking for. We take a look at our answer choices. We can eliminate option C and option D already, they do not have the correct magnitude. Now, the other question we needed to answer was the distance above the center of mass where this acts, we looked at the sum of the forces, but we can also think about the sum of the torques. And what we're gonna do is we're gonna take the torque about the center of mass. So that when we look at the distance in our center or in our torque calculation, it is relative to that center of mass. In order to do that, we first need to figure out what this lift force is. So let's use our summer forces in the Y direction as well to find that lift force and then we can move on to the torque. So the sum of the forces in the Y direction we know again will be equal to zero because we have a constant velocity in a positive direction. We have the lift force in the negative direction, the force of gravity. So F lift minus FG is equal to zero. This tells us that the lift force must be equal to the force of gravity, which we know is equal to the mass and multiplied by the acceleration due to gravity. G. In this case, that's gonna be eight multiplied by 10 to the exponent or kilograms multiplied by 9.8 m per second squared. And while we work this out, we get a lift force of 7.84 multiplied by 10 to the exponent five nos. OK. So we do have lift from the aircraft wings. Hey, we have all of our forces. Now we can think about the torque. So let's switch over to the torque. And again, because this airplane is not rotating about itself about its center of mass. OK. It's orbiting the earth, but it's not spinning around its own axis. The sum of the torques will be equal to zero. If we think about the torque again from that center of mass point or about that center of mass point, we have to think about the torque caused by all of the different forces. Now, the simplest one is the force of gravity, the force of gravity is acting exactly at the center of mass. And so it is not going to impart any torque, it's acting. The distance of zero. The other forces we are acting are the lift force and we can imagine this lift force and it is pushing up from the left side of our center of mass that's gonna cause clockwise rotation of our airplane. That's gonna be a negative torque. So we have negative the torque, the lift force, we have the torque due to the drag force. Now, this is acting a little bit above the center of mass pointing to the left. OK. So if we push that way on our airplane, we can imagine causing a counterclockwise rotation, that's gonna be a positive torque. So we get plus to D and finally, we have the torque due to the thrust FW. OK. Under the center of mass pointing to the right again causing counterclockwise rotation, which is gonna be a positive torque. So plus Tau W is equal to zero. Now, the torque recall can be written as the force multiplied by the distance to that rotational axis multiplied by sign of the angle between that distance vector and the force vector. So we got negative F left multiplied by our left multiplied by sign of the lift was the drag force multiplied by the distance RD multiplied by sine of theta D plus FWRWS of theta W that's all equal to zero. So we're just labeling R and theta with the subscript indicating what force they are corresponding to. Now, if we think about these angles, the, and if we imagine looking at the distance, say between the center of mass and the force FW. OK. And then the direction the force is acting, those are perpendicular, the will be 90 degrees. And that's the same for all of our forces. If we think about the distance between where they're acting in the center of mass and the force itself, we are gonna get 90 degrees. So all of the angle theta will be 90 degrees sign of 90 degrees is just equal to one. So all of those are going to go to one, we're gonna be left with minus the force of lift multiplied by R left plus FDRD plus FWRW is equal to zero. Now, we can substitute in the values for all of these forces. We know all of them and what we know about the distances as well. So we're gonna have negative 7.84 multiplied by 10 to the exponent five noons. Again, looking at our diagram, this lift force we know is acting 3.5 m to the left of the center of mass. So our lift will be 3.5 m. Similarly, our thrust force FW is acting 1.8 m below the center of mass. So RW will be 1.8 m. So our equation is gonna become negative 7.84 multiplied by 10 to the exponent five newtons multiplied by 3.5 m plus 7.2 multiplied by 10 to the exponent five newtons multiplied by RD. So that's the distance between the center of mass and where this force of drag is acting, that's the value we're looking for right now plus 7.2 multiplied by 10 to the exponent five newtons multiplied by 1.8 m. This is all equal to zero. If we move all of the constant terms to the right hand side and simplify, we can write that 7.2 multiplied by 10 to the exponent five newtons multiplied by RD is gonna be equal to 1.448 multiplied by 10 to the exponent six newtons dividing both sides by 7.2 multiplied by 10 to the exponent five. We get that the distance between the drag force in the center of mass RD is gonna be 2.011 m. OK? So that is that distance that we were looking for. We can go back to our answer choices and now we can see that the correct answer is going to be option A. OK. We found that the drag force is 7.2 multiplied by 10 to the exponent five newtons. And it has a distance from the center of mass of 2 m. Thanks everyone for watching. I hope this video helped see you in the next one.

Key Concepts

Newton's First Law of Motion

Force and Motion

Torque and Moment

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