A helicopter is moving horizontally to the right at a constant velocity. The weight of the helicopter is W = 58600 N. The lift force L generated by the rotating blade makes an angle of 21.0° with respect to the vertical.
(a) What is the magnitude of the lift force?
(b) Determine the magnitude of the air resistance R that opposes the motion.
Can anyone help lead me in the right direction?
2007-02-14 08:16:11 · 5 answers · asked by Anonymous in Science & Mathematics ➔ Physics
Since the lift force (L) is at an angle, it has both forward and upward components.
forward component Lx = L sin(21)
upward component Ly = L cos(21)
Because the velocity is constant, the upward and downward forces must be equal. This means the weight of the helicopter (a downward force) must equal the upward component of the lift.
58600 N = L cos (21)
Solving for L gives L = 62769 N (that's the answer to (a))
The air resistance must equal the forward component so that there's no acceleration.
R = L sin 21
R = 62769 N (.359)
R = 22500 N
2007-02-14 08:41:53 · answer #1 · answered by Thomas G 3 · 0⤊ 0⤋
This problem is familiar to me, perhaps from a Serway book? I'm pretty sure there is a figure that goes along with this question.
You can set this problem up with a free body diagram. What you will see is three force vectors acting on the helicopter. You can set this up with Newton's 2nd Law. Set up an equation for each direction. In the y direction, you should see all of the Weight, plus part of the Lift (either sin or cos, depending on the set up). Since the velocity is constant, the acceleration will be 0. At this point, you will be able to find the Lift force.
Using the Lift force, set up an Newton's Second Law equation in the x direction with the Lift and the Resistance. You will again need to use sin or cos for the lift, then solve for resistance. Again, since the velocity is constant, accleration will be 0
Hope this helps.
2007-02-14 08:39:02 · answer #2 · answered by The Chuck 3 · 0⤊ 0⤋
The lift has to be greater than the weight of the helicopter or it wouldn't be in the air. The angle is irrelevant; it would become relevant if we knew the resistance of the air and the horizontal speed or velocity of the helicopter, since some of the energy is being used to move the vehicle horizontally (a vector) instead of straight up and down. There is not enough information in your question to answer either question precisely and the last question at all.
Try the NASA site for a lot of help with these issues, including interactive tutorials.
2007-02-14 08:29:50 · answer #3 · answered by thylawyer 7 · 0⤊ 1⤋
Newton's First regulation says products in action will stay in action IF NO pressure is performing on them. For something to boost up, a pressure should be utilized. In a vacuum without air resistance (a pressure) and merchandise might want to save going at inspite of speed it replaced into going even as the pressure replaced into taken away. case in point, a container quickens even as it really is pushed in the approach the floor, even though it slows down and forestalls even as that pressure is bumped off. that's because of the forces of friction and air resistance. once again, in a vacuum, without those forces, the container might want to save shifting at inspite of speed it replaced into going once the frenzy replaced into taken away.
2016-11-03 11:08:31 · answer #4 · answered by ? 4 · 0⤊ 0⤋
giving me a headach
i hate math
2007-02-14 08:21:35 · answer #5 · answered by sunny s 1 · 0⤊ 0⤋