The helicopter in the drawing is moving horizontally to the right at a constant velocity. The weight of the helicopter is W = 51700 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.
2006-10-12 15:13:58 · 3 answers · asked by Anna G 1 in Science & Mathematics ➔ Physics
(a)
Break the Lift into horizontal and vertical components.
Vertical = Lift*sin(21.0)
Horizontal = Lift*cos(21.0)
Since the helicopter is ONLY moving in the horizontal direction, that means the forces acting on it in the vertical direction must = 0. There are only two forces acting on the helicopter, it's weight and the vertical Lift.
Set these equal to zero since they oppose each other, then you can solve for the Lift magnitude.
W = L*sin(21.0) ------------> L = W/sin(21.0) = 144,265 N
(b)
Since the helicopter is moving at a constant velocity horizontally, that means the acceleration is zero, meaning the air resistance is defined by the lift in the horizontal direction.
R = L*cos(21.0) = 144,265*cos(21.0) --------> R = 134,683 N
Hope this helps
2006-10-12 16:46:59 · answer #1 · answered by JSAM 5 · 0⤊ 1⤋
From a point on paper, draw a right triangle with the vertical component = 51700 units. Now from the original point make a line 21 degrees from vertical and extend it far enough so you can connect a horizontal line from the end of the original vertical line. Now it is a simple trigonometry problem. The lift force is the hypotenuse of the triangle. The air resistance R is the x (horizontal) magnitude of the triangle.
2006-10-12 15:25:22 · answer #2 · answered by craig p 2 · 1⤊ 0⤋
The vertical component of the lift counteracts the weight.
The horizontal component will be counteracting the drag.
2006-10-12 15:22:53 · answer #3 · answered by arbiter007 6 · 0⤊ 0⤋