For the airfoil and conditions in Problem 2.2, calculate the lift-to-drag ratio. Comment on its magnitude.

Physics

1 answer:

raketka [301]4 years ago

8 0

Answer:

L/D= 112

Explanation:

Aerodynamics can be defined as the branch of dynamics which deals with the motion of air, their properties and the interaction between the air and solid bodies.

Aerodynamics law explains how an airplane is able to fly. There are four forces of flight, and they are; lift, weight, thrust and drag. The amount of lift generated by a wing divided by the aerodynamic drag is known as the lift to drag ratio.

Lift increases proportionally to the square of the speed.

The solutions to the question is the file attached to this explanation.

Lift,L= qC(l). S---------------------------(1).

and,

Drag,D = qC(d).S ----------------------(2).

Hence, Lift to drag ratio,L/D= C(l)/C(d).

Therefore, we have to compute various angle of attack.(check attached file)...

Then, (L/D) will then be equal to 112.

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A 75.0kg bicyclist (including the bicycle) is pedaling to the right, causing her speed to increase at a rate of 2.20m/s^2, despi

malfutka [58]

1) 4 forces

2) 165 N

3) 225 N

Explanation:

There are in total 4 forces acting on the bicylist:

- The gravitational force on the byciclist, acting vertically downward, of magnitude $mg$ , where m is the mass of the bicyclist and g is the acceleration due to gravity

- The normal force exerted by the floor on the bicyclist and the bike, N, vertically upward, and of same magnitude as the gravitational force

- The force of push F, acting horizontally forward, given by the push exerted by the bicylist on the pedals

- The air drag, R, of magnitude R = 60.0 N, acting horizontally backward, in the direction opposite to the motion of the bicyclist

The magnitude of the net force on the bicyclist can be calculated by considering separately the two directions.

- Along the vertical direction, we have the gravitational force (downward) and the normal force (upward); these two forces are equal in magnitude, since the acceleration of the bicyclist along this direction is zero, therefore the net force in this direction is zero.

- Along the horizontal direction, the two forces (forward force of push and air drag) are balanced, since the acceleration is non-zero, so we can use Newton's second law of motion to find the net force on the bicylist:

$F_{net}=ma$