Posted by **boom** on Wednesday, March 27, 2013 at 8:22am.

A fully loaded Cessna-182 airplane of mass 1250 kg has an engine failure when flying with an airspeed of 129 km/h at an altitude of 2670 m on a calm day. It then glides at a constant glide angle (which is the direction of flight below the horizontal) towards a safe landing at this constant speed of 129 km/h experiencing a drag force of 1300 N that opposes the direction in which the plane is moving.

1.The lift force which acts perpendicular to the wings of the plane.

2.The rate with which the loaded plane is losing gravitational potential energy.

3.The rate with which the loaded plane is losing gravitational potential energy

- Physics Oasis!! Help!! -
**bobpursley**, Wednesday, March 27, 2013 at 10:13am
1. The lift force is mass*g if the aircraft is at constant speed downward gliding.

2. rate losing PE?

altitude= vertical speed*time=

= speedgiven*sinGlideAngle*time

rate of altitude loss= speedgiven*sinGlideAngle

rate of PE loss: mg*rateOfAltitudeLossAbove.

- Physics Oasis!! Help!! -
**Damon**, Wednesday, March 27, 2013 at 10:23am
Lift does not act perpendicular to the wings but to the oncoming air flow. The two angles differ by the angle of attack.

speed = 129*10^3/3600 = 35.8 m/s

weight = 1250 * 9.81 = 12,263 N

Path at angle T down from Horizontal

Vertical forces:

weight down = 12,263 N

Lift component up = L cos T

Drag component up = 1300 sin T

so

12,263 = L cos T + 1300 sin T

Horizontal forces:

weight not horizontal

Lift forward = L sin T

drag back = 1300 cos T

so

1300 cos T = L sin T

- Physics Oasis!! Help!! -
**bobpursley**, Wednesday, March 27, 2013 at 10:25am
I agree with Damon, I was thinking the drag force as horizontal.

- Physics Oasis!! Help!! -
**Damon**, Wednesday, March 27, 2013 at 10:26am
By the way, this means that for a glider:

Drag/Lift = tangent of steady glide angle

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