AVIA 245 QUIZ 4 LATEST QUESTION AND ANSWER
(LIBERTY UNIVERSITY) GRADED A+
An airplane is towing a glider to altitude. The tow rope is 20° below the horizontal
and has a tension force of 300 lb exerted on it by the airplane. Find the horizontal
drag of the glider and the amount of lift that the rope is providing to the glider. Sin
20° = 0.342; cos 20 °= 0.940. - ANSWER: Drag:
Cos 20° = D/T (Rearrange) D = Cos 20° X Tension
D = 0.940 X 300lbs
D = 282lbs
Lift:
Sin 20° = L/T (Rearrange) Lift = Sin 20° X Tension
Lift = 0.342 X 300 lbs
Lift = 102.6 lbs
A jet airplane is climbing at a constant airspeed in no‐wind conditions. The plane is
directly over a point on the ground that is 4 statute miles from the takeoff point and
the altimeter reads 15,840 ft. Find the tangent of the plane's climb angle and the
distance that it has flown through the air. - ANSWER: Find the tangent of the plane's
climb angle:
Convert Altitude Into Miles: 15,840 ft / 5280ft = 3sm
Tan = Height / Distance
Tan = 3sm / 4sm
Tan = 0.75
Find the Distance that it has flown through the air:
Da = √ Distance² + Height²
Da = √ 4sm² + 3sm²
Da = √ 4 x 4 + 3 x 3
, Da = √ 25sm
Da = 5sm
Find the distance (S) and the force (F) on the seesaw fulcrum shown in the figure.
Assume that the system is in equilibrium. - ANSWER: For the system to be in
equilibrium, the moments on the left and right of the fulcrum must be equal.
Find the Distance:
Moment = Force (W) X Distance
Weight 1 X S = Weight 2 X (Distance - S)
(20lbs)S = 10lbs (24ft - S)
20S = (10)(24) + (10)(-s)
+ 10s [20S= 240 - 10s] +10s
30S = 240
/30 [30s = 240] /30
S=8
Find the Fulcrum:
Since the system is in equilibrium, the force (F) is equal to the sum of the two
weights
F = 20 + 10
F = 30
The airplane in Problem 2 starts from a brakes‐locked position on the runway. The
airplane takes off at an airspeed of 200 fps. Find the time for the aircraft to reach
takeoff speed. - ANSWER: (Use answer from problem 2 as acceleration rate)
A = 12 fps²
Vo = 0
V1 = 200fps
V = Vo + at (Rearrange) T = V - Vo / A
T = 200fps - 0 fps / 12fps²
(LIBERTY UNIVERSITY) GRADED A+
An airplane is towing a glider to altitude. The tow rope is 20° below the horizontal
and has a tension force of 300 lb exerted on it by the airplane. Find the horizontal
drag of the glider and the amount of lift that the rope is providing to the glider. Sin
20° = 0.342; cos 20 °= 0.940. - ANSWER: Drag:
Cos 20° = D/T (Rearrange) D = Cos 20° X Tension
D = 0.940 X 300lbs
D = 282lbs
Lift:
Sin 20° = L/T (Rearrange) Lift = Sin 20° X Tension
Lift = 0.342 X 300 lbs
Lift = 102.6 lbs
A jet airplane is climbing at a constant airspeed in no‐wind conditions. The plane is
directly over a point on the ground that is 4 statute miles from the takeoff point and
the altimeter reads 15,840 ft. Find the tangent of the plane's climb angle and the
distance that it has flown through the air. - ANSWER: Find the tangent of the plane's
climb angle:
Convert Altitude Into Miles: 15,840 ft / 5280ft = 3sm
Tan = Height / Distance
Tan = 3sm / 4sm
Tan = 0.75
Find the Distance that it has flown through the air:
Da = √ Distance² + Height²
Da = √ 4sm² + 3sm²
Da = √ 4 x 4 + 3 x 3
, Da = √ 25sm
Da = 5sm
Find the distance (S) and the force (F) on the seesaw fulcrum shown in the figure.
Assume that the system is in equilibrium. - ANSWER: For the system to be in
equilibrium, the moments on the left and right of the fulcrum must be equal.
Find the Distance:
Moment = Force (W) X Distance
Weight 1 X S = Weight 2 X (Distance - S)
(20lbs)S = 10lbs (24ft - S)
20S = (10)(24) + (10)(-s)
+ 10s [20S= 240 - 10s] +10s
30S = 240
/30 [30s = 240] /30
S=8
Find the Fulcrum:
Since the system is in equilibrium, the force (F) is equal to the sum of the two
weights
F = 20 + 10
F = 30
The airplane in Problem 2 starts from a brakes‐locked position on the runway. The
airplane takes off at an airspeed of 200 fps. Find the time for the aircraft to reach
takeoff speed. - ANSWER: (Use answer from problem 2 as acceleration rate)
A = 12 fps²
Vo = 0
V1 = 200fps
V = Vo + at (Rearrange) T = V - Vo / A
T = 200fps - 0 fps / 12fps²
