AVIA 305 Test Prep UPDATED Exam
Questions and CORRECT Answers
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. -
CORRECT 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. - CORRECT 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. - CORRECT 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
Questions and CORRECT Answers
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. -
CORRECT 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. - CORRECT 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. - CORRECT 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
