The Wright brothers used a falling mass attached to their airplane by ropes and

Question

The Wright brothers used a falling mass attached to their airplane by ropes and pulleys to accelerate their plane down a rail (as shown). According to notes from Wilbur Wright for their initial test ight “A 600 lb. weight dropped 16-1/2 feet, pulled the aeroplane forward at a good clip” (http://www.thewrightbrothers.org/1904.html). The airplane (plus pilot) had a weight of 745 pounds. For this analysis you can neglect any eects from the mass of the pulleys and rope.

1. When the airplane is locked in place on the rail so that it is not moving, and its engines are not on, what is the tension in the tope connecting the plane to the falling mass?

2. When the airplane is released and accelerates down the track, is the tension in the rope less, the same or more than when it is locked in place?

3. Of course the falling mass alone did not accelerate the airplane, it also had an engine driving a pair of propellers providing constant thrust. These propellers push the air backwards with a force of magnitude F. How big is the forward force on the airplane due to these propellers? What causes it? Explain.

4. If the airplane was accelerated up to a take-o speed of 2.0×101 mph by the combination of the falling mass and the engine, how much thrust did the propellers provide if the coecient of kinetic friction for the plane sliding along the track is µk = 0.20?

Graded Assignment #6 Phys50 Th bout the Wright tthers 0 The Wright sed aaling mass attached to the azpiaue by opes and pueys to accelerate thei plaue down a ail 'as shoen. Accoding to uotes Iou Wibr Wright ko their initial test light "A G00 lb. weight dropped It-l2 eet, puled the aeoplaue forward at a good clip" http:ww. thewrightbrothere.org 1904 tlThe aiplane iplus mass i the pulleys and xope 1. When the airllare Iirla,d iri plan, on the TaLi that it is wit ltiiving·n"d its mirtirs, are rintrin, wiat is the te" won in tile top" omrKrting thr1a rin to tin, fa lil'K

Explanation / Answer

1) when the plane is locked and is not moving the tension on the rope will be equal to weight of the falling mass

since the hanging mass will also dont move so,

mg -T = 0

mg = T

mass of object attached to pulley = 600 lb or 272.155 kg

T = 272.155 * 9.8

T = 2667.119 N

tension in the rope = 2667.119 N

2) when airplane is released and accelerates down the track then equation of motion will become

mg - T = ma

T = mg - ma

since now a is positive ans having some value so new value of T will be less than mg so,

tension in the rope will be less than when it is locked in place

3) Force F will be of very large magnitude as it has to provice the lift enough to support the weight of the plane which is more than the weight of the falling mass, this force F comes from propellers which pushes the air backward like fan anf by newtons third law, aiir pushes propellers in reverse direction this giving plane constant thrust

4) plane is accelerated to 2 * 10^1 mph or 8.94 m/s in 16.5 ft or 5.03 m

by third equation of motion

v^2 = u^2 + 2as

8.94^2 = 0 + 2 * a * 5.03

a = 7.9447 m/s^2

equation of motion of falling mass

mg - T = ma

T = mg - ma

T = 272.155 * 9.8 - 272.155 * 7.9447

T = 504.929 N

equation of motion of plane

T + F - uk * mg = ma

mass of plane = 745 pounds or 337.926 kg

504.929 + F - 0.2 * 337.926 * 9.8 = 337.926 * 7.9447

F = 2842.1266 N

thrust provided by propeller = 2842.1266 N

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