How much acceleration does a 747 jumbo jet of mass 30000kg experience when the thrust for each of its four engines is 30000 N?

Answer 1

Ignoring air resistance, (although we know that

that's definitely not a valid assumption when we're dealing with an airliner! )

Total force = 4 x 30,000 = 120,000 newtons

F = M A

A = F / M = (120,000) / (30,000) = 4m/sec2 = about 0.41 of a 'G' .

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In actuality a fully loaded B747-200 airliner has a maximum takeoff weight of 990,000 Lbs which is 450,000 Kg mass.

Each General Electric CF6 Turbofan jet engine on the B747-200 has a maximum static thrust output of 63,300 Lbs which is equal 281,685 Newtons of thrust output per engine.

There are four (4) engines on the B747-200. Therefore at maximum static thrust output power there will be 1,126,740 Newtons of maximum static thrust on the Boeing 747-200 airliner.

The Drag Coefficient (Cd), Aircraft Projected Frontal Area (A), Air Density in Slugs per Cubic Foot (P) and Airspeed (V) are factors which when all combined give the Aerodynamic Drag which resists forward acceleration and counteracts thrust. Tire rolling resistance also resists forward acceleration as well when the airliner is rolling on takeoff during the ground.

But let's ignore Aerodynamic Drag & Tire Rolling Resistance in this case and focus purely on the acceleration that is incurred on this airliner at the given thrust output during runway roll acceleration only.

Let say a Boeing 747-200 airliner's acceleration is to be calculated during it takeoff roll with the given mass & static thrust information from rest (0 MPH) to 170 MPH within 30 seconds of runway roll time.

mass = 450,000 kg

force (thrust) = 1,126,740 N

Acceleration = [(Force) / (Mass)]

Acceleration = [(Distance) / (Time x Time)]

Acceleration = [(Speed / Time)]

1 MPH = 1.467 feet per second

1 meter = 3.28 feet

170 MPH x 1.467 = 249.39 feet per second

249.39 feet per second / 3.28 = 76 meters/sec

Acceleration = [(76 meters per second) / (30 seconds)] = 2.5 meters per second squared.

Acceleration = [(1,126,740 Newtons) / (450,000 Kg)] = 2.5 meters per second squared.

The acceleration is calculated at: 2.5 m/s^2

The horizontal G-Force is a ratio between the horizontal acceleration & gravitational acceleration plus a reference Horizontal G Force of 1.0.

The gravitational acceleration on Earth is: 9.8 m/s^2.

Horizontal G-Force during runway roll acceleration: [(1) + (2.5 / 9.8)] = +1.25 Horizontal Gs'

The Horizontal G-Force is calculated at: +1.25 Horizontal Gs'

During cruise flight the airliner's speed is held at a relatively constant rate so therefore the acceleration is zero.

Acceleration is a substantial factor on an airliner during runway takeoff roll more than any other time. Even during landing procedures, there is a gradual deceleration but not the extent of the takeoff acceleration where the airliner moves from 0 MPH to 170 MPH (0 m/s to 76 m/s) in a time period of 30 seconds!

Answer 2

The total thrust from all four engines is 120,000 N. To find the acceleration, we divide the total thrust by the mass of the jet: 120,000 N / 30,000 kg = 4 m/s^2. Therefore, the 747 jumbo jet would experience an acceleration of 4 m/s^2.