Please explain engine thrust and aircraft weight

Question

This is probably really easy for someone... But... My non-engineer head just cant seem to get an answer to this question in terms that are simple to digest.

A new Boeing 777X has a max take off weight of 775,000 lbs. It has 2 GE9X engines capable of delivering 110,000 lbs of thrust. (220,000 total)

So (and forgive my stupidity) how does 220,000 lbs of thrust get a 750,000 lb aircraft into the air - never mind doing it when one engine has failed and now only has 110,000 lb of thrust.

I'm probably not comparing apples with apples here, but I've never really had this explained to me.

Answer 1

Unless an aircraft takes off vertically the thrust it needs is much less than its weight, roughly proportional to the sin of take-off angle. Or sometimes a plane can fly with zero trusts. such as on the top end of a sharp climb or at the pull-up of an aerobatic loop. In both cases, its speed is enough to keep the wind on the wings and maintain the lift.

Just to get a visual on the forces involved at the takeoff, if I remember correctly ( I am a private pilot), at Bob Hope airport near Los Angeles because of the dense residential neighborhood the minimum take-off profiles are steep depending on the choice of the runway (R 8 or R33). They range between 3.75% to 9% for R33.

Say we consider the R33 with 9% if we draw the force vectors triangle like a ramp we see the small glide angle like a gentle slope calls for a small fraction of the weight even taking into account the large angle of attack at take-off causing large induced drag.

Usually, passenger jets have a thrust of approximately 1/3 of their weigh e.g. Boeing 737-300 the ratio is 30%. For small planes, Like Cessna 172 the ratio is 25%.

Here is the FAA's diagram of the take off forces.

Answer 2

The wings do the heavy lifting. Which is why there is a minimum take off speed.

The airfoil section of the wing creates lift - based on Bernoulli...