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At $560$ metric tons, the airbus A-$380$ is the world's largest airliner. Whats the upward force on an A-$380$ when the plane is flying (a) at a constant altitude and (b) accelerating upward at $1.1 \: m/s^2$

So I know $f=ma$

So for part a I did

$f= 560 * 10^3$ kg ($9.8 \: m/s^2$)

$f=5,488,000 N$

So for part b I did

$f=560 * 10^3$ kg ($1.1 \: m/s^2$)

$f=616,000 N$

Are these answers correct?

JKnecht
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Lil
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3 Answers3

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Think about it for a moment.

You say that the force when the airplane is flying at a constant altitude is $5,488,000$ newtons, but when the airplane is accelerating upwards, the force is $616,000$ newtons.

That means that if I am pushing the plane upward so it is flying at a constant altitude, then decrease the upward force bt $4,872,000$ newtons, the airplane will suddenly start moving upward?

5xum
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If you consider the numbers you got, you should be able to tell you have made a mistake somewhere - surely to accelerate upwards it takes more force. Your error lies in the fact that you used only an acceleration of $1.1\text{m/s}^2$ for part b, you forgot to consider gravity.

Dom
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  • ok makes sense. So in order to find the force upward do I multiple the mass by 9.8m/s^2 and 1.1 m/s^2? – Lil Sep 30 '14 at 13:08
  • @Lil Draw a plane and all forces acting on it. Then calculate the sum of the forces. The sum of all forces on the plane is equal to the mass of the plane times the acceleration of the plane. – 5xum Sep 30 '14 at 13:09
  • At zero acceleration the force is mg = m* 9.8 correct? So if the acceleration is 1.1 m/s^2 then the new force is m* (9.8 +1.1) = 56010^3 (9.8+1.1)/1.1 – Lil Sep 30 '14 at 13:12
  • @Lil: Why are you dividing by 1.1? – Dom Sep 30 '14 at 13:13
  • not quite sure.. trying to mimic another problem in my textbook. Should it just be 560*10^3(9.8+1.1) that way I just add the two forces acting on the object and multiply it by the mass? – Lil Sep 30 '14 at 13:15
  • You should have that sum of the force to balance, and to accelerate (that is to accelerate it if there was no gravity), so just $m(9.8+1.1)$. – Dom Sep 30 '14 at 13:17
  • @Dom makes sense. Thank you very much for your help! – Lil Sep 30 '14 at 13:18
  • @lil A more general advice is: don't just mimic thigs you see in textbooks. Draw diagrams and think about the problem. Only when you understand the problem can you start to put it down in equations. Here, you must understand that there is a constant force of gravity equal to $g\cdot m$ working on the plane, and if it flies at a level altitude, that means the sum of all forces is $0$, meaning there is another force, equal in magnitute, that is countering gravity. – 5xum Sep 30 '14 at 13:28
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You answer for part a is right.

For the second you are wrong since it is accelerating at 1.1 units. The other force acting on it is gravity which is opposite so it has to oppose this and is then also accelerating at 1.1 units so the net force is m*(9.8+1.1) units where m is mass.

So the answer is 6,104,000

Jasser
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  • Actually, the first answer is correct, since the upward force must match the force of gravity in order for the plane to fly at a constant altitude. – 5xum Sep 30 '14 at 13:08
  • There are certainly forces acting on the plane when it is holding it's altitude. The forces are merely balanced. – Dom Sep 30 '14 at 13:08
  • If it it is moving with constant height without moving up or down it must not accelerate @5xum – Jasser Sep 30 '14 at 13:11
  • Which implies that the enforce is zero. – Jasser Sep 30 '14 at 13:12
  • @Dom yes I was talking about the net force acting on the plane – Jasser Sep 30 '14 at 13:14
  • Ok fine in the question he is asking about the upward force.... not the net force. – Jasser Sep 30 '14 at 13:16