Force exerted by aircraft dive brakes- where did I go wrong?

In summary, an aircraft of 5000kg mass with a speed of 500 knots dives vertically downwards and reduces its speed to 325 knots with the use of dive brakes. The average air resistance during the deceleration is 15KN. To calculate the force exerted by the dive brakes, the equation V2=u2+2as is used after converting the velocities to m/s. The resulting acceleration is -6.3696m/s2. By applying Newton's 2nd law, the drag force is calculated to be 66,000N, which is significantly different from the answer given in the book. Further analysis shows that the book is incorrect and the correct answer is 66,000N.
  • #1
telebender
3
0

Homework Statement


An aircraft of 5000kg mass is diving vertically downwards at a speed of 500 knots. The pilot operates the dive brakes at a height of 10 000m and reduces the speed to 325 knots at 7000m. If the average air resistance of the remainder of the aircraft during the deceleration is 15KN what average force must be exerted by the dive brakes? (assume the engine is throttled back and is not producing any thrust)


Homework Equations



V2=u2+2as
where V= final velocity
u=original velocity
a=acceleration
s=distance

The Attempt at a Solution


First convert velocities into m/s. 1 knot=0.5145 m/s so:

u=500 knots=257.25 m/s
v=325 knots=167.21m/s

transposing above formula for a: a=(V2-u2)/2s

=-6.3696m/s2

The force causing the plane to acelerate downwards is mass*gravity (mg) so to stop the acceleration the drag force would have to be equal to mg. 5000kg * 9.81m/s2=49050N As the drag force acts opposite to acceleration call it -49050N

To acelerate the plane (in this case opposite to direction of travel) there must be an additional force of mass * aceleration (ma). 5000kg*-6.3696m/s2=-31848

The total drag force would be: (force equal to mg)+ma

-49050N+-31848N=-80898N

These figures are -ve because they are based on a negative aceleration so the actual drag force is +80898N

Take away from this the air resistance without dive brakes of 15000N=

65898N



The answer is given as 16.62KN, I have given this a lot of thought but can't work out where I've gone wrong. Any help please?
 
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  • #2
Welcome to PF!:smile:

Don't take any shortcuts, draw a free body diagram to identify all the forces, then use Newton 2, Fnet = ma. The air resistance force is the drag force and is given. You have the weight force. Solve for the dive brake force, using Newton 2 and the correct value of the acceleration you determined (the acceleration and the net force must be in the same direction...which direction is that?).

Edit: In checking your work, you appear to have arrived at the correct answer using an unorthodox method.
 
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  • #3
Hello,
Thank you for your reply.

The answer of 16.62KN from my last post is from the book I'm working out of. The answer I arrived at was 65.9KN which is why i posted here.

https://www.physicsforums.com/attachments/35372
I have attached my attempt at a free body diagram here.

The Fnet and acceleration must be in an upwards direction as the plane is going straight down and is slowing.

From reading your reply I think I just need to multiply the mass of 5000kg by the acceleration of 6.3696m/s2 which gives 31848N. Subtracting the 15000N of air resistance gives 16.848KN. That seems quite close to the correct answer but I've somehow gained 228N

I think where I went wrong with my first attempt was that I assumed the plane was still accelerating due to gravity when the dive brakes were deployed. If this was the case would there be a force of mg reqiired to halt the acceleration in addition to the force accelerating the plane upwards as I thought?

Hope I have made my thoghts clear. Apologies if not.
thanks again
ed
 

Attachments

  • drag fbd.JPG
    drag fbd.JPG
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  • #4
What I meant is that you were right and the book was wrong.

Fnet = ma
Brake drag + other drag - mg = ma
Brake Drag + 15,000 - 5000(9.8) = 5000(6.37)
Brake Drag - 34,000 = 31,850
Brake Drag = 66,000 N

You had it right the first time...but be meticulous when using Newton's 2nd law.
 
  • #5
Many thanks for your help on this Jay, I didn't even consider that the book might be wrong!
Now that's cleared up I can move on without worrying that I've misunderstood the basics.
really appreciate it,
ed
 

1. What is the purpose of dive brakes on aircraft?

The primary purpose of dive brakes on aircraft is to aid in reducing the speed of an aircraft during a dive. This is achieved by increasing the drag on the aircraft, allowing it to slow down more quickly and safely.

2. How do dive brakes work?

Dive brakes typically consist of panels or flaps that are extended from the wings or fuselage of an aircraft. When these panels are extended, they create additional surface area and increase the drag on the aircraft. This slows down the aircraft by reducing its forward momentum.

3. How much force is exerted by dive brakes on an aircraft?

The amount of force exerted by dive brakes on an aircraft can vary depending on the design and size of the brakes, as well as the speed and weight of the aircraft. However, on average, dive brakes can exert up to several thousand pounds of force.

4. Can dive brakes cause damage to an aircraft?

Although dive brakes are designed to withstand the forces exerted on them during flight, they can potentially cause damage to an aircraft if not used properly. It is important for pilots to follow proper procedures and limitations when using dive brakes to avoid any potential damage.

5. What could have gone wrong if an aircraft experienced difficulty with its dive brakes?

There are several factors that could contribute to issues with dive brakes on an aircraft. This could include mechanical failures, incorrect usage by the pilot, or even external factors such as weather conditions. It is important for proper maintenance and training to be in place to prevent any potential problems with dive brakes.

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