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A ship, on which airplanes land, has an airstrip of 70 m. The ship uses three different techniques to stop airplanes when they are landing.

The first technique is to use a cable reel of radius 0.4 m, around which a non-elastic rope is coiled. The reel is to be rotated about its axle - which is a fixed point (doesn't move) - opposite to a friction force cause by two brake pads in contact with the outer radius of the reel that is 0.6 m. The coefficient of static and kinetic friction between a brake pad and the reel are 0.45 and 0.41, respectively.

The second technique is to use an elastic rope that restrains the airplane in a way that it exerts a force, on the airplane, proportional to the rope extension.

The third technique is to use a piston - located inside a cylindrical water tank which is lying horizontally - such that when an airplane lands, the piston pumps water out, through a perforation at the end of the tank, by means of a non-elastic rope between the airplane and the piston. In such a case, the pressure acting against the piston is proportional to V^2 (velocity squared) of the piston.

If the airplane has a mass of 3000 kg, and a velocity of 48.9 m/s as it touches the ship. And if the airplane braking system is not used, but rather, a technique (of the three) starts to decelerate the airplane as soon as it touches the ship.

(1)

F=a(b)^c

Use this form to write expressions for the force supplied by the three techniques.

Knowing that F is the force, a and c are constants, and b is replaced by: velocity or acceleration or time or displacement.

(2)

For the first and second techniques, find the maximum acceleration experienced by the airplane and determine at what stage this acceleration is experienced.

(3)

If the braking system of the airplane is used, where it applies a total horizontal force of 6000 N to slow the airplane down, as soon as it touches the ship. Find the maximum acceleration experienced by the airplane when it is landing, if the third technique is used.

MY attempt:

For the first branch:

Tech. 1 : F=ma so b=a (deceleration) and a=mass=3000 kg and c=1

Now, to determine the deceleration value, we use : V^2=(Vi)^2 - 2 (a) x which gives us a=17.07 ms^-2

Conclusion F=(3000)(17.07) N And here comes the confusion, since the force is constant, so how would it be written in the given form !?

Tech. 2 : F= kx so b=x (displacement) and a=k and c=1

Now, to determine k , we use 0.5mv^2=0.5kx^2 which leads to k= 1464 N/m

Conclusion, F= 1464x

Tech. 3 : Given that P ∝ v^2 ==> F/A ∝ v^2 ==> F= (cA).v^2 where cA is a constant, let it be K

Conclusion, F= K.v^2 where a=K and b=v and c=2

The second branch : the occurrence of the maximum acceleration can be found from the velocity-time graph, which shows that the max. acceleration occurs, for the first tech. at the beginning, and for the second tech. at the end.

the value of each is :

For tech. 2 : we have F=ma = 3000a and F=kx=(1464)(70)=102480 N

equating both, 3000(a)= 102480 N ==> a= 34.16 ms^-2

For Tech.1 I can not find it, since the force at the very beginning is highest, since the coefficient of static friction of the brake pads is used, which is larger than the kinetic. However, the effect of this static friction force of the braking system is instantaneous i.e at the first instant only. So, if I neglect this effect, the acceleration is constant, which is a=17.07 ms^-2 (found above). But I believe it shouldn't be like this !

For the third branch: we have F=K.v^2 , and since the braking system applies a force of 6000 N, then,

6000+F=K.v^2 , but here I have two unknowns; F and K.

Again, from the velocity-time graph I can see that the acceleration is maximum at the beginning, but I am unable to work it out, (I tried, but got wrong answers)

Hello Soregno,

For the 1st branch:

For Tech 1, I think you are meant to think the way the designer of the reel + braking system had to think in determining that his system could generate the required force, 3000*17.07 N = 51,240 N. It appears that the cable never places more than one layer on the reel, so the tension on the cable always has the radius of 0.4 m as its lever arm. The brakes operate at the 0.6 m radius, so 0.6 m is the lever arm for the brakes. Therefore the Mechanical Advantage is 0.6/0.4 = 1.5.

So the force that the brakes generate translates to 1.5 times as much force decelerating the plane.

F = 1.5*muk*N

At first I thought that is the form F = a(b)^c for this technique. Except b is required to be V, a, t, or x. As you said F = ma. You worked out a for this case which has constant acceleration and you know m. So you can say that

m*a = 1.5*muk*N

N = m*a / (1.5*muk) = (m/(1.5*muk)) * a

So m/(1.5*muk) is the constant a in F = a(b)^c, b is the acceleration a, and c is 1. In this technique, b is also a constant.

For techniques 2 and 3, I agree with what you did.

For the 2nd branch:

For Tech 1, the braking force is constant throughout except temporarily when getting the reel to start turning. For that brief moment the braking is higher by a factor of 0.45/0.41. You know the value of the constant acceleration throughout the plane's run to the end of the runway. So it is briefly higher by the factor 0.45/0.41.

I agree with you for tech 2.

For the 3rd branch:

This part seems less that totally clear about what the situation is. I will assume that the system has not been changed, it has the same performance as when we assumed the plane's brakes were not used. So there is an additional 6000 N slowing the plane and it will stop with extra safety margin. The ship's system will apply greatest force at the moment of touchdown. We need to know that value of force. For that I think we need integration. Once that value is determined, the force in this branch will be greater by 6000 N.

You said in branch 1 that the relationship between force and velocity was

F= K.v^2

The original kinetic energy is 1/2mV² = (0.5)(3000)(48.9)² = 3,586,815 J. By the principle of conservation of energy, that amount of work has to be done to stop the plane. And work is Force*distance. It is because force changes with speed and therefore with distance, that integration is needed.

Work = the integral of K.v^2 dx from touchdown until the end of the runway.

Hmmm. I seem to be stuck. I'll work more on this. Meanwhile, you can look at this much. When I add to this, you will be notified.

Sorry, but I can't get this one. You probably see the problem. I had

Work = the integral of K.V^2 dx

I tried replacing V^2 with Vi^2 + 2*a*x. The problem there is that a is not constant. It varies with x but I can't find an expression in terms of x to replace it with.

In case you can overcome this problem, I'll explain what would come next. Knowing what the total work has to be, you should be able to solve for K. So then you should be able to determine the initial force in the case of not using the plane's brakes. And then you just increase the force by 6000 N and calculate the acceleration.

I hope I have been some help,

Steve

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Comment | Thanks a lot for the explanation. There seems to be something missing in the question, that the last branch led to no answer. I tried to solve it, but I guess it will lead to a diffirential equation of the second order, non-linear, which is very difficult to solve. Many thanks. |

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