How Do Different Techniques Stop an Airplane on a Ship's Short Airstrip?

[alirami]

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. This question has 13 branches, I solved all, except for these 3..

For the first branch, I am not sure if the constants are to be replaced or not. And about the first technique, I think that the force is dependent on time in a way (the coefficient of static friction)

For the second branch, I know that the maximum acceleration occurs ; for tech. 1 at the very begging , and for tech. 2 when velocity=0 , but I am unable to find it.

For the third branch, I don't know how/where to start

I will be thankful if you can explain.

 

[Simon Bridge]

Welcome to PF;
Please show your working.

(1) you are expected to write the expressions for the force on the aircraft from each method in the form given - this means that you replace as many of the constants as you can. Probably a good idea to at least rename the one's you can't so they relate to the system in question

(2) Please show your working.
How are you trying to find the the max acceleration, and how do you know where this occurs in advance?

(3) Start with a free body diagram.

 

[alirami]

Thank you for you for your consideration.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= 1464xTech. 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=2The 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^-2For 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)Any help will be really really appreciated. Thank you in advance.

 

[Simon Bridge]

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 !?

You are using the wrong equation.
F=ma is applicable to all three technologies.
You have been asked to find a relation to replace F on the RHS of that equation, so that Newton's Law becomes: $F=ab^c=m\ddot x$

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 !

Looks like you need to take another look at how friction works.
How much speed is lost overcoming the static friction?

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)

You do not appear to be combining the equations properly.
Newton's Law says: the sum of the forces = mass tomes acceleration ... where is the "ma" in your equation?

 

[paranormal]

(1) The expressions for the force supplied by each technique are as follows:

First technique: F = μsN - μkN
where μs is the coefficient of static friction, μk is the coefficient of kinetic friction, and N is the normal force (which is equal to the weight of the airplane, mg).

Second technique: F = kx
where k is the spring constant and x is the displacement of the rope.

Third technique: F = ρAv^2
where ρ is the density of water, A is the cross-sectional area of the piston, and v is the velocity of the piston (which is proportional to the velocity of the airplane).

In all three expressions, the force (F) is dependent on either velocity or displacement, but not time.

(2) For the first technique, the maximum acceleration occurs when the force is at its maximum, which is when the airplane is just about to start moving (μs = 0.45). The maximum acceleration can be calculated using Newton's Second Law, F = ma, where F is the maximum force (0.45N) and m is the mass of the airplane (3000kg). Therefore, the maximum acceleration is a = F/m = 0.45N/3000kg = 0.00015 m/s^2.

For the second technique, the maximum acceleration occurs when the force is at its maximum, which is when the airplane reaches its maximum displacement (x). The maximum acceleration can be calculated using the equation a = kx/m, where k is the spring constant and m is the mass of the airplane. To find the maximum displacement, we can use the equation F = kx, where F is the maximum force (6000N). Therefore, x = F/k = 6000N/0.00015 m/s^2 = 40,000 m. Plugging this into the equation for acceleration, we get a = (0.00015 m/s^2)(40,000 m)/3000kg = 0.002 m/s^2.

(3) If the braking system is used, the total horizontal force applied to the airplane is 6000N. This force is acting in the opposite direction of the airplane's motion, so it will cause the airplane to decelerate. The maximum acceleration can be found using the same equation as in the second technique, a = F/m. Therefore,