An airliner makes an emergency landing with its nose wheel locked in a position perpendicular to its normal rolling position. The forces acting to stop the airliner arise from friction due to the wheels and from the breaking effort of the engines in reverse thrust mode. The force of the engine on the plane is constant, Fengine = -F0. The sum of the horizontal forces on the airliner can be written as F(t)=-F0+(t/ts-1)*F1, from touchdown at time t=0s to the final stop time at ts=28s the mass of the plane is M=80tonnes, F0=254kN, F1=40kN Neglect all air drag and friction forces, except the one stated in the problem.
a) Find the speed of the plain at touchdown
b) horizontal acceleration at time a(ts) a(0)
c) what distance s does the plane go between touchdown and the final stop time ts
d) What work do the engines during the reverese thrust mode during the landing
e) How much heat energy is absorbed by the wheels during the emergency landing?
a) To find the speed of the plane at touchdown, we can use the equation of motion:
F(t) = Ma(t)
At touchdown, the forces acting on the plane are only due to the engine in reverse thrust mode, so we have:
-F0 = M * a(0)
Solving for a(0), we find:
a(0) = -F0 / M
Now, we know that the initial velocity (v(0)) is 0, so we can use the equation of motion:
v(t) = v(0) + a(0) * t
v(ts) = 0 + a(0) * ts
Substituting the values, we get:
v(ts) = -(F0 / M) * ts
So, the speed of the plane at touchdown is -(F0 / M) * ts.
b) At time t = ts, the plane comes to a stop. So, the final velocity (v(ts)) is 0. We can use the equation of motion to find the acceleration:
v(ts) = v(0) + a(0) * ts
0 = 0 + a(0) * ts
Solving for a(ts), we get:
a(ts) = -F0 / M
The horizontal acceleration at time t = ts is -F0 / M.
At time t = 0, the plane has an acceleration a(0) = -F0 / M as we found in part (a).
c) To calculate the distance traveled by the plane between touchdown and the final stop time ts, we can use the equation of motion:
s(t) = v(0) * t + (1/2) * a(t) * t^2
At touchdown, the initial velocity (v(0)) is 0, so we have:
s(0) = (1/2) * a(0) * 0^2
s(0) = 0
At time t = ts, the final distance (s(ts)) is the total distance traveled by the plane. So, we have:
s(ts) = v(0) * ts + (1/2) * a(0) * ts^2
Substituting the values, we get:
s(ts) = 0 + (1/2) * (-F0 / M) * ts^2
So, the plane travels a distance of (1/2) * (-F0 / M) * ts^2 between touchdown and the final stop time ts.
d) The work done by the engines during reverse thrust mode is equal to the force applied by the engines multiplied by the distance over which the force is applied. So we can calculate it using the equation:
work = force * distance
The force applied by the engines is constant and equal to -F0. The distance over which the force is applied is the same as the distance traveled by the plane between touchdown and the final stop time ts, which we calculated in part (c). So we have:
work = -F0 * s(ts)
Substituting the values, we get:
work = -F0 * [(1/2) * (-F0 / M) * ts^2]
e) The heat energy absorbed by the wheels can be found using the work-energy principle. The work done on an object is equal to the change in its kinetic energy. Since the plane comes to a stop at ts, the change in kinetic energy is equal to its initial kinetic energy. The initial kinetic energy is given by:
KE(0) = (1/2) * M * v(0)^2
Substituting v(0) = -(F0 / M) * ts from part (a), we get:
KE(0) = (1/2) * M * (-(F0 / M) * ts)^2
= (1/2) * M * (F0^2 / M^2) * ts^2
= (1/2) * F0^2 / M * ts^2
So, the heat energy absorbed by the wheels is (1/2) * F0^2 / M * ts^2.
To solve this problem, we need to use the equations of motion and the concept of work and energy. Let's go through each part of the problem step by step.
a) Find the speed of the plane at touchdown:
The speed of the plane at touchdown can be found by using the equation of motion:
v = u + at
where:
v = final velocity (which is zero since the plane stops),
u = initial velocity (unknown),
a = acceleration (which we will find later),
t = time (which is the time of touchdown, which is 0s in this case).
Since the final velocity is zero, we can rewrite the equation as:
0 = u + a(0)
Simplifying the equation, we get:
u = 0
Therefore, the speed of the plane at touchdown is 0.
b) Horizontal acceleration at time a(ts), a(0):
To find the horizontal acceleration at time a(ts) and a(0), we need to differentiate the equation of motion with respect to time, t.
a = dv/dt
Differentiating the equation F(t) = -F0 + (t/ts - 1)*F1 with respect to time, we get:
a = (dF(t)/dt) = d/dt (-F0 + (t/ts - 1)*F1)
Since F0 and F1 are constants, their derivatives are zero, and we're left with:
a = d/dt (t/ts)*F1
Simplifying further:
a = F1/ts
Therefore, the horizontal acceleration at time a(ts) is F1/ts, and at time a(0) is also F1/ts.
c) What distance s does the plane go between touchdown and the final stop time ts:
To find the distance traveled by the plane between touchdown and the final stop time ts, we can use the equation of motion:
s = ut + (1/2)at^2
Substituting u = 0 and a = F1/ts, and t = ts, we get:
s = 0 + (1/2) * (F1/ts) * (ts)^2
Simplifying the equation:
s = (1/2) * F1 * ts
Therefore, the distance traveled by the plane between touchdown and the final stop time ts is (1/2) * F1 * ts.
d) What work do the engines during the reverse thrust mode during the landing:
To find the work done by the engines during the reverse thrust mode, we need to calculate the force exerted by the engines and multiply it by the distance traveled. The force exerted by the engines is equal to -F0 (since it's constant and in the opposite direction of motion).
The work done is given by the equation:
Work = Force * Distance
Substituting the values, we get:
Work = -F0 * s
Substituting the value of s from part c), we get:
Work = -F0 * (1/2) * F1 * ts
Therefore, the work done by the engines during the reverse thrust mode is -F0 * (1/2) * F1 * ts.
e) How much heat energy is absorbed by the wheels during the emergency landing:
To find the heat energy absorbed by the wheels, we need to consider the work done against friction. The force of friction can be calculated using the equation of motion:
Frictional Force = Mass * Acceleration
Substituting the given values, We have:
Mass = M = 80 tonnes = 80,000 kg
Acceleration = F1/ts (as calculated in part b)
Frictional Force = 80,000 kg * F1/ts
The work done against the frictional force is given by:
Work = Frictional Force * Distance
Substituting the value of the frictional force and the distance traveled (as calculated in parts c), we get:
Work = (80,000 kg * F1/ts) * (1/2) * F1 * ts
Simplifying further:
Work = (40,000 kg * F1^2/ts)
Therefore, the heat energy absorbed by the wheels during the emergency landing is (40,000 kg * F1^2/ts).