A jet plane is flying with a constant speed along a straight line, at an angle of 27.0° above the horizontal, as Figure 4.30a indicates. The plane has weight whose magnitude is 86500 N, and its engines provide a forward thrust of 103000 N. In addition, the lift force (directed perpendicular to the wings) is 77100 N and the air resistance is 63700 N. Suppose that the pilot suddenly jettisons 2550 N of fuel. If the plane is to continue moving with the same velocity under the influence of the same air resistance , by how much does the pilot have to reduce the thrust and the lift?
well, at constant speed, thrust= friction and retarding forces. I do not see any connection with retarding forces and mass.
Now for lift, break it into a vertical component and a horizontal component. So the vertical is 77100N*sin27=3.5E4N
now look at the vertical component of thrust: it has to provide an upward force to keep the plane stable, so 86500-3.5E4= thrust upward=5.1E4N
if the pilot jettisons 2.55E4 fuel, then the new upward component is 2.60E4N of thrust, and at the same flight angle, thrust = 2.60E4N/sin27=5.7E4
So, 86500-change-3.5E4=5.7E4 or change is 5500N of thrust. check my calcs.
To maintain the same speed AND climb angle, the net vertical force and the net horizontal force must both remain zero.
The net change in vertical lift and drag components must balance the weight loss.
The net change in horizontal lift and drag components must be zero
T2*sin27 + L2*cos27 - T1*sin27 + L1*cos27 = -2550
T2*cos27- L2*sin27 = T1*cos27 - L2*27
The only unknowns are L2 and T2. Solve for them.
You don't need to know the air resistance. It is redundant information.
To find out by how much the pilot needs to reduce the thrust and the lift, we first need to analyze the forces acting on the plane and determine their components in the vertical and horizontal directions.
Let's break down the forces acting on the plane:
1. Weight (W): The weight of the plane acts vertically downwards. Its magnitude is given as 86500 N.
2. Thrust (T): The forward thrust provided by the engines helps propel the plane forward. Its magnitude is given as 103000 N.
3. Lift (L): The lift force acts perpendicular to the wings of the plane and helps counteract gravity. Its magnitude is given as 77100 N.
4. Air Resistance (R): The air resistance opposes the motion of the plane and acts in the opposite direction to its velocity. Its magnitude is given as 63700 N.
The angle of the plane with respect to the horizontal is 27.0°.
To maintain the same velocity after jettisoning fuel, the net force in the horizontal direction must remain zero. This means that the thrust force should be equal to the air resistance force.
T = R
To find out the new value for thrust, we need to subtract the force jettisoned from the original thrust:
New Thrust = Old Thrust - Jettisoned Fuel Force
= 103000 N - 2550 N
= 100450 N
Now, let's calculate the lift force required to maintain the same velocity. Since the plane is at an angle with the horizontal, we need to consider the vertical component of the lift to counteract the weight of the plane.
Vertical Component of Lift (L_v) = L * cos(angle)
L_v = 77100 N * cos(27.0°)
L_v ≈ 69476.66 N
To find out the new value for lift, we need to subtract the component due to the jettisoned fuel force:
New Lift = Old Lift - Jettisoned Fuel Vertical Component
= 69476.66 N - 2550 N
= 66926.66 N
Therefore, the pilot needs to reduce the thrust by approximately 100450 N and the lift by approximately 66926.66 N to maintain the same velocity under the influence of the same air resistance after jettisoning 2550 N of fuel.