A jet plane is flying with a constant speed along a straight line, at an angle of 26.5° 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 77412 N and the air resistance is 64404 N. Suppose that the pilot suddenly jettisons 2850 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 each of the following?

(a) the thrust
N

(b) the lift

To find out how much the pilot needs to reduce the thrust and lift in order for the plane to continue moving with the same velocity under the influence of the same air resistance after jettisoning 2850 N of fuel, we need to analyze the vertical and horizontal forces acting on the plane.

Let's break down the given information:
Weight of the plane (W) = 86500 N
Thrust of the engines (T) = 103000 N
Lift force (L) = 77412 N
Air resistance (R) = 64404 N
Change in fuel weight (F) = 2850 N

We are given that the plane is flying at an angle of 26.5° above the horizontal. Based on this information, we can divide the forces into vertical and horizontal components.

Vertical forces:
The weight of the plane acts vertically downwards.
W = 86500 N

The lift force acts perpendicular to the wings, which can be split into vertical and horizontal components.
Vertical component of lift force:
L_vertical = L * cos(26.5°)

Horizontal forces:
The thrust of the engines and air resistance act horizontally.
Thrust component:
T_horizontal = T * cos(26.5°)

Air resistance component:
R_horizontal = R

Now, let's calculate the initial total vertical force (F_v_initial) and the total horizontal force (F_h_initial) without considering the change in fuel weight:

Initial vertical force:
F_v_initial = W + L_vertical

Initial horizontal force:
F_h_initial = T_horizontal - R_horizontal

After jettisoning the fuel, the total weight of the plane reduces:
New weight of the plane:
W_new = W - F

Now, we can calculate the new vertical and horizontal forces (F_v_new and F_h_new) using the new weight of the plane:

New vertical force:
F_v_new = W_new + L_vertical

New horizontal force:
F_h_new = T_horizontal - R_horizontal

Since the plane needs to continue moving with the same velocity, the net vertical and horizontal forces should remain the same after jettisoning the fuel.

Setting up the equation for vertical forces:
F_v_initial = F_v_new

W + L_vertical = W_new + L_vertical

Since L_vertical appears on both sides, it cancels out. We can simplify the equation:

W = W_new

Substituting the values, we have:

86500 N = W_new

Therefore, the weight of the plane remains the same after jettisoning the fuel.

Setting up the equation for horizontal forces:
F_h_initial = F_h_new

T_horizontal - R_horizontal = T_horizontal - R

Since T_horizontal appears on both sides, we can simplify the equation:

-R_horizontal = -R

Since -R is a constant, we can conclude that the air resistance does not change after jettisoning the fuel.

Now, let's answer the given questions:

(a) To find out how much the pilot needs to reduce the thrust (T), we can calculate the difference between the initial thrust and the horizontal component of thrust without considering the change in fuel weight:

Thrust reduction:
T_reduction = T_horizontal_initial - T_horizontal_new

Since the horizontal component of thrust remains the same:
T_reduction = T_horizontal - T_horizontal

Therefore, the pilot does not need to reduce the thrust.

(b) Since the lift force does not change after jettisoning the fuel, the pilot does not need to reduce the lift. Therefore, the reduction in lift force is 0 N.

To summarize:
(a) The pilot does not need to reduce the thrust.
(b) There is no reduction in lift force.