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.
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?
2 answers
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.
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.
1 answer