I really need help on this Q, don;t know what to do with multtiple complex forces:
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?
101.952 N
To solve this problem, we need to consider the forces acting on the plane and find the net force in the horizontal and vertical directions separately.
Let's start with the horizontal direction. The only force acting in the horizontal direction is the air resistance, which is opposing motion. The plane is moving with constant speed, so the net force in the horizontal direction must be zero. Therefore, the forward thrust must be equal to the air resistance:
Thrust = Air resistance = 63700 N
Next, let's consider the vertical direction. The forces acting in the vertical direction are the weight, lift, and the vertical component of the thrust. Since the plane is flying at an angle of 27.0° above the horizontal, the vertical component of the thrust is given by:
Vertical component of Thrust = Thrust * sin(27.0°)
Now, we can find the net force in the vertical direction. It should be zero because the plane is flying with constant velocity:
Net force (vertical) = Lift + Vertical component of Thrust - Weight = 0
Plugging in the given values, we have:
77100 N + Thrust * sin(27.0°) - 86500 N = 0
Now, we can solve this equation for the thrust:
Thrust * sin(27.0°) = 86500 N - 77100 N
Thrust = (86500 N - 77100 N) / sin(27.0°)
Lastly, to find the reduced thrust and lift required after jettisoning fuel, we need to subtract the weight of the fuel from the weight and thrust. The reduced thrust and lift values are:
Reduced Thrust = Thrust - 2550 N
Reduced Lift = Lift - 2550 N
So, by reducing the thrust by (86500 N - 77100 N) / sin(27.0°) and the lift by 2550 N, the plane will continue moving with the same velocity under the influence of the same air resistance.