Vectors. Calculating Weight (N) of Plane with force and angle of take off

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SUMMARY

The discussion focuses on calculating the weight of a plane during takeoff, where an upward force of 8445 N acts at an angle of 50.2° above the horizontal. Given that the acceleration due to gravity is 9.8 m/s², the problem requires determining the weight of the plane in Newtons. The vertical component of the force remains constant, indicating that the vertical acceleration is zero, which is crucial for solving the problem. The net force equations and vector components are essential for deriving the weight of the plane.

PREREQUISITES
  • Understanding of Newton's laws of motion
  • Basic knowledge of vector decomposition
  • Familiarity with force calculations in physics
  • Concept of constant velocity and its implications
NEXT STEPS
  • Study vector decomposition techniques in physics
  • Learn how to apply Newton's second law (F=ma) in various scenarios
  • Explore the concept of forces acting at angles and their components
  • Investigate the relationship between weight and gravitational force
USEFUL FOR

Students studying physics, particularly those learning about forces, vectors, and motion dynamics during takeoff scenarios in aerodynamics.

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Homework Statement


On takeoff” refers to after the plane has left the ground.
On takeoff, the combined action of the air around the engines and wings of an airplane exerts a(n) 8445 N force on the plane, directed upward at an angle of 50.2◦ above the hori- zontal. The plane rises with constant velocity in the vertical direction while continuing to accelerate in the horizontal direction.
The acceleration of gravity is 9.8 m/s2 . What is the weight of the plane? Answer in units of N.

8445 N on take off. 50.2 degrees on take off. Fg=9.8


Homework Equations


Fnet=ma Fvector 1+Fvecotr 2=Fvector 1 +Fvector 2

The Attempt at a Solution



To get to vertical, the plane would have to travel 39.8 degrees higher (90). I do not know what to do from here.
 
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Hey, I'm kind of new to physics, but it seems to me that the fact that the vertical component of velocity is constant tells you something about the acceleration in the vertical direction. I think that should help.
 

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