Homework Help is Here – Start Your Trial Now!
Math
Advanced Math
Given the following information, calculate the Thrust vector in a straight and steady power-on descent with an angle of 30°. Weight: 11350 kg Lift: 10320 kg Drag: 11010 kg A 5335 kg 1181 kg C 11010 kg D 4645 kg Thrust Refer to figure. The balance of forces in steady climb show thrust is partly acting upwards and weight is partly adding to the drag. Because the thrust is now assisting lift, there is less lift required than in level flight. For steady speed, the thrust and two effects of aerodynamic drag and weight must be equal. T= Thrust W= Weight D= Drag T= DW sin(y) This means that the climb angle (sin y) depends on excess thrust and weight. In descent the formula will become T = D - W sin(y) The load factor is the relationship between lift and weight. As the lift is now decreased and the load factor in straight and level flight is 1, the load factor will now be less than 1. Solving this excercise: = Thrust drag- (W * sin(angle)) Thrust 11010- (11350 * sin (30)) Thrust 5335

Given the following information, calculate the Thrust vector in a straight and steady power-on descent with an angle of 30°. Weight: 11350 kg Lift: 10320 kg Drag: 11010 kg A 5335 kg 1181 kg C 11010 kg D 4645 kg Thrust Refer to figure. The balance of forces in steady climb show thrust is partly acting upwards and weight is partly adding to the drag. Because the thrust is now assisting lift, there is less lift required than in level flight. For steady speed, the thrust and two effects of aerodynamic drag and weight must be equal. T= Thrust W= Weight D= Drag T= DW sin(y) This means that the climb angle (sin y) depends on excess thrust and weight. In descent the formula will become T = D - W sin(y) The load factor is the relationship between lift and weight. As the lift is now decreased and the load factor in straight and level flight is 1, the load factor will now be less than 1. Solving this excercise: = Thrust drag- (W * sin(angle)) Thrust 11010- (11350 * sin (30)) Thrust 5335

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
See similar textbooks
icon
Related questions
Question

How do I solve this question I don't know what steps I need to take to get 5335 as a answer I don't understand this formula 

Given the following information, calculate the Thrust vector in a straight and steady power-on descent with an angle of 30°. Weight: 11350 kg Lift: 10320 kg Drag: 11010 kg A 5335 kg 1181 kg C 11010 kg D 4645 kg
Transcribed Image Text:Given the following information, calculate the Thrust vector in a straight and steady power-on descent with an angle of 30°. Weight: 11350 kg Lift: 10320 kg Drag: 11010 kg A 5335 kg 1181 kg C 11010 kg D 4645 kg
Thrust Refer to figure. The balance of forces in steady climb show thrust is partly acting upwards and weight is partly adding to the drag. Because the thrust is now assisting lift, there is less lift required than in level flight. For steady speed, the thrust and two effects of aerodynamic drag and weight must be equal. T= Thrust W= Weight D= Drag T= DW sin(y) This means that the climb angle (sin y) depends on excess thrust and weight. In descent the formula will become T = D - W sin(y) The load factor is the relationship between lift and weight. As the lift is now decreased and the load factor in straight and level flight is 1, the load factor will now be less than 1. Solving this excercise: = Thrust drag- (W * sin(angle)) Thrust 11010- (11350 * sin (30)) Thrust 5335
Transcribed Image Text:Thrust Refer to figure. The balance of forces in steady climb show thrust is partly acting upwards and weight is partly adding to the drag. Because the thrust is now assisting lift, there is less lift required than in level flight. For steady speed, the thrust and two effects of aerodynamic drag and weight must be equal. T= Thrust W= Weight D= Drag T= DW sin(y) This means that the climb angle (sin y) depends on excess thrust and weight. In descent the formula will become T = D - W sin(y) The load factor is the relationship between lift and weight. As the lift is now decreased and the load factor in straight and level flight is 1, the load factor will now be less than 1. Solving this excercise: = Thrust drag- (W * sin(angle)) Thrust 11010- (11350 * sin (30)) Thrust 5335
Expert Solution
steps

Step by step

Solved in 2 steps with 1 images

SEE SOLUTIONCheck out a sample Q&A here
Blurred answer
Recommended textbooks for you
Advanced Engineering Mathematics
Advanced Engineering Mathematics
Advanced Math
ISBN:
9780470458365
Author:
Erwin Kreyszig
Publisher:
Wiley, John & Sons, Incorporated
Numerical Methods for Engineers
Numerical Methods for Engineers
Advanced Math
ISBN:
9780073397924
Author:
Steven C. Chapra Dr., Raymond P. Canale
Publisher:
McGraw-Hill Education
Introductory Mathematics for Engineering Applicat…
Introductory Mathematics for Engineering Applicat…
Advanced Math
ISBN:
9781118141809
Author:
Nathan Klingbeil
Publisher:
WILEY
Mathematics For Machine Technology
Mathematics For Machine Technology
Advanced Math
ISBN:
9781337798310
Author:
Peterson, John.
Publisher:
Cengage Learning,
Basic Technical Mathematics
Basic Technical Mathematics
Advanced Math
ISBN:
9780134437705
Author:
Washington
Publisher:
PEARSON
Topology
Topology
Advanced Math
ISBN:
9780134689517
Author:
Munkres, James R.
Publisher:
Pearson,
SEE MORE TEXTBOOKS