Circular Motion with Newton's Laws

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SUMMARY

The discussion focuses on analyzing the forces acting on a model airplane of mass 7.50 kg flying in a horizontal circle with a speed of 35 m/s, tethered by a 60.0 m control wire. The aerodynamic lift acts at an angle of 20 degrees West of North, while the tension in the wire acts at 20 degrees South of East. The scalar equations for the x and y components are established as Fx = -Tcos20 - Fsin20 = mv^2/r and Fy = -Tsin20 + Fcos20 - mg = 0, with the solution for tension being 12.8 N. The discussion emphasizes the importance of correctly positioning the object in force diagrams to solve such problems effectively.

PREREQUISITES
  • Understanding of Newton's Laws of Motion
  • Knowledge of circular motion dynamics
  • Familiarity with vector decomposition
  • Basic trigonometry for angle calculations
NEXT STEPS
  • Study the application of Newton's Laws in circular motion scenarios
  • Learn about vector decomposition techniques in physics
  • Explore the concept of centripetal force and its calculations
  • Investigate the effects of aerodynamic lift on moving objects
USEFUL FOR

This discussion is beneficial for physics students, educators, and anyone interested in understanding the dynamics of circular motion and the application of Newton's Laws in real-world scenarios.

rmarkatos
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A model airplane of mass 7.50 with a speed 35m/s flies in a horizontal circle at the end of 60.0m control wire. Aerodynamic lift acts on the plane at an angle of 20 degrees West of North.

In the picture the book has the plane on the right. The weight is acting straight down, the tension is acting at angle of 20 degrees South of East and the aerodynamic lift is acting 20 degrees west of north.

Can someone set up the x and y scalar equations please. The answer is 12.8N. I have set it up 5 different ways and i can't seem to get the right answer. My teacher says always put the object on the left when doing these types of problems but it doesn't seem to work

Fx -Tcos20 - Fsin20 = mv^2/r
Fy -Tsin20 + Fcos20 - mg = 0 where T is the tension and F is the aerodynamic lift

Those are the equations based on the picture described.
 
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Fx -Tcos20 - Fsin20 = mv^2/r
Fy -Tsin20 + Fcos20 - mg = 0 where T is the tension and F is the aerodynamic lift

Those are the equations based on the picture described.
 

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