Circular Motion and Newton's Laws

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SUMMARY

The discussion focuses on the application of Newton's Laws to a model airplane in circular motion, specifically analyzing the forces acting on the airplane. The airplane, with a mass of 7.50 kg and a speed of 35 m/s, is tethered by a 60.0 m control wire. The tension and aerodynamic lift forces are acting at angles of 20 degrees South of East and 20 degrees West of North, respectively. The scalar equations derived are Fx - Tcos20 - Fsin20 = mv^2/r and Fy - Tsin20 + Fcos20 - mg = 0, leading to a calculated tension of 12.8 N.

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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.
 
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You should show some work and post it in the homework help section next time. Now if you'll excuse me, I got to jet.

~Ciao
 
Looks like this was moved to homework help, but rmarkatos had already posted there.

May be best to continue in the other thread.
 
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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