Air resistance on a suspended object in motion

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SUMMARY

The discussion focuses on calculating the force of air resistance acting on a 625 kg water bucket suspended from a helicopter flying at a constant speed of 36.6 m/s, with the cable making a 35.3° angle with the vertical. The key equations utilized include F=ma and Fnet=0, indicating that the net force in both the x and y directions equals zero due to the constant speed. Participants emphasize the relationship between tension, weight, and air resistance, suggesting that the horizontal and vertical components of tension can be resolved using trigonometric functions to find the air resistance force.

PREREQUISITES
  • Understanding of Newton's laws of motion
  • Familiarity with trigonometric functions in physics
  • Knowledge of force equilibrium concepts
  • Ability to resolve forces into components
NEXT STEPS
  • Study the application of Newton's second law in two dimensions
  • Learn about vector resolution and force components
  • Explore the concept of equilibrium in static and dynamic systems
  • Investigate the effects of air resistance on moving objects
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Students studying physics, particularly those focusing on mechanics and force analysis, as well as educators seeking to enhance their understanding of force equilibrium in real-world applications.

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Homework Statement


A fire helicopter carries a 625 kg empty water bucket at the end of a cable 21.7 m long. As the aircraft flies back from a fire at a constant speed of 36.6 m/s, the cable makes an angle of 35.3° with respect to the vertical. Calculate the force of air resistance on the bucket.

Homework Equations


F=ma.
Fnet=0


The Attempt at a Solution


I started with the rationalization that constant speed implies 0 acceleration, meaning Fnet=0 in both the x and y direction.
Dividing the forces up into Fnety=T(tension in the cable)-W(mg)
and Fnetx is where I get stuck. If the acceleration is 0, then which force is opposing air resistance?
Should I be calculating the horizontal component of the displacement from the vertical?
Help, please!
 
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You don't need to get into wire tension. The weight (W) acts straight down, and the horiz air resistance force (Fx) acts left. These horiz and vert forces are trig related , using the given angle. The resultant of these 2 forces, W and Fx, must lie in the direction of the cable, from equilibrium consideration.
 
If you want to use tension you can and then just cancel it out by dividing the x forces equation by the y forces equation.

You'll have the upward Ty = F wt and the horizontal Tx = F air resistance
The y component of T is of course Tcos of your vertical angle and the x component, Tsin of the vertical angle.

When you divide T's cancel and you have the same equation as the one you'll get if you use PJ's faster method.
 

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