How Do Resultant Velocity Vectors Determine Sailboat Movement?

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SUMMARY

The discussion focuses on determining the resultant velocity vectors affecting a sailboat's movement in wind conditions. The sailboat, weighing 700 kg, experiences a wind force defined by the equation Fsail = (550 N/(m/s))Vwind, with a 20-knot wind blowing east and the boat moving north at 23 knots. The correct interpretation of the wind's direction reveals that it is indeed blowing east, making the resultant wind velocity vector southeast relative to the boat. The solution involves using the Pythagorean theorem to find the magnitude of the resultant velocity and inverse tangent to calculate the angle south of east.

PREREQUISITES
  • Understanding of vector components and their graphical representation
  • Familiarity with the Pythagorean theorem for calculating resultant vectors
  • Knowledge of basic trigonometry, specifically inverse tangent
  • Concept of relative motion in physics, particularly in sailing dynamics
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  • Study vector decomposition in physics to better understand resultant forces
  • Learn about the principles of sailing dynamics and how wind affects sailboats
  • Explore advanced trigonometric applications in physics problems
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Physics students, sailing enthusiasts, and anyone interested in the dynamics of wind and water interactions affecting boat movement.

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[SOLVED] Resultant Velocity Vectors

Here is the problem:
The force exerted by the wind on a sailboat is approximately perpendicular to the sail and proportional to the component of the wind velocity perpendicular to the sail. For the 700 kg sailboat shown in Figure P4.54, the proportionality constant is given below.

Fsail = (550 N/(m/s))Vwind

Water exerts a force along the keel (bottom) of the boat that prevents it from moving sideways, as shown in the figure. Once the boat starts moving forward, water also exerts a drag force backwards on the boat, opposing the forward motion. If a 20 knot wind (1 knot = 0.514 m/s) is blowing to the east, and the sailboat is heading directly north at a speed of 23 knots, determine the magnitude and directions of the wind velocity as measured on the boat.

p4-54.gif


Answer boxes:
_____ knots at _____ degrees south of east

What is the component of the wind velocity in the direction parallel to the motion of the boat?
______ knots south

----------------

OK, I figured out the components of the wind vector to be 17 on the x and 10 on the y (Make a right triangle b/c of the perpendicular vector to the sail, other angle becomes 60 which is complementary to 30 in another right triangle that actually makes up the wind vector). so I tried putting in 10 for the second part of the question and it was wrong. I don't see at all how the anything in the wind velocity is south- it's going northeast.

I don't see how the velocity goes south of east either. It already gave you the velocity.

Also, the Force stuff doesn't seem to be useful at all.

Very confused. Thanks for any help. BTW, this is due by 8:30 Wednesday morning.

ADD: Ok, so I think I misinterpreted the wind: I guess it is going due east instead of into the sail- that is just the force. I will keep trying.
 
Last edited:
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Ok, I was making the question way to complicated. In case this problem comes up in the future in a search or something I will be a good poster now: :-)

The wind is going east and since the boat is moving north, the wind is going in a southeast direction relative to the boat. So make a right triangle so that your hypotenuse "points" southeast with the wind velocity on the x and the boat velocity on the y. Solve for the hypotenuse with Pythagorean theorem to get the magnitude of the velocity and then use inverse tangent to get the angle south of east.

Then your component of velocity that is parallel to the boat is the boat speed, the vector you put on the y-axis of your right triangle.

Sorry for the needless post. Had to talk myself through it.
 

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