[SOLVED] Resultant Velocity Vectors


by ace214
Tags: resultant, solved, vectors, velocity
ace214
ace214 is offline
#1
Sep17-07, 05:30 PM
P: 29
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.



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.
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ace214
ace214 is offline
#2
Sep17-07, 05:49 PM
P: 29
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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