Calculating Acceleration of a Sailboat Using Wind and Water Forces

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The discussion focuses on calculating the acceleration of a sailboat based on wind and water forces. The force exerted by the wind on the sailboat is proportional to the wind velocity component perpendicular to the sail, with the formula F(sail) = 550V(wind). Participants clarify that the drag force from water opposes the boat's motion, and the total force must be considered using the equation ΣF = ma. The correct approach involves breaking down the wind force into its components and factoring in the drag to find the net force. Ultimately, the calculated initial acceleration of the sailboat is 2.312 m/s².
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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 900 kg sailboat shown in Figure P4.54, the proportionality constant is given below.

F(sail) = 550V(wind)

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 17 knot wind (1 knot = 0.514 m/s) is blowing to the east, what is the initial acceleration of the sailboat?


There is a given diagram with the Force of the keel pointing west and the Force of the sail is pointing northeast with an angle of 30. The force of the sail is 4805.9 N. I was thinking that the force of the keel is negligible because the wind counters its effect. So I use the formula F=ma to solve for the acceleration. a = 4805.9 / 900 = 5.33 m/s^2
However that is inccorect. Any tips?
 
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The force of the wind does not point exlusively along the x-axis; you'll need to break it into its components.

Also, the problem says there's drag pointing in the opposite direction of the wind, so that will reduce acceleration: remember, it's not that F=ma, its that \Sigma F=ma, sigma being the sum of all forces (along a given axis, which gives acceleration along a given axis)
 
Without the figure it's hard to tell, but where is the cos(30) or sin(30) (depending on the diagram)?
 
the angle 30 is between the y-axis of the boat and the sail. the boat is pointing north. that if u draw a horizontal line between the boat, the sail is southeast of the point of origin.
 
I'm guessing this is vital information "to the component of the wind velocity perpendicular to the sail." Since the wind velocity is east, we have to find the wind velocity perpendicular. V = 10.089. Then I plug it into the given formula and got F(sail) = 5549.4 N.

The components are Fy = 2774.7 and Fx = 4805.9. The only problem I have is how do I solve for the force the water is exerting on the boat. Help please.
 
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The wording of the problem is unclear to me. Is the velocity of the wind due east or at that 30 degree angle in the diagram?
Also, I don't suppose the text gives any additional info about drag? (After all, why would you withold part of the question? You don't want to pop out at us yelling, "Surprise!" You want to solve the problem. Shame. Well, hope springs eternal...)
 
I think the drag force is not required because the question is asking for the initial acceleration... but yea that's all the info that is given. Ok, I'll attach the diagram and Force of wind on sail equation. How do I get F(keel) or the force exerted by water. Better yet, how do I solve this problem...
 

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My thinking is this, I fouind the wind velocity perpendicular to the sail. With that velocity, I plugged it into the equation. I obtained the Fsail. Then I separated Fsail into X and Y components. The problem is with Fkeel and whether or not the drag force is required. I hope someone can help me finish this or at least guide me towards the answer, i'll be back soon, class now. Thanks

edit: Ok, I got the answer but I don't understand the steps in acquiring the answer. Ans is 2.312 m/s^2
 
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