Finding the angle of the traveling direction of a sailboat

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SUMMARY

The discussion focuses on calculating the angle between the direction of force F1, acting on a sailboat, and due north. The sailboat travels 55 m north while a breeze exerts force F1 at an angle west of north, equivalent in magnitude to force F2, which would do the same work over a distance of 42 m. The work equations W1 = F1(cosθ) * 55 m and W2 = F2 * 42 m lead to the conclusion that cosθ = 0.76, resulting in an angle θ of approximately 40.54° west of north. The solution emphasizes the importance of using trigonometric relationships to solve physics problems involving forces.

PREREQUISITES
  • Understanding of basic trigonometry, specifically sine and cosine functions.
  • Familiarity with the concept of work in physics, defined as W = Fd.
  • Knowledge of vector components and how to resolve forces into their respective directions.
  • Ability to interpret and analyze physics problems involving forces and motion.
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  • Study the principles of vector resolution in physics to better understand force components.
  • Learn about the work-energy theorem and its applications in mechanics.
  • Explore trigonometric identities and their use in solving real-world physics problems.
  • Investigate the mechanics of sailing and how forces interact with sailboats in motion.
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Students studying physics, particularly those focusing on mechanics, as well as educators and anyone interested in understanding the dynamics of sailing and force interactions.

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Homework Statement


As a sailboat sails 55 m due north, a breeze exerts a constant force F1 on the boat's sails. This force is directed at an angle west of due north. A force F2 of the same magnitude directed due north would do the same amount of work on the sailboat over a distance of just 42 m. What is the angle between the direction of F1 and due north?


Homework Equations


Don't know if this pertains to adding the forces together.


The Attempt at a Solution


I assume that trigonometry is involved here to solve this problem. Other than that, that's how far I have gotten into the problem.
 
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Well W=Fd, where the force F is in the same direction as the distance d.

You know that F1=F2 from the problem.
And W1=W2

And, from the second case the work W2=F2 x 42m

But as the sailboat is traveling due north, and a breeze is acting on the sails with a force F1 at a certain angle, west of due north. Here, the work W1 on the sailboat would be the component of F1 in the north direction and the distance 55m.
 
256bits said:
Well W=Fd, where the force F is in the same direction as the distance d.

You know that F1=F2 from the problem.
And W1=W2

And, from the second case the work W2=F2 x 42m

But as the sailboat is traveling due north, and a breeze is acting on the sails with a force F1 at a certain angle, west of due north. Here, the work W1 on the sailboat would be the component of F1 in the north direction and the distance 55m.

So with this, W=F1(cosθ)d

And plugging in the equations,
F1(cos\Theta)55m = F2 X 42m (being that cos0°=1)

divide F on both sides,
cosθ55m = 42m

divide 55m with 42m, and I assume that in the end,
cosθ = .7(63 repeating)°
cosθ = .76°

Am I close?
 
divide 55m with 42m, and I assume that in the end,
cosθ = .7(63 repeating)°
cosθ = .76°

Am I close?

0.76° --> the little dot goes on the angle but not on the cos of the angle. Sine, cosine, tangent come from the ratios of the sides of a traingle and so dimensionless.

Close,
What angle θ will give cosθ=0.76.
They ask for the angle, and, I presume, you have to state whether or not it is east of north or west of north angle.
If you draw a diagram of the sailboat and forces you should be able to figure that out visually.

Hint;
to check if your answer is correct or not, put the values back into the equation.
ie W2 = F2 x 42

W1 = F1 cosθ x 55 = F1 0.76 x 55 = F1 x 42

Since F1 = F2, then W1 = W2 and correct.

You could do that for all problems you encounter as a check.
 
I get the impression that the author of this problem doesn't understand how sailboats work.
 

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