Finding the angle of the traveling direction of a sailboat

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Homework Help Overview

The problem involves determining the angle between the direction of a force exerted on a sailboat and due north, given two scenarios of work done by forces acting on the boat. The context is within the subject area of physics, specifically focusing on forces and work in mechanics.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the relationship between work and force, noting that the work done by the forces is equal. There is an exploration of the trigonometric relationships involved in calculating the angle, with some participants attempting to derive the angle using cosine functions. Questions arise regarding the assumptions made about the forces and their directions.

Discussion Status

The discussion is ongoing, with participants sharing their reasoning and calculations. Some guidance has been offered regarding the use of trigonometric functions to find the angle, and there is an acknowledgment of the need to clarify the direction of the angle relative to north. Multiple interpretations of the problem are being explored, and no consensus has been reached yet.

Contextual Notes

There are indications that some participants question the assumptions made in the problem setup, particularly regarding the understanding of how sailboats operate and the implications of the forces involved.

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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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