Distance and direction of one leg of a sailboat sail.

[cdornz]

Homework Statement

A sailboat sails 2km east, then 5km at 40° west of north, then sails off in a third direction. After the third leg of the trip, the sailboat returns to its starting point by sailing 15 km at 60° west of north. How far and in what direction did the sailboat sail during the third leg of its trip?

Leg 1: 2km to the right (east) 0°
Leg 2: 5km up and to the left (northwest) 40°
Leg 3: unknown
Leg 4: 15km up and to the left (northwest) 60°

Since I was told a direction, I assumed the kilometers would be the force in these equations. (This could be wrong, but I didn't know what else to try.)

Homework Equations

I thought I could use $\Sigma$F=0
F₁ + F₂ +F₃ + F₄ = 0
F₃ = -F₁ - F₂ - F₄

then I could use pythagoreans theorem and arc tangent to solve for magnitude and direction.

The Attempt at a Solution

F₁ + F₂ +F₃ + F₄ = 0
F_3x = -2cos0 - 5cos40 - 15cos60
F_3x = -13.33


F₁ + F₂ +F₃ + F₄ = 0
F_3y = -2sin0 - 5sin40 - 15sin60
F_3y = 9.77

From here you would draw two arrows on a set of coordinate axis then use F_3x and F_3y to find F₃. But the answer I come out to isn't equalling the answer in the book. I just have no idea what I'm doing wrong or how to go about this question correctly. Any help would be very much appreciated.

 

[haruspex]

Since I was told a direction, I assumed the kilometers would be the force

It isn't about forces, but it is about vectors, so although it's quite wrong to say these are forces it should give the right answer.

F_3x = -2cos0 - 5cos40 - 15cos60

It would help if you were to state your choice of co-ordinate system. I.e. is the +x direction N, S, E or W? Whichever, I think your equation above takes an inconsistent view.

 

[cdornz]

Ok, so a vector has both a magnitude and direction. I'm given direction of each of the vectors. In the book I'm working with states that an equation I could use is:

Vector_x = (vector magnitude)cos$\Theta$
Vector_y = (vector magnitude)sin$\Theta$

This leads me to the idea (if i set up the axis like the compass, i.e. north is +y, east is +x) that I can do the distance(cos$\Theta$) So I would count the angle from the +x axis to the vector:

V₁ 2cos0 = 2
V₂ 5cos130 = -3.21
V₄ 15cos300 = 7.5

So V₁ + V₂ + V₄ = V₃ ?

But when I follow this, I don't get the correct distance.

 

[haruspex]

V₁ 2cos0 = 2
V₂ 5cos130 = -3.21
V₄ 15cos300 = 7.5
So V₁ + V₂ + V₄ = V₃ ?

A few problems there.
First, I don't think the 300 is right - check that.
Second, need to be careful with the signs. It's probably less confusing to write V₁ + V₂ + V₃ + V₄ = 0 and use a different angle in the V₄ equation.
Third, what is the equation for V₃? There is an angle involved again.
Fourth, you will need an equation for the NS direction too.

 

[cdornz]

For the 300, it is listed that the boat went 15km in the SE direction..if it didn't then the 3rd direction wouldn't be correct. So since it was listed at 60°, counting from +x to that point would be 300, unless I'm missing some way in figuring out that angle.

So if I have V₁ + V₂ + V₃ + V₄ = 0, in what way am I supposed to calculate the equation for V₃ if I'm not given any information about it? Unless I am assuming it is 30°; 90° (SE corner) - 60° (listed from leg 4).

How would I figure out an equation for the NS direction, if nothing is specifically given on this? I feel like I am just missing a piece of information.

 

[haruspex]

For the 300, it is listed that the boat went 15km in the SE direction

No, it says 60^o W of N. From the x+ axis that's 150^o.

So if I have V₁ + V₂ + V₃ + V₄ = 0, in what way am I supposed to calculate the equation for V₃ if I'm not given any information about it?

By rearranging as V₃ = -(V₁ + V₂ + V₄)

How would I figure out an equation for the NS direction, if nothing is specifically given on this?

The above is a vector equation, so you can use it to produce two scalar equations, one NS and one EW, with sines and cosines.

 

[cdornz]

No, it says 60^o W of N. From the x+ axis that's 150^o.

Ah, of course, I was consider the directions specifically based on a compass, thank you for that!

By rearranging as V₃ = -(V₁ + V₂ + V₄)

You're right, this equation was much simpler to deal with.

So after finding the components of V₁, V₂ and V₄ I ended with:

V_3x = 14.203
V_3y = -11.33

I then redrew this out on the coordinate plane, used the Pythagorean theorem then arc tangent for the angle. It came out to the correct answer. Thank you soo much for your help on this problem!