Calculating Distance Using Vector Methods

[Crush1986]

Homework Statement

You are canoeing on a lake. Starting at your camp on the shore, you travel 240m in the direction 32 degrees south of east to reach a store to purchase supplies. You know the distance because you have located both your camp and the shore on a map of the lake. On the return trip you travel distance B in the direction 48 degrees north of west, distance C in the direction of 62 degrees south of west, and then you are back at your camp. You measure the directions of travel with your compass, but you don't know the distances. Since you are curious to know the total distance you rowed, use vector methods to calculate distances B and C.

Homework Equations

A/Sin(a) = B/sin(b)=C/sin(c)

The Attempt at a Solution

I obtained the correct answer by drawing out the vectors with the angles given. After some reasoning I found that the vectors made a triangle with angles 94, 16, and 70 degrees with a side of 240m. I used the law of sin's and found that B = 255m and C = 70m, I saw that I was correct after looking up the answer. Is this the only way to do this problem? Or did I miss a concept somewhere that had more to do with resolving the X and Y components of the 240m vector and somehow moving on from there?

I thank you for your help.

 

[Doc Al]

Good for you in solving it that way.

The way I suggest is to break each vector into components. You know the resultant displacement must be zero. Use that fact to get two equations, which will allow you to solve for your two unknowns, the distances B and C.

 

[Crush1986]

Good for you in solving it that way.

The way I suggest is to break each vector into components. You know the resultant displacement must be zero. Use that fact to get two equations, which will allow you to solve for your two unknowns, the distances B and C.

This was my initial idea. I'm having trouble coming up with two equations with the same two variables without introducing new variables.

Something like Bx = -Ax - Cx and Bx = |B|sin(theta), is something I keep coming up with (I don't think that will work). I'm jarring my brain trying to have an epiphany here lol.

 

[Doc Al]

You are on the right track. Do this. Express the x and y components of each of the three vectors. Be careful with signs.

 

[Crush1986]

You are on the right track. Do this. Express the x and y components of each of the three vectors. Be careful with signs.

Ok, yeah I think I got it.

Eq. 1 Ax = -|B|Cos138 - |C|Sin242

Eq. 2 Ay = -|B|Sin138 - |C|Cos242

Solve for Magnitude of C |C| = -((Ax + |B|Sin138)/Cos242))

Plug that into Eq. 1 for |C| and solve for |B|.

I think that is all correct. I didn't get the right answer though. I may just need a walk and a cookie though to recharge my batteries. If there is an issue with my reasoning and math though I'd LOVE to hear it.

 

[Doc Al]

Where did you get the 138 degrees? Also, what do Ax and Ay equal?

(But right idea.)

 

[Crush1986]

Where did you get the 138 degrees? Also, what do Ax and Ay equal?

(But right idea.)

I tried just using the angles that were made with the axis' I had defined. Ax = 203.5 and Ay= -127.18

I'm not getting the correct answer though... I tried it again with the angles under 90 degrees and appropriate signs.

At the end I thought I had it. My final equation for |B| was...

|B| = (Ay*cos(28)-Ax*sin(28))/(-sin(48)*cos(28)-cos(48)*sin(28))

My approach this newest time was only slightly different. I solved for |C|from the Ax = |B|*cos(48) + |C|*cos(28) and plugged it into the equation Ay = -|B|sin(48) + |C|sin(28) (both of the components of the right side I found were positive because |B| and |C| both have negative x components. I solved for |B| and.. wrong answer. I must have maybe some angles off... or something. I need to go to sleep now lol been up for like 30 hours... When I wake up I think i'll work on this more. Hopefully knock it out as soon as I look at it.

 

[Doc Al]

I tried just using the angles that were made with the axis' I had defined. Ax = 203.5 and Ay= -127.18

OK.

I'm not getting the correct answer though... I tried it again with the angles under 90 degrees and appropriate signs.

At the end I thought I had it. My final equation for |B| was...

|B| = (Ay*cos(28)-Ax*sin(28))/(-sin(48)*cos(28)-cos(48)*sin(28))

Why are you using 28 degrees?

I used this method and was able to obtain the same answers you gave in your first post.

Get some sleep!

 

[Crush1986]

OK.


Why are you using 28 degrees?

I used this method and was able to obtain the same answers you gave in your first post.

Get some sleep!

Ok! Woke up and got it super quick. I looked at last nights prior work and I noticed that one of my Ax's turned into an Ay by mistake... Yikes.

Thank you for your help! I'm disappointed in myself for thinking of that route to solve this problem but failing to see that it was in fact viable.

Lesson learned... Get more sleep, take maybe a few more breaks lol. I got up last night and both legs were pretty weak haha.