Vector Magnitudes given only direction-help

  • Thread starter Thread starter jcwhitts
  • Start date Start date
  • Tags Tags
    Magnitudes Vector
AI Thread Summary
To solve for the distances B and C, the x and y components of the vectors must sum to zero since the journey returns to the starting point. The equations can be set up using the cosine and sine of the angles for each vector. Specifically, the x-component equation is 230*cos(31) - B*cos(43) - C*cos(60) = 0, while a similar equation for the y-components can be developed. The user initially struggled with the law of sines and manipulating angles but realized the importance of setting up a system of equations instead. This approach clarified the problem, leading to a more straightforward solution.
jcwhitts
Messages
9
Reaction score
0
Vector Magnitudes given only direction--help!

Homework Statement


You leave your campsite to go get supplies at a store. You travel 230M at an angle 31 degrees south of east to get there(I'll call this vector A for simplicity's sake). On the way back to camp, you go a distance B at 43 degrees north of west and a distance C at 60 degrees south of west. Find the distances B and C.



Homework Equations


Not sure. I've tried the law of sins: sin(a)/A=sin(b)/B=sin(c)/C. I can't think of anyway to use the dot product or cross product to solve this problem, though that's what the section is supposedly about.



The Attempt at a Solution


I used the law of sins first. Initially, I used the angles provided and got a wrong answer, as I didn't manipulate the measures of the angles. Later, I tried making the X-axis go along A and adjusting the angle between A and B to 12 degrees and the angle between B and C as 103 degrees. This also gave a wrong answer of 213.95. (I had the sin(103)/230=sin(65)/B)). I got the 65 degree based on triangles having 180 degrees.

I've also attempted setting up a system of equations for the components of B and C, as I know that the X and Y components of B and C must add up to be the opposite of the X and Y components of vector A, but I hit a wall there as well.

I haven't attempted finding the distance of C until I am sure B is correct.

I have no idea why the dot or cross product is pertinent in the case, if it is at all.

I have a feeling I am either missing a very simple concept, making a math error in the angles I am using, or just making the problem harder than it needs to be.

Please point me in the right direction!

Thanks for any help, it's really appreciated.
 
Physics news on Phys.org


jcwhitts said:

Homework Statement


You leave your campsite to go get supplies at a store. You travel 230M at an angle 31 degrees south of east to get there(I'll call this vector A for simplicity's sake). On the way back to camp, you go a distance B at 43 degrees north of west and a distance C at 60 degrees south of west. Find the distances B and C.

Homework Equations


Not sure. I've tried the law of sins: sin(a)/A=sin(b)/B=sin(c)/C. I can't think of anyway to use the dot product or cross product to solve this problem, though that's what the section is supposedly about.

The Attempt at a Solution


I used the law of sins first. Initially, I used the angles provided and got a wrong answer, as I didn't manipulate the measures of the angles. Later, I tried making the X-axis go along A and adjusting the angle between A and B to 12 degrees and the angle between B and C as 103 degrees. This also gave a wrong answer of 213.95. (I had the sin(103)/230=sin(65)/B)). I got the 65 degree based on triangles having 180 degrees.

I've also attempted setting up a system of equations for the components of B and C, as I know that the X and Y components of B and C must add up to be the opposite of the X and Y components of vector A, but I hit a wall there as well.

I haven't attempted finding the distance of C until I am sure B is correct.

I have no idea why the dot or cross product is pertinent in the case, if it is at all.

I have a feeling I am either missing a very simple concept, making a math error in the angles I am using, or just making the problem harder than it needs to be.

Please point me in the right direction!

Thanks for any help, it's really appreciated.

Lets reserve comment on the laws of sin. But that said you will need to solve for both of your unknowns with what you do know.

Since you end up back at the same place - camp - then the x,y components of all the vectors must add to 0. This should allow you to construct 2 equations - one for the sum of x components in terms of angles and unknowns and likewise for all the y components. Then you can solve for B and C.

For instance 240*Cos31 - B*Cos43 - C*Cos60 = 0
Develop the equation for Y and then you can calculate for B and C.
 
Last edited:


Thanks a lot! I was making this question way too hard... I forgot how to solve simple systems of equations. I kept getting 0=0 because when I tried it before, I substituted the variable I solved for into the same equation rather than whichever I hadn't solved initially.

Thanks again!
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top