Calculating Vector Sum: 40 ft and 14 degrees south of west - Vectors J and K

  • Thread starter Thread starter gurmeet97
  • Start date Start date
  • Tags Tags
    Vectors
AI Thread Summary
Vector J has a magnitude of 28 ft at 34 degrees west of north, while Vector K has a magnitude of 48 ft at 20 degrees south of east. To find their resultant, it's essential to resolve both vectors into their x and y components and then sum these components. The initial calculations presented were incorrect, particularly in treating the vectors as perpendicular, which is not the case. The correct approach involves using standard angle conventions, where east is 0 degrees, and careful vector addition to determine the accurate magnitude and direction of the resultant vector.
gurmeet97
Messages
1
Reaction score
0

Homework Statement


Vector J has a magnitude of 28 ft and a direction of 34 degrees west of north. Vector K has a magnitude of 48 ft and a direction of 20 degrees south of east. Find the magnitude and direction of their sum.
a) 30 ft and 13 degrees north of east
b) 36 ft and 54 degrees north of east
c) 38 ft and 27 degrees south of west
d) 40 ft and 14 degrees south of west


Homework Equations





The Attempt at a Solution


Square root of 28^2 plus 48^2= 55 ft

34 degrees West of North=304 degrees
20 degrees south of east=160 degrees
304 plus 160= 464
464/2= 232 degrees
38 degrees west of south
 
Physics news on Phys.org
Hi gurmeet97. Welcome to Physics Forums!

Resolve each of the two vectors J and K into x and y components, and then sum the components. This will give you the x and y components of their resultant.

Chet
 
gurmeet97 said:

Homework Statement


Vector J has a magnitude of 28 ft and a direction of 34 degrees west of north. Vector K has a magnitude of 48 ft and a direction of 20 degrees south of east. Find the magnitude and direction of their sum.
a) 30 ft and 13 degrees north of east
b) 36 ft and 54 degrees north of east
c) 38 ft and 27 degrees south of west
d) 40 ft and 14 degrees south of west


Homework Equations





The Attempt at a Solution


Square root of 28^2 plus 48^2= 55 ft

34 degrees West of North=304 degrees
20 degrees south of east=160 degrees
304 plus 160= 464
464/2= 232 degrees
38 degrees west of south

It's not clear what convention you are using to measure angles. The standard convention is that east is zero degrees (or 360 degrees after a full circuit) and angles are measured positive counterclockwise, so that north is 90 degrees, west is 180 degrees, and south is 270 degrees.

You should draw your vectors and the resultant to check your answer. (which isn't one of the choices anyway).
 
Try to break down both vectors in their x and y components.

"Square root of 28^2 plus 48^2 = 55 ft"
Wrong, this would only be valid if the vectors were perpendicular to each other.
 
SteamKing said:
It's not clear what convention you are using to measure angles. The standard convention is that east is zero degrees (or 360 degrees after a full circuit) and angles are measured positive counterclockwise, so that north is 90 degrees, west is 180 degrees, and south is 270 degrees.

You should draw your vectors and the resultant to check your answer. (which isn't one of the choices anyway).

My prof used to do that too. 0 is still east, but all your angles aren't necessarily give off the east direction, its just an extra step.
 
Oh wait, I see what you mean. I didn't catch ops "solved" directions. It appears as though op is starting west and measuring ccw which will give them -1 of what they want.
 
I picked up this problem from the Schaum's series book titled "College Mathematics" by Ayres/Schmidt. It is a solved problem in the book. But what surprised me was that the solution to this problem was given in one line without any explanation. I could, therefore, not understand how the given one-line solution was reached. The one-line solution in the book says: The equation is ##x \cos{\omega} +y \sin{\omega} - 5 = 0##, ##\omega## being the parameter. From my side, the only thing I could...
Back
Top