Breaking up vectors into components

[fd25t6]

A surveyor walks 120 meters from a beginning point at 0 degrees along the east boundary line. he then turns 90 degrees north for aprox. 260 meters. on the way back he then walks 325 meters west at 155 degrees. determine the magnitude and direction of the resultant vector using the head to tail method and analytically using trig

Homework Equations

vx= v(cosine of theta)
vy= v(sine of theta)

theta= vy totals over vx totals

magnitude = square root of vx^2 = vy^2


Hi everyone, i am really new to physics and i understand this is probably extremely simple stuff, however i been doing pretty well but ran into a jam with this one. for some reason my analytical resultant and my graphical one do not match up at all. my analytical is about 100 meters higher then what i am getting from my graph and i am not too sure which one is correct. anyways here is my work... and sorry this is so long but i want to make sure i see what I am doing wrong here.. thank u in advance.

vector 1- 120m @ o degrees east
vector 2- 260m @ 90 degrees north
vector 3- 325m @ 155 degrees west

v1x= 120(cos0)= 120
v1y= 120(sin 0)= 0

v2x= 260(cos90)= 0
v2y= 260(sin90)= 260

v3x= -325(cos155)= 294.5 (negative 325 because the vector is on the negative x positive y quadrant correct?)
v3y= 325(sin155)= 137.4

total x components= 414.5 total y components= 397.4

so the theta = 397.4/414.5; tan-1 ans = 43.7 degrees
magnitude = square root of 414.5^2+397.4^2 which = 574.2 meters.

so i get a magnitude of 574.2 meters analytically and 442 meters when i graph it. I've done this over and over for the past three hours and i know that I am probably missing a small detail... can not figure out what it is and it is not letting me sleep at night. any help is greatly appreciated. thanks in advance.

 

[PhanthomJay]

A surveyor walks 120 meters from a beginning point at 0 degrees along the east boundary line. he then turns 90 degrees north for aprox. 260 meters. on the way back he then walks 325 meters west at 155 degrees. determine the magnitude and direction of the resultant vector using the head to tail method and analytically using trig

Homework Equations

vx= v(cosine of theta)
vy= v(sine of theta)

theta= vy totals over vx totals

magnitude = square root of vx^2 = vy^2


Hi everyone, i am really new to physics and i understand this is probably extremely simple stuff, however i been doing pretty well but ran into a jam with this one. for some reason my analytical resultant and my graphical one do not match up at all. my analytical is about 100 meters higher then what i am getting from my graph and i am not too sure which one is correct. anyways here is my work... and sorry this is so long but i want to make sure i see what I am doing wrong here.. thank u in advance.

vector 1- 120m @ o degrees east
vector 2- 260m @ 90 degrees north
vector 3- 325m @ 155 degrees west

v1x= 120(cos0)= 120
v1y= 120(sin 0)= 0

v2x= 260(cos90)= 0
v2y= 260(sin90)= 260

good so far...

v3x= -325(cos155)= 294.5 (negative 325 because the vector is on the negative x positive y quadrant correct?)

this your error...the x component points in the neg x direction, and should be minus ...325 is the magnitude of the vector taken as a positive number

v3y= 325(sin155)= 137.4

total x components= 414.5 total y components= 397.4

so the theta = 397.4/414.5; tan-1 ans = 43.7 degrees
magnitude = square root of 414.5^2+397.4^2 which = 574.2 meters.

so i get a magnitude of 574.2 meters analytically and 442 meters when i graph it. I've done this over and over for the past three hours and i know that I am probably missing a small detail... can not figure out what it is and it is not letting me sleep at night. any help is greatly appreciated. thanks in advance.

Redo the calc with the noted correction. It is sometimes best to look at your sketch and note that the x comp is the projection onto the x-axis and the y comp is the projection onto the y axis. So that the 325 m vector has an x component of 325 cos 155, or , from the sketch, 325 cos 25 pointing in the negative x direction (minus).

 

[fd25t6]

thank you so much for your help... for some reason i had gotten away with just simply throwing a negative sign in front of the vector on all my practice problems up until this one. i thought i was going crazy. once again your help is very much appreciated...