# Still having problems with vectors!

Vector V1 is 8.94 units long and points along the -x axis. Vector V2 is 4.13 units long and points at +35.0° to the +x axis.
(a) What are the x and y components of each vector?
V1x =
V1y =
V2x =
V2y =
(b) Determine the sum V1 + V2.
Magnitude
________
Direction
______° (counterclockwise from the +x axis is positive)

so i know that V1x=-8.94 and V1y = 0. i did the other components, but i got them wrong. i solved for V2y by 4.13cos(35) = 3.38. then solved for V2x by 4.13sin(35) = 2.37.

So for part B i did Vx = 8.94+2.37=11.31 and Vy=0+3.38=3.38. Then found the magnitude by V=the square root of (11.31^2) + (3.38^2) = 11.80. then for the direction i did tan (angle) = 3.38/11.31=.299. the inverse of that is 16.63 degrees.

Can anyone tell me what i did wrong

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jtbell
Mentor
so i know that V1x=-8.94 and V1y = 0. i did the other components, but i got them wrong. i solved for V2y by 4.13cos(35) = 3.38. then solved for V2x by 4.13sin(35) = 2.37.

So for part B i did Vx = 8.94+2.37=11.31 [...]
You lost the minus sign on -8.94. Minus signs count when you're adding components.

ohh ok thanks! but what about the V2x and V2y components. what did i do wrong? because i got those components wrong too so i cant do part B until i finish part A...help please!

Päällikkö
Homework Helper
Draw the triangle, with V2 as the hypotenuse.
Can you see now why V2y is not 4.13cos(35) but 4.13sin(35) ?

still having trouble. so heres what i got..but i got them wrong!

V1x= -8.94
V1y = 0
V2x= 4.13cos(35) = 3.38
V2y= 4.13sin(35) = 2.37

Vx= -8.94 +3.38 = -5.56
Vy= 0 + 2.37 = 2.37

V = [the square root of (-5.56)^2 +(2.37)^2] = 6.04

tan (angle)= 2.37/-5.56 = -.426

inverse tan (-.426) = -23.09

so what did i do wrong. I got the V1x & V1y correct, but everything else wrong. PLEASE HELP!!!

jtbell
Mentor
Your magnitude for V looks OK to me. What is it supposed to be?

As for the direction (angle) of V, draw a diagram that shows the x and y axes, and has Vx and Vy pointing in the proper directions, based on their signs. Look for a right triangle that has Vx and Vy as its two sides, and see where your angle of 23.09 degrees fits in (don't worry about the - sign here). From the diagram you should now be able to read off the angle that you were asked for, which is the angle measured counterclockwise from the +x axis to V.