Still having problems with vectors

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Homework Help Overview

The discussion revolves around vector addition and component analysis in a two-dimensional space. The original poster presents vectors V1 and V2, along with their lengths and directions, and seeks assistance in determining their components and resultant vector.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the calculation of vector components, specifically questioning the use of trigonometric functions for V2's components. There is an exploration of the signs in vector addition and the implications for the resultant vector's magnitude and direction.

Discussion Status

Some participants have provided feedback on the calculations, particularly regarding the signs of the components and the correct application of trigonometric functions. There is ongoing clarification about the correct approach to finding the components and resultant vector.

Contextual Notes

Participants express confusion over the calculations and the use of trigonometric functions, indicating a need for further clarification on the setup of the problem and the assumptions made regarding angles and directions.

confusedaboutphysics
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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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confusedaboutphysics said:
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 can't do part B until i finish part A...help please!
 
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 here's 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!
 
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.
 

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