Vector addition and subtraction

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Vector A has a magnitude of 14 m at a 26-degree angle with the positive x-axis, while Vector B has a magnitude of 13 m directed along the negative x-axis. The x and y components for Vector A are calculated as 12.6 m and 6.14 m, respectively, while Vector B contributes -13 m in the x direction and 0 m in the y direction. The resultant vector A + B is found using the formula for vector addition, yielding a magnitude of 6.16 m and an angle of -86.3 degrees. Corrections were made regarding the use of components in the arctan equation, clarifying that the vertical component should indeed be 6.14 m. The discussion highlights the importance of accuracy in calculations and component analysis in vector operations.
steve7
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1. Vector A has a magnitude of 14 m and makes an angle of 26o with the positive x axis. Vector B also has a magnitude of 13 m and is directed along the negative x axis. Enter your answers in distance then angle(in degrees).
2. Find A + B
Find A - B
3. 14cos(26)=12.6 x component for A
14sin(26)=6.14 y component for A

-13 x component for B
0 y component for B

A+B
sqrt( (Ax+Bx)^2 + (Ay+By)^2 )

sqrt( (12.6-13)^2 + (6.14)^2 )

A+B=6.16

tan-1(y/x)
tan-1(6.14/-.4)=-86.3*

I haven't attempted A-B yet.
Tell me where I'm going wrong. Any help is appreciated.
 
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in the arctan equation .4 should be -.4, also why are you using 6.15 when the only vertical component is 6.14?
 
Asphyxiated said:
in the arctan equation .4 should be -.4, also why are you using 6.15 when the only vertical component is 6.14?

it was a typo. I corrected them.

thanks
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

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