# This is a vector displacement (direction and magnitude) problem, but there is no compass directions. This measurement on the coordinate plane is based on 360 degrees starting at 0 degrees east...

This is a vector displacement (direction and magnitude) problem, but there is no compass directions. This measurement on the coordinate plane is based on 360 degrees starting at 0 degrees east going counter-clockwise.

Find the Magnitude and direction of the resultant using:

A= 63 m @ 11 degrees

B= 76 m @ 296 degrees

C= 87 m @ 222 degrees

I have the picture but I need a thorough algebraic explanation. Thank you.

(I normaly use a chart to find the answer but this degree format makes it confusing)

valentin68 | College Teacher | (Level 3) Associate Educator

Posted on

If you take the x axis direction from W to E and the y axis from S to N (a normal Cartesian system) you will end that what you have are just normal vectors absolute values with the elevation angles of the vectors measured in the trigonometric direction from the x axis. This is the so called standard notation.

A: 63 m @ 11 degree is equivalent to

`A_x =63*cos(11) =61.84 m`

`A_y = 63*sin(11) = 12.02 m`

B: 76 m@296 degree is equivalent to

`B_x =76*cos(296) =33.32 m`

`B_y =76*sin(296) =-69.31 m`

C: 87m @222 degree is equivalent to

`C_x =87*cos(222) =-64.65 m`

`C_y =87*sin(222)= -58.21 m`

Now to find the resultant all you have to do is to add the x and y components

` ``R_x = A_x +B_x +C_x =61.84+33.32-64.65 =`

`=30.51 m`

`R_y =A+y +B_y +C_y =12.02-69.31-58.21 =`

`=-115.5 m`

Fina` `ly you sum `R_x` and `R_y` to find the amplitude of resultant

`R = sqrt(R_x^2 +R_y^2) =119.46 m`

The angle the vector R is making with x axis comes from

`alpha= arctan(R_y/R_x) =arctan(-115.5/30.51) =75.20 degree`

The resultant is R: 119.46 m @75.20

steveschoen | College Teacher | (Level 1) Associate Educator

Posted on

Hi, persi,

These can be confusing, I do agree.  All vectors have 2 things, magnitude and direction.  They may not be given as such; there are different ways to show them.  Here, the magnitude should be alright, the first parts of each, i.e. 63 m.  As for the angles, those represent the direction.  What they are, most likely, are angles as measured from the positive x axis.  I attached a picture to show how some standard angles would be measured.  They all start from the positive x axis and circle counterclockwise for a positive angle.

So, an 11 degree angle would end up in the first quadrant.

A 296 degree angle would end up in the 4th quadrant.

A 222 degree angle would end up in the 3rd quadrant.

All following how the angles in the picture I made are measured.

I hope this helps, persi.  Let me know anything.

Till Then,

Steve

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