Adding vectors

[swami59]

What is the approximate sum of the following two vectors? Magnitude 23 at 62° and magnitude 17 at 10°.

[Warrengarber]

Do you remember any of the choices? If you draw the vectors estimating the angles and the vector lengths you could quickly estimate the angle between 62 and 10 degrees with and intensity of around 30. During the test you really do not have time to work out these types of problems using sin and cos so you really have to look at the problem and rule out some of the answers and then use an educated guess. I will work it out and get an exact answer shortly. In the mean time there is a paper on Test guy on vectors that may help also.

https://www.mathsisfun.com/polar-cartesian-coordinates.html


What is the approximate sum of the following two vectors? Magnitude 23 at 62° and magnitude 17 at 10°.

[Warrengarber]

I can not take credit for this answer it was posted by Kalbi_Rob on another post with the same question about a week ago.

First you have to convert to rectangular (Cartesian)

x=r*cosθ and y=r*sinθ

x1=20 cos(30) = 17.32
y1= 20 sin(30) = 10
x2=10 cos(60) = 5
y2= 10 sin(60) = 8.66

Now add them together:

x3= x1+x2 = 22.32
y3= y1+y2= 18.66

Now convert back to polar:

r=sqrt((x^2)+(y^2)) tanθ=(y/x)

r=sqrt((22.32)^2+(18.66)^2) = 29.09

θ=tan^(-1) (18.66/22.32) = 39.9

Thus we end up with an answer of:
29.09<39.9o

So, in general, you can add/subtract rectangular forms.
You can also multiply/divide polar magnitudes(angles are added when multiplying and subtracted when dividing).

Great references for this math below:

https://www.mathsisfun.com/polar-car...ordinates.html
https://www.allaboutcircuits.com/tex...er-arithmetic/



Do you remember any of the choices? If you draw the vectors estimating the angles and the vector lengths you could quickly estimate the angle between 62 and 10 degrees with and intensity of around 30. During the test you really do not have time to work out these types of problems using sin and cos so you really have to look at the problem and rule out some of the answers and then use an educated guess. I will work it out and get an exact answer shortly. In the mean time there is a paper on Test guy on vectors that may help also.

https://www.mathsisfun.com/polar-cartesian-coordinates.html

[Warrengarber]

So you see you really do not have time to work a problem out like this. I would draw the two vectors on top of one another estimating my intensities and angles. With the 10 degree on top of the 62 degree you would see that the intensity would be greater then the 23 and the final angle would fall somewhere in the middle of the two. It is all about an educated guess after that.


I can not take credit for this answer it was posted by Kalbi_Rob on another post with the same question about a week ago.

First you have to convert to rectangular (Cartesian)

x=r*cosθ and y=r*sinθ

x1=20 cos(30) = 17.32
y1= 20 sin(30) = 10
x2=10 cos(60) = 5
y2= 10 sin(60) = 8.66

Now add them together:

x3= x1+x2 = 22.32
y3= y1+y2= 18.66

Now convert back to polar:

r=sqrt((x^2)+(y^2)) tanθ=(y/x)

r=sqrt((22.32)^2+(18.66)^2) = 29.09

θ=tan^(-1) (18.66/22.32) = 39.9

Thus we end up with an answer of:
29.09<39.9o

So, in general, you can add/subtract rectangular forms.
You can also multiply/divide polar magnitudes(angles are added when multiplying and subtracted when dividing).

Great references for this math below:

https://www.mathsisfun.com/polar-car...ordinates.html
https://www.allaboutcircuits.com/tex...er-arithmetic/

[Kalbi_Rob]

So you see you really do not have time to work a problem out like this. I would draw the two vectors on top of one another estimating my intensities and angles. With the 10 degree on top of the 62 degree you would see that the intensity would be greater then the 23 and the final angle would fall somewhere in the middle of the two. It is all about an educated guess after that.

Actually, NETA has upgraded the calculator provided for test taking to the TI-36x pro. The nice thing about the TI-36x (also the top calculator available for use on FE and PE exams) has the function built in. So, all you have to do is punch the numbers in and it does it for you. I highly recommend purchasing a TI-36x for use on jobs, and to get familiar with it for the exams. It doesn't do graphing functions so much cheaper than the TI-89 series.

It would also be nice if testguy admin would upgrade the calculator emulator to the TI-36x vices the old calculator used on the NETA exams for the quizzes.

$18 on Walmart website right now:
https://www.walmart.com/ip/Texas-Instruments-TI-36X-Pro-Scientific-Calculator/20564365?wmlspartner=wlpa&selectedSellerId=0&wl13=5343&adid=22222222227014893211&wl0=&wl1=g&wl2=c&wl3=40940417912&wl4=pla-78877143872&wl5=9008234&wl6=&wl7=&wl8=&wl9=pla&wl10=8175035&wl11=local&wl12=20564365&veh=sem&gclid=CjwKCAjw4NrpBRBsEiwAUcLcDPQhQuClfsSQ1twOQnQR j7zoceK-cJyVoinl8GPS4lHkrIo25AbgAhoCmIkQAvD_BwE

[Warrengarber]

Thank you for the update on the calculator. Nice of them to let us know prior to the test. I was just on the NETAWorld website yesterday and they still reference the TI30x. I guess I will be going to walmart tomorrow because my test is on Friday morning. Thanks again for the great info.


Actually, NETA has upgraded the calculator provided for test taking to the TI-36x pro. The nice thing about the TI-36x (also the top calculator available for use on FE and PE exams) has the function built in. So, all you have to do is punch the numbers in and it does it for you. I highly recommend purchasing a TI-36x for use on jobs, and to get familiar with it for the exams. It doesn't do graphing functions so much cheaper than the TI-89 series.

It would also be nice if testguy admin would upgrade the calculator emulator to the TI-36x vices the old calculator used on the NETA exams for the quizzes.

$18 on Walmart website right now:
https://www.walmart.com/ip/Texas-Instruments-TI-36X-Pro-Scientific-Calculator/20564365?wmlspartner=wlpa&selectedSellerId=0&wl13=5343&adid=22222222227014893211&wl0=&wl1=g&wl2=c&wl3=40940417912&wl4=pla-78877143872&wl5=9008234&wl6=&wl7=&wl8=&wl9=pla&wl10=8175035&wl11=local&wl12=20564365&veh=sem&gclid=CjwKCAjw4NrpBRBsEiwAUcLcDPQhQuClfsSQ1twOQnQR j7zoceK-cJyVoinl8GPS4lHkrIo25AbgAhoCmIkQAvD_BwE

[Kalbi_Rob]

Thank you for the update on the calculator. Nice of them to let us know prior to the test. I was just on the NETAWorld website yesterday and they still reference the TI30x. I guess I will be going to walmart tomorrow because my test is on Friday morning. Thanks again for the great info.

NETA hasn't confirmed this, but it was the emulator I was given on my last test taken in April Cycle. I also took it at a Pearson Vue test center, which may be their default calculator.

[randyw8483]

What is the approximate sum of the following two vectors? Magnitude 23 at 62° and magnitude 17 at 10°.

Break them into their vector parts:

j=sqrt(-1) Denotes the complex portion of the vector

23 @ 62° = 23*cos(62°) + j*23*sin(62°)

23*cos(62°) = 10.798

j*23*sin(62) = j*20.308

Therefore, you have: 10.798 + j*20.308

17 @ 10° = 17*cos(10°) + j*17*sin(10°)

17*cos(10°) = 16.742

j*17*sin(10°) = j*2.952

Therefore, you have: 16.742 + j*2.952

Adding these together:

Add the real parts,

10.798 + 16.742 = 27.54

Add the complex parts,

j*20.308 + j*2.952 = j*23.26

Now join them,

27.54 + j*23.26

Get them back into the original form:

27.54^2 + (j*23.26)^2 = 758.452 + 541.028) = 1299.48

Now we have to take the square root here, this is really just Pythagorean Theorem:

sqrt(1299.48) = 36.04

Now to get the resultant angle:

tan^-1(complex part / real part) = tan^1(23.26/27.54) = 40.18°

So, our answer has become 36.04@ 40.18°