How Do You Calculate Vector Magnitudes and Directions from Given Angles?

[mahadasgar]

Homework Statement

Three vectors, A, B and C each have a magnitude of 50 units. Their
directions relative to the positive direction of the x-axis are 20°, 160° and
270°, respectively. Calculate the magnitude and direction of each of the
following vectors.

1.) A + B + C
2.) A-B+C
3.) 2(A+C)

Homework Equations

a^2 + B^2=C^2

The Attempt at a Solution

1.)

Vector A
Ax=50cos20=46.985
Ay=50sin20=17.101
Vector B
-Bx=50cos20=-46.985
By=50sin20=17.101
Vector C
Cx=0
Cy=-50

Ax+Bx+Cx=46.985+(-46.985)+0=0
Ay+By+Cy=17.101+17.101+(-50)=-15.798

0^2 +(-15.798)^2=c^2
c=15.798

tan^-1= 0 degrees

2.
Ax-Bx+Cx=46.985+(--(makes positive) 46.985) + 0=93.97
Ay-By+Cy=17.101+(-17.101)+(-50)=-50

(-50)^2+(93.97)^2=106.444

tan^-1=-50/93.97=28.02 degrees below x axis

3.

2(Ax+Cx)=2(46.985+0)=93.97
2(Ay+Cy)=2(17.101+(-50))=-65.798

(-65.798)^2+(93.97)^2=c^2
C=114.72

tan^-1(-65.798/93.97)
=35.0 degrees below x axis

Did I do this correct?

 

[Simon Bridge]

You chose to break the vectors down into components in the x and y directions, added the components to get the resultant vector then found the magnitude and angle from that. This is a solid approach.

You should get used to checking your own answers - you gain confidence that way.
The best way to check if you got this right is to sketch out the vectors on some axis.
Compare the directions and relative sizes of components in your sketch with the ones you calculated.
Compare the final angles with the final angle in the sketch (it is common for the calculator to give angles in the 1st quadrant only).

Note: since all the vectors are the same length and the angles are fairly nice, the different combinations should give fairly easy shapes - you could have used geometry instead. The method you chose is very popular for students because it can be done almost automatically - for this reason, it is common for problems to be set that are actually faster to do by some other method - thus rewarding students who think outside the box.

i.e. A+B should end up pointing in the +y direction - so A+B+C will point in the -y direction with length < 50 units. If you drew it out first, you could have just written down |A+B+C|=50-100sin(20), angle=270deg right away. i.e. for number 1 you got the sign and the angle wrong.

 

[mahadasgar]

I am still slightly confused... How did you come to get the angle and sign on number 1?

You chose to break the vectors down into components in the x and y directions, added the components to get the resultant vector then found the magnitude and angle from that. This is a solid approach.


You should get used to checking your own answers - you gain confidence that way.
The best way to check if you got this right is to sketch out the vectors on some axis.
Compare the directions and relative sizes of components in your sketch with the ones you calculated.
Compare the final angles with the final angle in the sketch (it is common for the calculator to give angles in the 1st quadrant only).

Note: since all the vectors are the same length and the angles are fairly nice, the different combinations should give fairly easy shapes - you could have used geometry instead. The method you chose is very popular for students because it can be done almost automatically - for this reason, it is common for problems to be set that are actually faster to do by some other method - thus rewarding students who think outside the box.

i.e. A+B should end up pointing in the +y direction - so A+B+C will point in the -y direction with length < 50 units. If you drew it out first, you could have just written down |A+B+C|=50-100sin(20), angle=270deg right away. i.e. for number 1 you got the sign and the angle wrong.

 

[Simon Bridge]

I drew the diagram and then used geometry.
You should do the same.

To do A+B - draw them head-to-tail
A is 20deg anticlockwise from the +x axis.
B is 20deg clockwise from the -x axis.
They have the same length of 50 units, so they have equal but opposite x components and the same y component. So when you add them together, the x components will cancel out.

Thus, the resultant of A+B must point straight up - A+B is 100sin(20) units long pointing along the +y axis.

C is 50 units long pointing straight down along the -y axis.

|A+B+C| is |A+B|=100sin(20) units upwards, plus |C|=50 units downwards, which is a net 50-100sin(20) units downwards. That's length and angle.

 

[MD777]

I drew the diagram and then used geometry.
You should do the same.

To do A+B - draw them head-to-tail
A is 20deg anticlockwise from the +x axis.
B is 20deg clockwise from the -x axis.
They have the same length of 50 units, so they have equal but opposite x components and the same y component. So when you add them together, the x components will cancel out.

Thus, the resultant of A+B must point straight up - A+B is 100sin(20) units long pointing along the +y axis.

C is 50 units long pointing straight down along the -y axis.

|A+B+C| is |A+B|=100sin(20) units upwards, plus |C|=50 units downwards, which is a net 50-100sin(20) units downwards. That's length and angle.

Hello Mr. Bridge,

thank you for your assistance with this problem

When I draw everything out the resultant of A + B + C is 15.8 pointing in the downward direction. The angle that I observe from my drawing is 270 degrees. is this correct?

thank you in advance!

Ab

 

[Nathanael]

Hello Mr. Bridge,

thank you for your assistance with this problem

When I draw everything out the resultant of A + B + C is 15.8 pointing in the downward direction. The angle that I observe from my drawing is 270 degrees. is this correct?

thank you in advance!

Ab

This is correct but next time you need to create a new thread of your own if you have questions. (This thread is over 2 years old.)