- #1

- 83

- 0

Determine the vector sum of the displacements d1= 2.4 m [32 degrees S of W]; d2=1.6 m

do i add one and subract the rest i dont get it.

What am i supposed to do? I also have a second question but i will post it later.

- Thread starter F.B
- Start date

- #1

- 83

- 0

Determine the vector sum of the displacements d1= 2.4 m [32 degrees S of W]; d2=1.6 m

do i add one and subract the rest i dont get it.

What am i supposed to do? I also have a second question but i will post it later.

- #2

- 159

- 0

The other way it to graph these on coordinate planes, and treat each vector as if it were the hypotenuse of a right triangle, where the horizontal leg is the x component of the vector and the vertical leg is the y component. The sum of the x components of the vectors will be the x component of the resultant vector. The sum of the y components will be the y component of the resultant vector, and you can use Pythagorean’s theorem and trig ratios to solve the magnitude and direction of the resultant.

- #3

- 83

- 0

drx= d1+d1+d3

=2.4cos32 - 1.6 - 4.9cos63

=2.08 - 1.6 - 2.22

= -1.79

dry= d1 + d2 + d3

=2.4sin32 - 1.6 - 4.9sin63

=1.27 - 1.6 - 4.365

= -4.695

I dont know whether i was supposed to add these or subtract them. If this is right what do i do from here??

- #4

- 83

- 0

Help me please

- #5

Diane_

Homework Helper

- 390

- 0

One possible approach not mentioned is to resolve all three vectors into north/south and east/west components, add those, then determine the resultant. In that case, some of the components may well be negative - you'll still add them, but adding a negative is subtraction (she said, trying not to sound too stupid saying it).

Doing the tip-to-tail method will give you a four-sided figure (including the resultant). If you're going to do it that way, I would suggest you pick two of them and add them (you'll have a triangle, and basic trigonometry will help immeasurably. Don't forget the Law of Sines and the Law of Cosines), then add the resultant to the third. Working with four of them at once in polar notation usually results in a quick trip to the nearest asylum for the terminally bewildered. Converting the polar coordinates you have to Cartesian coordinates is functionally equivalent to my first suggestion.

- #6

HallsofIvy

Science Advisor

Homework Helper

- 41,833

- 961

"Sum"F.B said:

drx= d1+d1+d3

=2.4cos32 - 1.6 - 4.9cos63

=2.08 - 1.6 - 2.22

= -1.79

dry= d1 + d2 + d3

=2.4sin32 - 1.6 - 4.9sin63

=1.27 - 1.6 - 4.365

= -4.695

I dont know whether i was supposed to add these or subtract them. If this is right what do i do from here??

Do you have much experience with drawing graphs? On a standard map, East is to the right, West left; North is up, South down.

On a standard graph, positive x is right, negative x left; positive y is up, negative down.

The components of the first vector are (-2.4 cos(32), -2.4 sin(32)): both negative because the vector is "south of west". You had everything right except the sign.

The components of the second vector are (0, -1.6):

The components of the third vector are (4.9 cos(27), -4.9 sin(27)). (Take a look at your triangle. You said the vector was "27 degrees S of E". Using 63 degrees is measuring the angle from the South and then you swap sine and cosine.) Now the x-component is positive since the vector is "E" and the y-component is positive because the vector is "S".

The "sum" is (-2.4 cos(32)+ 0+ 4.9 cos(27), -2.4 sin(32)- 1.6-4.9 sin(27)).

It's not a matter of "adding" or "subtracting", it's a matter of whether the numbers themselves are positive or negative- and that is based on "North positive, South negative; East positive, West negative".

Last edited by a moderator:

- Last Post

- Replies
- 1

- Views
- 657

- Replies
- 2

- Views
- 2K

- Last Post

- Replies
- 5

- Views
- 9K

- Last Post

- Replies
- 11

- Views
- 972

- Last Post

- Replies
- 1

- Views
- 2K

- Last Post

- Replies
- 6

- Views
- 16K

- Last Post

- Replies
- 1

- Views
- 3K

- Last Post

- Replies
- 11

- Views
- 4K

- Last Post

- Replies
- 4

- Views
- 2K

- Last Post

- Replies
- 8

- Views
- 14K