Express Sum of Vectors in Unit Vector Notation

[dolpho]

Homework Statement

Referring to the vectors in the figure, express the sum A + B + C in unit vector notation.

http://i.imgur.com/ajMkK.png

Homework Equations

I'm not sure if it wants me to add Dy, I've tried it both times and masteringphysics won't take either answer.

The Attempt at a Solution

I'm assuming we need Ax, Ay, Bx, By, Cy, Cx and add Dy.

Ay= sin40 x 1.5 = .96
Ax= Cos40 x 1.5 = 1.15
Bx= Cos19 x 2.0 = 1.89
By= Sin19 x 2.0 = .65
Cx= Cos25 x 1 = .906
Cy= sin25 x 1 = .42
Dy= 1.5

The sum would be (3.946, 3.53)?

Would appreciate any help on this problem, I'm not really quite sure what I'm missing or if I did the trig wrong.

 

[CAF123]

Start by writing vector equations for each of the given vectors. In your attempt, you have mixed up the A_x and A_y components. Also, be careful with signs. Choose which direction you want to be positive x and positive y.

 

[dolpho]

Start by writing vector equations for each of the given vectors. In your attempt, you have mixed up the A_x and A_y components. Also, be careful with signs. Choose which direction you want to be positive x and positive y.

Ahh, I'm sorry I wrote it wrong on the post. So let's say on the graph I linked we start at Ax and Ay which are both positive. (1.15, .96)

On Bx and By, we should subtract the Y value because we are heading towards the X-axis and keep the X value positive.

For Cx, Cy the vector is moving towards the Y value and upwards. So we should subtract the X value and keep the y value positive? So the correct way would be.

Dx = 1.15 + 1.89 - .906 = 2.134

Dy = .96 - .65 + .42 = .73

Am I on the right track kind of?

 

[CAF123]

Ahh, I'm sorry I wrote it wrong on the post. So let's say on the graph I linked we start at Ax and Ay which are both positive. (1.15, .96)

On Bx and By, we should subtract the Y value because we are heading towards the X-axis and keep the X value positive.

For Cx, Cy the vector is moving towards the Y value and upwards. So we should subtract the X value and keep the y value positive?

If I understand you correctly, yes this is right but as I said above, it is probably better to express things in vector notation. I.e for vector A, we would have \vec{A} = 1.5(\cos(40)\hat{x} + \sin(40)\hat{y}) and then add your x and y components.

So the correct way would be.

Dx = 1.15 + 1.89 - .906 = 2.134

Dy = .96 - .65 + .42 = .73

Am I on the right track kind of?

I don't understand this - the projection of D along x-axis is clearly zero

 

[dolpho]

If I understand you correctly, yes this is right but as I said above, it is probably better to express things in vector notation. I.e for vector A, we would have \vec{A} = 1.5(\cos(40)\hat{x} + \sin(40)\hat{y}) and then add your x and y components.

I don't understand this - the projection of D along x-axis is clearly zero

Ahh sorry, I'm still stuck in the notation like one section behind lol. Just ignore that part! Do the unit signs look correct to you?

So the Sum of A, B, C vectors for the X value = 2.134, Y value = .73

(2.134, .73)

 

[CAF123]

Ahh sorry, I'm still stuck in the notation like one section behind lol. Just ignore that part! Do the unit signs look correct to you?

So the Sum of A, B, C vectors for the X value = 2.134, Y value = .73

(2.134, .73)

Yes, the resultant vector (taking into account only A,B,C) is a vector with the above x and y components. Now take into consideration D and you are done.
Edit: I see from the question it only wants you to consider A,B,C.

 

[dolpho]

Yes, the resultant vector (taking into account only A,B,C) is a vector with the above x and y components. Now take into consideration D and you are done.
Edit: I see from the question it only wants you to consider A,B,C.

Cool, couldn't have done it without you! Thanks a bunch :D