How do I make the resultant vector equal zero?

[Ohoneo]

Homework Statement

Three forces are applied to an object, as indicated in the drawing. Force 1 has a magnitude of 33.0 Newtons (33.0 N) and is directed 30.0° to the left of the +y axis. Force 2 has a magnitude of 26.0 N and points along the +x axis. What must be the magnitude and direction (specified by the angle θ in the drawing) of the third force 3 such that the vector sum of the three forces is 0 N?

There is a drawing included, that I can post if it's wanted. Otherwise, Force 3 extends left along the x-axis as well as left along the y (making it negative?)

Homework Equations

I'm guessing just Ax = A cos theta and Ay = A sin theta. Oh and tan^-1 = Ay/Ax

The Attempt at a Solution

First, Force 1 = A, Force 2 = B and Force 3 = C.
So, I found the combinant vectors for A and B.
For A:
Ax = 33 cos 30 = -28.57 (negative because it is heading left along the x-axis)
Ay = 33 sin 30 = 16.5

For B:
Since B simply extends right along the x-axis, Bx = 26 and By = 0
So, I found that Rx (resultant vector of A+B) was -2.57, and Ry = 16.5
So, basically, Cx would have to be 2.57 and Cy would have to be -16.5.
Is this right so far?

However, it asks me for the total magnitude, and without the angle, how do I find that?
Then, how do I find the angle? I could use the formulas above but it appears, to me, that I'm missing a piece of information.

Any help is much appreciated :)

 

[cupid.callin]

sorry we can't understand the question.

is the force 30^o towards X or Z axis.

well i can tell you that write eqns in form of x(i) + y(j) + z(k)

and the vector opposite the resultant of vectors given in question is the answer

 

[Ohoneo]

sorry we can't understand the question.

is the force 30^o towards X or Z axis.

well i can tell you that write eqns in form of x(i) + y(j) + z(k)

and the vector opposite the resultant of vectors given in question is the answer

sorry, it's hard to understand you. Let me show you a picture of the problem, and the exact text I was given:

Three forces are applied to an object, as indicated in the drawing. Force 1 has a magnitude of 33.0 Newtons (33.0 N) and is directed 30.0° to the left of the +y axis. Force 2 has a magnitude of 26.0 N and points along the +x axis. What must be the magnitude and direction (specified by the angle θ in the drawing) of the third force 3 such that the vector sum of the three forces is 0 N?

[URL]http://imgur.com/jwS54[/URL]

I tried the problem again and got this:
For ease, we will say Force 1 = A, Force 2 = B and Force 3 = C

Ax = A cos (theta)
Ax = 33 cos (30) = 29, but we make it negative because the drawing shows that Force 1 extends left along the x-axis, so Ax = -29
Ay = 33 sin 30 = 17

From the drawing, we can conclude that Bx = 26 and By = 0, because Force 2 only goes along the x-axis and does not move vertically.

The resultant vector components of those two forces is Rx = -3 and Ry = 17.
Is this right so far?

Now, in order to make the magnitude zero, wouldn't the x and y components of C (force 3) simply be the same as the components of the resultant vector, but with opposite signs?) What I mean is, wouldn't Cx = 3 and Cy = -17?
I think this logically makes sense, but, according to the drawing Force 3 is negative along the x- and y-axis. Also, when I used Pythagoras (sqrt(rx^2+ry^2) t find the magnitude, I apparently got the wrong answer.
So, how much of this did I get wrong, and what do I do to fix it?

 

[cupid.callin]

Ax = A cos (theta)
Ax = 33 cos (30) = 29, but we make it negative because the drawing shows that Force 1 extends left along the x-axis, so Ax = -29
Ay = 33 sin 30 = 17

Dont you think that you identified compononts A opposite? shouldn't Ax = Asinθ and Ay = Acosθ where θ=30degree ?

 

[rwisz]

Your approach is correct so far. To find the total magnitude of the third force, you can use the Pythagorean theorem:

C = √(Cx^2 + Cy^2)

To find the angle θ, you can use trigonometric identities:

θ = tan^-1(Cy/Cx)

Alternatively, you can use vector addition to find the magnitude and direction of the third force. Since the resultant vector must be zero, the third force must be equal in magnitude and opposite in direction to the resultant vector of the first two forces. This can be written as:

C = -Rx

And the direction of the third force would be:

θ = tan^-1(Ry/Rx)

I hope this helps!