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How do I make the resultant vector equal zero?

  • Thread starter Ohoneo
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  • #1
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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 :)
 

Answers and Replies

  • #2
1,137
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sorry we cant understand the question.

is the force 30o 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
 
  • #3
22
0
sorry we cant understand the question.

is the force 30o 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?
 
Last edited by a moderator:
  • #4
1,137
0
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? shouldnt Ax = Asinθ and Ay = Acosθ where θ=30degree ???
 

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