# Understanding Stability and Equilibrium

1. Oct 1, 2007

### Fluff

And probably the rest of the lessons I missed.

We were asked to research stability and equilibrium. So far, I know that equilibrium is a state an object is in when no unbalanced forces are acting upon it.

http://www.physicsclassroom.com/Class/vectors/U3L3c.html

I got lost when I read after the scale diagram. I searched some more and I found some problems

(1. An object acted on by three forces moves with constant velocity. One force acting on the object is in the positive x direction and has a magnitude of 6.5 N; a second force has a magnitude of 4.4 N and points in the negative y direction. Find the magnitude of the third force acting on the object. http://www.physics247.com/physics-homework-help/simple-equilibrium.php)

that I really had no idea how to solve because all I knew was that equilibrium was that balance!

So could someone please help me understand? I know it looks like I haven't tried hard enough but I really did. Don't be mad.

2. Oct 1, 2007

### kateman

hey, welcome

now dont fret
think about it in the form of a graph. the x axis is horizontal and the y axis is vertical
now, if you pull and object in a "Y" shape (that is on either side of the two pointy parts of the "Y"), it goes straight forward.

but in this question, you have two axis, and if both were of equal force, they would go 45 degrees at a gradient of 1.

however, the question says that you have two different forces! so the third force would not be straight down the middle, rather towards the larger force as it has a larger pull.

3. Oct 2, 2007

### learningphysics

For the first problem, you need the net force on the object to be zero. Have you studied i, j vectors?

You have three forces.
$$\vec{a} = 6.5\vec{i}$$.
$$\vec{b} = -4.4\vec{j}$$.
You want to get $$\vec{c}$$.

You know that:
$$\vec{a} + \vec{b} + \vec{c} = 0$$.

Solve for $$\vec{c}$$. what is its magnitude?

4. Oct 2, 2007

### Fluff

We did the first condition of equilibrium. I'm not familiar with i, j vectors. I'm not too familiar with magnitude. Basically, I entered a three month old physics class and I just... fell into this.

5. Oct 2, 2007

### learningphysics

Do you know how to get the sum of two vectors... ie the resultant?

6. Oct 3, 2007

### Fluff

Oh yes. Just add the two vectors. Like 5N east + 6N east = 11 N east.

7. Oct 3, 2007

### learningphysics

cool. How about 5N north + 6N east?

8. Oct 4, 2007

### Fluff

Oh yeah. I just learned that.

5N north + 6N east =

25 +36 = square root of 61 north east

9. Oct 4, 2007

### learningphysics

Cool. the magnitude is correct. and you're right it is north east. but you should also be able to calculated the exact angle. so you have a right triangle with one side 5, the other side 6. and the hypoteneuse sqrt(61).

You need the angle between the 5 and the sqrt(61). so you can use trig to get this angle. you can use arctan(6/5) = 50.19 degrees. So the direction is north 50.19degrees east.

Now for your problem, do the same type of thing. Get the magnitude of the sum of the 6.5N force which is along the x-axis and the 4.4N force which is along the -y axis.

Last edited: Oct 4, 2007