# Static Equilibrium (beginner level)

1. Mar 16, 2009

### Oscar Wilde

1. The problem statement, all variables and given/known data
Three forces are applied to a tree sapling to stabilize it. F1= 310 N and F2= 425 N
The forces are 110 degrees apart, solve for the magnitude and direction of the third force applied

2. Relevant equations
perhaps law of sine/cosine , this is really just solving for vector quantities

3. The attempt at a solution
I drew it with the first force parallel with the x axis, and the second force making a 70 degrees angle with the horizontal. I then solved for the "vertical" force exerted, and it came out to ~400N, which would equal the stabilizing force, but this is wrong because the answer is 430N. All help appreciated

2. Mar 16, 2009

### LowlyPion

The desired result is that you want Σ F = 0.

I find it clearer to just add up the i,j components

F1 + F2 + F3 = 0

So you are looking for an F3 that = -1 * (F1 + F2)

Where F1 = 310 i + 0 j

... and so on.

3. Mar 16, 2009

### murrdpirate0

I think you only balanced the vertical force that remained unbalanced and neglected the remaining horizontal imbalance.

You have 310 N pulling at 0 degrees and 425 N pulling at 110 degrees. You need to break this one down into its components so you can add it to the other force. You correctly found that this 425 N force has a vertical component of 400 N, however it also has a horizontal component of about 165 N, pointing in the opposite direction of the 310 N force.

So you are correct that your force needs a vertical component of 400 N pointing down to balance the 400 N component pointing up. Now you need to find the horizontal component. You have 310 pointing to the right at 0 degrees and 165 pointing to the left at 180 degrees. This results in 145 N pointing towards the right at 0 degrees.

So your force needs 400 N pointing down at 270 degrees and 145 N pointing to the left at 180 degrees. The magnitude of this force is F = (4002 + 1452)1/2 and the direction can be found by the tangent between the magnitudes of these two components.

These problems are really easy if you just remember to break vectors down into their components and then combining them. You can only add vectors that point in the same direction, that's why you gotta break em down.

4. Mar 16, 2009

### Oscar Wilde

it's become a little clearer. I get F3= -1 * (310 i + 399 j+145 i)= -(455 i + 399 j)

I'm still unclear as to where to take these numbers, it does appear as though I will get 430 as a value if I plug these numbers in to the pythagorean theorem, or triangulate them otherwise.

5. Mar 16, 2009

### LowlyPion

Not quite. Your i component has a positive x and the other has a negative x so your i component is really the difference.

F1 = 310 i + 0 j
F2 = -145.35 i + 399.25 j

6. Mar 16, 2009

### Oscar Wilde

The moment of comprehension! I see now

I insert my values of forces into their respective sides. I get F3= -164.65 i and -399.25 j
I calculate my third side to equal ~430 N at 248 degrees (-112 degrees from F1)

Thanks guys!