# Is this static equilibrium problem even solvable?

Gold Member
Okay, me and my friend have been working on this problem for a while and it just been a headache catalyst for us...your help is called upon, people of the internet.

## Homework Statement

A system of weights is in equilibrium.

Calculate the distance "s" where
WD = 5[N]
WF = 5 [N]
WE = 8 [N]
a = 4 [m]

((The answer is 5.33 according to the book, and the clue is to do sum of all forces on Y axis...but nothing comes up to me))

http://img526.imageshack.us/img526/3186/71090239.jpg [Broken]

((The diagram is a little askew because of the scanjob it's supposed to be straight))

None provided.

## The Attempt at a Solution

My attempts led nowhere as we have 1 distance in meters, and the force that's applied on the string (in Newtons). We have no angle, no nothing! Is this even solvable? We're supposed to be studying basics of technical mechanics (sum of all moments/forces+sin/cos trignometry)...nothing too fancy... so if there is something too fancy it's probably out of our league..but let me know how to go about this please.

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tiny-tim
Homework Helper
Hi Dory! We have no angle …
That's because the question asks you to find the angle!! (Well, to find s, and you know the tangent of the angle is a/s) …

Go! Gold Member
Hello tiny-tim!

Yes, we're supposed to find the angle, BUT we only have 1 length! The other component is of force, not length. Can't convert force to meters. Stuck. Did you actually manage to get to the answer or are you just saying what seems logical?

tiny-tim
Homework Helper
uhhh? they give you three forces, and you need them to be in equilibrium …

(and if they'd specifically asked you for the angle, they wouldn't even need to give you the length "a")

… get on with it!​

Gold Member
Ah...I get it... so the hypotenuse is the side weights (5+5)....the opposite is the middle weight (8)... I was being stupid...got it....

Arcsin (8/10) = 53.13 degrees

tan(53.13) x 4 = 5.33 [m]

:)

tiny-tim
Homework Helper
Correct! but you didn't need to find the angle at all, you could just have used Pythagoras, and still got (4/3) x 4 …

try it! Gold Member
I just noticed I forget to say thanks...thanks A LOT!! You're incredible t-tim. Second problem you've helped me figure out the answer to!

Pythagoras eh...you mean like this?:

8^2 + X^2 = 10^2

X = 6

4 x (8/6) = 5.33

Ah...yesss :) like that. Indeed. Either way, I got that triangle in my pocket!