Is this static equilibrium problem even solvable?

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Discussion Overview

The discussion revolves around a static equilibrium problem involving a system of weights. Participants are attempting to calculate the distance "s" given specific forces and a length, while grappling with the lack of angles and additional information. The context is a homework problem related to technical mechanics.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant expresses frustration over the problem's solvability due to the absence of angles and the relationship between forces and distances.
  • Another participant suggests that the angle can be derived from the given lengths and forces, indicating that the tangent of the angle relates to the known length "a".
  • A different participant questions the logic of finding the angle when only one length is provided, emphasizing the challenge of converting force to distance.
  • One participant concludes that the hypotenuse corresponds to the combined weights, while the opposite side relates to the middle weight, leading to a calculation of the angle using arcsin.
  • Another participant points out that the angle was not necessary for the solution, proposing an alternative method using Pythagorean theorem to arrive at the same distance "s".
  • A participant acknowledges the help received and confirms understanding of the problem after applying the Pythagorean theorem.

Areas of Agreement / Disagreement

Participants exhibit a mix of agreement and disagreement. While some agree on the methods to solve the problem, there is contention regarding the necessity of finding the angle versus using the Pythagorean theorem directly. The discussion does not reach a consensus on the best approach.

Contextual Notes

Participants note the limitations of the problem, including the lack of angles and the challenge of relating forces to distances without additional information. The problem's requirements and the assumptions made by participants are not fully resolved.

Femme_physics
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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]
Ignore radius of pulley

((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

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


Homework Equations



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.
 
Last edited by a moderator:
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Hi Dory! :wink:
Dory said:
We have no angle …

That's because the question asks you to find the angle! :rolleyes:

(Well, to find s, and you know the tangent of the angle is a/s) …

Go! :smile:​
 
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?
 
uhhh? :confused:

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!​
 
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]

:)
 
Correct! :smile:

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! :wink:​
 
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!
 

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