- #1
Determine maximum allowable weight of a load supported by chains
Click For Summary
Discussion Overview
The discussion revolves around determining the maximum allowable weight of a load supported by chains, focusing on the application of static equilibrium principles in a homework context. Participants explore the relationships between forces acting on the load and the chains, including vector components and the incorporation of multiple forces.
Discussion Character
- Homework-related
- Mathematical reasoning
- Technical explanation
Main Points Raised
- One participant requests assistance in solving a problem related to the maximum load supported by chains.
- Another participant suggests considering vector components as a hint for solving the problem.
- A participant expresses frustration over their initial calculations, indicating that their derived weight is too small and seeks guidance on incorporating additional forces into their equations.
- A later reply provides a modified approach, clarifying the relationship between the forces and suggesting that the force due to the load (Fad) is equal to the weight (W).
- The same reply emphasizes the importance of breaking down the problem into simpler parts and correctly applying trigonometric functions in the equations.
Areas of Agreement / Disagreement
Participants do not reach a consensus on the correct approach to incorporate all forces into the equations, and there remains uncertainty regarding the calculations and relationships between the forces involved.
Contextual Notes
The discussion includes unresolved mathematical steps and assumptions regarding the forces acting on the load and the chains, particularly concerning the role of the third force (Fad) and its relationship to the weight (W).
- #2
MechanicalMan
- 25
- 0
You're supposed to take some effort in solving homework problems, but I'll give you one hint: vector components.
- #3
Lithlyrian
- 3
- 0
I have taken a lot of effort and was very frustrated when posting; sorry I did not include my work.
I am at a loss as to how to include the chain that's attached to the weight.
Here is what I am getting; I know it is wrong because the weight is far too small a value.
[tex]\Sigma[/tex]Fx = -Fab(sin 90 degrees) + Fac (sin 60 degrees) = 0
[tex]\Sigma[/tex]Fy = Fab (cos 90 degrees) + Fac (cos 60 degrees) - W = 0
setting Fab to max load, Fab = 450 lb
Thus Fac = 519.62 lb and W = 259.8076 lb
I am not including Fad so I know this is incorrect. Please tell me how to incorporate Fad into this equation properly.
I know the weight of the engine can be much greater than 260 pounds if there are three chains that can each support at the very least 400 lb...
I am at a loss as to how to include the chain that's attached to the weight.
Here is what I am getting; I know it is wrong because the weight is far too small a value.
[tex]\Sigma[/tex]Fx = -Fab(sin 90 degrees) + Fac (sin 60 degrees) = 0
[tex]\Sigma[/tex]Fy = Fab (cos 90 degrees) + Fac (cos 60 degrees) - W = 0
setting Fab to max load, Fab = 450 lb
Thus Fac = 519.62 lb and W = 259.8076 lb
I am not including Fad so I know this is incorrect. Please tell me how to incorporate Fad into this equation properly.
I know the weight of the engine can be much greater than 260 pounds if there are three chains that can each support at the very least 400 lb...
- #4
peterblais
- 8
- 0
I'm not sure how far along you are, and the fact this topic is 2 weeks old, you probably already figured this out, or moved on... Regardless I thought I would shed some light on the situation just for future reference.
For a stater, I modified your jpeg, to show the relationship between Fad, and W, which you missed. This is done by cutting the link in order to evaluate the forces on it.
Here is the critical relationship confusing you:
Fad = W
Using this:
Fx = -Fab(sin 90 degrees) + Fac (sin 60 degrees) = 0
Fx = -Fab(1) + Fac(sin60) = 0
Fx = -Fab + Fac(sin60)
Fy = Fab(cos90) + Fac(cos60) - Fad = 0
Fy = 0 + Fac(cos60) - Fad = 0
Fy = Fac(cos60) - W = 0
Remember, the trick with statics is to cut up the problem to make it into simpler problems. Amazingly complex stuff can be conquered by doing this, but if you try to attack it all at once, you will pull your hair out. Also, pay attention to those cos90's and sin90's, that stuff simplifies down right away and eliminates terms and such- after a while and a bit of practice you don't even really need to put them in there.
You seem to be on the right track.
For a stater, I modified your jpeg, to show the relationship between Fad, and W, which you missed. This is done by cutting the link in order to evaluate the forces on it.
Here is the critical relationship confusing you:
Fad = W
Using this:
Fx = -Fab(sin 90 degrees) + Fac (sin 60 degrees) = 0
Fx = -Fab(1) + Fac(sin60) = 0
Fx = -Fab + Fac(sin60)
Fy = Fab(cos90) + Fac(cos60) - Fad = 0
Fy = 0 + Fac(cos60) - Fad = 0
Fy = Fac(cos60) - W = 0
Remember, the trick with statics is to cut up the problem to make it into simpler problems. Amazingly complex stuff can be conquered by doing this, but if you try to attack it all at once, you will pull your hair out. Also, pay attention to those cos90's and sin90's, that stuff simplifies down right away and eliminates terms and such- after a while and a bit of practice you don't even really need to put them in there.
You seem to be on the right track.
Attachments
- #5
peterblais
- 8
- 0
PS: Pardon my aweful MSpaint skills and the crappy default font.
Similar threads
- · Replies 15 ·
- · Replies 8 ·
- · Replies 12 ·
- · Replies 5 ·
- · Replies 8 ·