- #1
Determine maximum allowable weight of a load supported by chains
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SUMMARY
The discussion focuses on determining the maximum allowable weight of a load supported by chains, specifically using vector components in static equilibrium equations. The user initially calculated forces incorrectly, leading to an underestimated weight of 259.81 lb. Key insights include the critical relationship Fad = W, which clarifies how to incorporate the force from the attached chain into the equations. The correct approach involves breaking down the problem into simpler components and applying trigonometric functions accurately.
PREREQUISITES- Understanding of static equilibrium principles
- Knowledge of vector components in physics
- Familiarity with trigonometric functions (sine and cosine)
- Basic skills in problem-solving for engineering mechanics
- Study the application of static equilibrium in engineering mechanics
- Learn about resolving forces using vector components
- Explore trigonometric identities and their applications in physics
- Practice solving complex static problems by breaking them into simpler components
Engineering students, mechanical engineers, and anyone involved in load analysis and static equilibrium problems will benefit from this discussion.
- #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.
\SigmaFx = -Fab(sin 90 degrees) + Fac (sin 60 degrees) = 0
\SigmaFy = 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.
\SigmaFx = -Fab(sin 90 degrees) + Fac (sin 60 degrees) = 0
\SigmaFy = 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.
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