Determine maximum weight and angle for equilibrium

AI Thread Summary
The discussion focuses on determining the weight of a crate and the angle θ for equilibrium using the maximum load capacity of cords ABC and BD, each supporting 100 lb. Participants emphasize that the forces along the cords must balance, and they suggest that assuming the cord forces are equal simplifies the problem. The relevance of the 100 lb limit is debated, with some asserting it is essential for calculating the maximum weight of the crate. The conversation highlights the need to understand vector relationships and equilibrium conditions to derive the solution. Ultimately, the problem requires finding the maximum weight the crate can have without exceeding the cords' limits.
reigner617
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Homework Statement



The cords ABC and BD can each support a maximum load of 100 lb. Determine the weight of the crate, and the angle θ for equilibrium.

Homework Equations


∑F= 0
∑Fx=0
ΣFy=0
Setting derivative of an expression to zero To find critical point
Second derivative test for determining whether min or max at a point

The Attempt at a Solution


-FBDcosθ + FBCsin22.62 = 0
-mg -FBCcos22.62 + FBDsinθ
 

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I don't think your approach will give you a meaningful answer. I would suggest to assume that the cord forces are all equal to 100lbs.
 
paisiello2 said:
I don't think your approach will give you a meaningful answer. I would suggest to assume that the cord forces are all equal to 100lbs.
If we are to take that approach, then would the crate have to be 100 lb?
 
Sorry, I meant assume cord BD is 100lbs.
 
paisiello2 said:
Sorry, I meant assume cord BD is 100lbs.
So from that, we get.FBD=100cosθi + 100sinθj

The system is in equilibrium so FBD=FBA+ FBC

F
BA=-mgj
F
BC= FBCsin22.62i - FBCcos22.62j

Would I be correct in assuming that the magnitude of the force on BA and BC add to 100 lb?

If so, where would I go from there?
 
One of the tricks here is to realize that BA and BC must be equal. Otherwise the pulley wheel would start to rotate.
 
This is a rather simple problem. You get the force along AB from the weight of the crate. Then consider the relationship between the forces along AB and CB. Tip: Imagine the wheel's axle as fixed. The angle ABC can be read off directly from the diagram. With these, you can calculate the vector sum of the forces BA and BC. From this, you immediately get BD and θ. No need to fiddle with component vectors. :cool:

If you want to challenge yourself, do this one in your head. I just did.
 
Last edited:
reigner617 said:
Would I be correct in assuming that the magnitude of the force on BA and BC add to 100 lb?
It only makes sense to add magnitudes of vectors if they are in the same direction. As vectors they must add to be equal and opposite to the pull from BD. I assume you know how to add vectors.
Btw, the 100 lb is irrelevant. You don't need to know any actual force values to solve the question.
 
F0ckForcedLogin said:
This is a rather simple problem. You get the force along AB from the weight of the crate. Then consider the relationship between the forces along AB and CB. Tip: Imagine the wheel's axle as fixed. The angle ABC can be read off directly from the diagram. With these, you can calculate the vector sum of the forces BA and BC. From this, you immediately get BD and θ. No need to fiddle with component vectors. :cool:

If you want to challenge yourself, do this one in your head. I just did.
But how do we calculate the weight of the crate?
 
  • #10
haruspex said:
Btw, the 100 lb is irrelevant. You don't need to know any actual force values to solve the question.
I disagree since the 100lbs is the given limiting value you must use it to solve the problem.
 
  • #11
paisiello2 said:
I disagree since the 100lbs is the given limiting value you must use it to solve the problem.
My mistake - I was only looking at finding the angle. Didn't notice that we're asked for the weight of the crate too.
Note that it makes the wording a little strange. It doesn't ask for the max weight of the crate, but for the actual weight. So, correspondingly, it should state the maximum actual tension experienced by any of the strings in the set-up, but instead it says 100 lbs is the max they can support.
Maybe it lost something in the retelling.
 
  • #12
reigner617 said:
But how do we calculate the weight of the crate?
It is not entirely explicit in the problem statement, but what is asked for is the maximum weight the crate may have, without breaking the ropes. We have three force vectors that point along pieces of rope: BA, BC, BD. Any of these forces is obviously proportional to the weight of the crate: Double the weight of the crate, and each of the rope forces will double. You should know by now the relationship between the forces BA and BC, and between BA+BC and BD. Set the highest of the three rope forces to the maximum allowed, find the corresponding crate weight, and you're done.
 
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