Recent content by reigner617
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Determine maximum weight and angle for equilibrium
But how do we calculate the weight of the crate?- reigner617
- Post #9
- Forum: Introductory Physics Homework Help
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Determine maximum weight and angle for equilibrium
So from that, we get.FBD=100cosθi + 100sinθj The system is in equilibrium so FBD=FBA+ FBC FBA=-mgj FBC= 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?- reigner617
- Post #5
- Forum: Introductory Physics Homework Help
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Determine maximum weight and angle for equilibrium
If we are to take that approach, then would the crate have to be 100 lb?- reigner617
- Post #3
- Forum: Introductory Physics Homework Help
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Determine maximum weight and angle for equilibrium
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...- reigner617
- Thread
- Angle Equilibrium Maximum Weight
- Replies: 11
- Forum: Introductory Physics Homework Help
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Optimization of the Area of a Triangle
I used 4cos(2x)+4 for the height so A=4sin2x(4+4cos2x). I derived that and still got 30 for the angle. Substituting that into the Area formula, I get 12(sqrt(3)) which is the same as the answer I got when I solved in terms of h- reigner617
- Post #23
- Forum: Calculus and Beyond Homework Help
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Optimization of the Area of a Triangle
Ah thank you for pointing that out. It works now. The 1/2 was put there by mistake because I was thinking fo 1/2 (bh). Thank you for all your help- reigner617
- Post #21
- Forum: Calculus and Beyond Homework Help
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Optimization of the Area of a Triangle
So A= 1/2 (4cos(2θ))(4sin(2θ)) = 8cos(2θ)sin(2θ) deriving this, we get A'= -16(sin(2θ))2 + 16(cos(2θ))2 setting it to zero, we get θ= 30 degrees substituting this for θ we get 2(sqrt(3)) However, another part of the question was solving the area in terms of h (I used A=...- reigner617
- Post #19
- Forum: Calculus and Beyond Homework Help
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Optimization of the Area of a Triangle
Ahhh never mind. I think I see it now- reigner617
- Post #17
- Forum: Calculus and Beyond Homework Help
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Optimization of the Area of a Triangle
That would be sine. So if I were to use that, I would get sin(2θ)= [sqrt(16-(h^2))]/4. But I don't understand what this would accomplish since we're trying to look for max area, and that involves base and height- reigner617
- Post #16
- Forum: Calculus and Beyond Homework Help
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Optimization of the Area of a Triangle
Well since h=4cos2θ, couldn't we just substitute it into the expression to get sqrt(16-(4cos2θ)^2) for the base?- reigner617
- Post #13
- Forum: Calculus and Beyond Homework Help
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Optimization of the Area of a Triangle
That would be cos2θ=h/4 4cos2θ=h- reigner617
- Post #11
- Forum: Calculus and Beyond Homework Help
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Optimization of the Area of a Triangle
Are these labeled correctly? If those are correct, can I use the law of sines such that h/(sin90-2θ) = 4/sin90 h = 4(sin90-2θ) h=4(1-2θ) h=4-8θ- reigner617
- Post #9
- Forum: Calculus and Beyond Homework Help
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Optimization of the Area of a Triangle
If I solve for h, that would give me Ccosθ-4=h. Is it ok to have an extra variable (C) in there?- reigner617
- Post #7
- Forum: Calculus and Beyond Homework Help
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Optimization of the Area of a Triangle
How did you determine that angle to be 2θ? And to answer your question, I used the pythagorean theorem to find one of the side lengths- reigner617
- Post #5
- Forum: Calculus and Beyond Homework Help
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Optimization of the Area of a Triangle
for convenience, I'll say c= sqrt(8h+32). Would it be correct if I said the height was c(sinθ) and the base of the largest triangle 2c(cosθ)?- reigner617
- Post #3
- Forum: Calculus and Beyond Homework Help