Understanding Force Relationships: F_A & F_B

AI Thread Summary
The discussion revolves around understanding the relationship between forces F_A and F_B, particularly why F_B must act perpendicular to F_A. It highlights that the forces are components of a resultant vector, forming a right triangle, which is crucial for solving the problem. Participants suggest that exploring various angles can reveal that the minimum tension occurs at a 90-degree angle, emphasizing the importance of considering both x and y components in vector analysis. A misunderstanding arises regarding the use of y components, which leads to confusion about the conditions for minimum force. Ultimately, recognizing the perpendicular relationship is key to solving the problem intuitively.
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Homework Statement



http://img221.imageshack.us/img221/604/yesor.jpg

My question is: how would I know that F_B has to act perpendicular to F_A? Are there any rules or is there any logic that I am supposed to know?
 
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Just trig - work out what the force is for different relative angles.

note: cos(30) = sin(60) = (√3)/2, cos(60) = sin(30) = 1/2.

if x=Fa and y=Fb; you need (√3)x/2 + ycos(θ)=10kN, and x/2 = ysin(θ)

... which is how you'd intuitively do the problem.
So solve for y as a function of theta, and find the minimum.

This is the same as saying that x and y are components of the result - where x forms an angle of 30 degrees. Since they are components, they must for a rt-angled triangle.
 


fee said:

Homework Statement



My question is: how would I know that F_B has to act perpendicular to F_A? Are there any rules or is there any logic that I am supposed to know?
In my opinion, No, you should not be expected to know that. However, setting up the problem using an arbitrary value for θ. Then the solution should show that the minimum tension occurs when θ = 90°.
 
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I'd agree with SammyS - unless it was explicitly in your course-notes ... maybe earlier in the book the problem comes from?

Some teachers like to give problems that students have to explore then show the short-cut in the model answers. Much like this does in fact.
There is a logic to it - if the two forces are not components of the result, then more of each force gets "wasted".
But you'll only "see" it by playing around with the forces.
 
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Thank you very much, that was extremely helpful.
 


I don't think I understand how the conclusion in the picture is intuitive. I will post my chain of reasoning given my interpretation of the original post's problem and ask whether or not someone could find the precise flaw.

∵ We need the sum of the x components of vector U and vector V to equal a value C, and we would like for V to be at the least possible value.
∵ We cannot change the magnitude or angle of vector U,
∵ U·cos(30°) is the x component of U,
∵ V·cos(θ) is the x component of V,
∴ U·cos(30°) + V·cos(θ) = C,
∴ V·cos(θ) = C - U·cos(30°).
∵ Let D = C - U·cos(30°),
∴ V·cos(θ) = D,
∴ V = D/cos(θ),
∴ V = D·sec(θ).
∵ sec(θ) is at its minimum when θ = 0° with 0° ≤ θ < 90°,
∴ V is at its minimum when θ = 0° with 0° ≤ θ < 90°.

However, this contradicts the conclusion reached by the post. Could somebody assist me?
 


TMO said:
I don't think I understand how the conclusion in the picture is intuitive. I will post my chain of reasoning given my interpretation of the original post's problem and ask whether or not someone could find the precise flaw.

∵ We need the sum of the x components of vector U and vector V to equal a value C, and we would like for V to be at the least possible value.
∵ We cannot change the magnitude or angle of vector U,
∵ U·cos(30°) is the x component of U,
∵ V·cos(θ) is the x component of V,
∴ U·cos(30°) + V·cos(θ) = C,
∴ V·cos(θ) = C - U·cos(30°).
∵ Let D = C - U·cos(30°),
∴ V·cos(θ) = D,
∴ V = D/cos(θ),
∴ V = D·sec(θ).
∵ sec(θ) is at its minimum when θ = 0° with 0° ≤ θ < 90°,
∴ V is at its minimum when θ = 0° with 0° ≤ θ < 90°.

However, this contradicts the conclusion reached by the post. Could somebody assist me?
I don't see anywhere in your working of the problem that you used the y components of the vectors.

U·sin(30°) - V·sin(θ) = 0
 


SammyS said:
I don't see anywhere in your working of the problem that you used the y components of the vectors.

U·sin(30°) - V·sin(θ) = 0

Oh! I completely forgot that there wasn't supposed to be any acceleration in the y component. My apologies.
 
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