# Homework Help: Statics: how do I know a force has to be perpendicular for it to be a minimal force?

1. Dec 19, 2011

### fee

1. The problem statement, all variables and given/known data

http://img221.imageshack.us/img221/604/yesor.jpg [Broken]

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?

Last edited by a moderator: May 5, 2017
2. Dec 19, 2011

### Simon Bridge

Re: Statics: how do I know a force has to be perpendicular for it to be a minimal for

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.

3. Dec 19, 2011

### SammyS

Staff Emeritus
Re: Statics: how do I know a force has to be perpendicular for it to be a minimal for

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°.

Last edited: Dec 19, 2011
4. Dec 19, 2011

### Simon Bridge

Re: Statics: how do I know a force has to be perpendicular for it to be a minimal for

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.

Last edited: Dec 19, 2011
5. Dec 20, 2011

### fee

Re: Statics: how do I know a force has to be perpendicular for it to be a minimal for

Thank you very much, that was extremely helpful.

6. Dec 20, 2011

### TMO

Re: Statics: how do I know a force has to be perpendicular for it to be a minimal for

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?

7. Dec 20, 2011

### SammyS

Staff Emeritus
Re: Statics: how do I know a force has to be perpendicular for it to be a minimal for

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

8. Dec 20, 2011

### TMO

Re: Statics: how do I know a force has to be perpendicular for it to be a minimal for

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