Force Vectors Problem - finding the smallest resultant force

[jonnburton]

I have been doing some more questions on vectors from a different (more advanced textbook) in order to practice this topic more, before re-covering moments again.

There is one question which I am unsure about and was wondering if anyone could point me in the right direction with this?


Determine the magnitude of force F so that the resultant of the three forces is as small as possible.



It is clearly impossible to make the resultant force zero so they are in equilibrium: no triangle involving these forces would match up. (And this isn't the solution of the book, either).

So the only thing which I could think to do was the following:


Resolving in the x direction:

$\displaystyle -14 cos 30 + 8 - Fcosx$

$\displaystyle -4.12 - 0.7F$


Resolving in the y direction:

$\displaystyle 14 sin 30 - F sin45$

$\displaystyle 7 + 0.7F$


I can't see this leading anywhere, though; it doesn't make sense that:

$\displaystyle 0.71F = 7$ and $\displaystyle 0.71 F = 4.12$



Could anyone tell me the direction I should be heading in? (incidentally, the book's answer for the magnitude of F is 2.03KN)

 

[Deleted member 4993]

I have been doing some more questions on vectors from a different (more advanced textbook) in order to practice this topic more, before re-covering moments again.

There is one question which I am unsure about and was wondering if anyone could point me in the right direction with this?


Determine the magnitude of force F so that the resultant of the three forces is as small as possible.

View attachment 3086

It is clearly impossible to make the resultant force zero so they are in equilibrium: no triangle involving these forces would match up. (And this isn't the solution of the book, either).

So the only thing which I could think to do was the following:


Resolving in the x direction:

$\displaystyle -14 cos 30 + 8 - Fcosx$

$\displaystyle -4.12 - 0.7F$


Resolving in the y direction:

$\displaystyle 14 sin 30 - F sin45$

$\displaystyle 7 + 0.7F$


I can't see this leading anywhere, though; it doesn't make sense that:

$\displaystyle 0.71F = 7$ and $\displaystyle 0.71 F = 4.12$



Could anyone tell me the direction I should be heading in? (incidentally, the book's answer for the magnitude of F is 2.03KN)

View attachment 3086
What is the expression for the magnitude of the resultant forces (sum of the three forces)?

By the way, the answer in the book matches mine.

 

[DrPhil]

I have been doing some more questions on vectors from a different (more advanced textbook) in order to practice this topic more, before re-covering moments again.

There is one question which I am unsure about and was wondering if anyone could point me in the right direction with this?


Determine the magnitude of force F so that the resultant of the three forces is as small as possible.

View attachment 3086

It is clearly impossible to make the resultant force zero so they are in equilibrium: no triangle involving these forces would match up. (And this isn't the solution of the book, either).

So the only thing which I could think to do was the following:


Resolving in the x direction:

$\displaystyle -14 cos 30 + 8 - Fcosx$

$\displaystyle -4.12 - 0.7F$


Resolving in the y direction:

$\displaystyle 14 sin 30 - F sin45$

$\displaystyle 7 + 0.7F$

I can't see this leading anywhere, though; it doesn't make sense that:

$\displaystyle 0.71F = 7$ and $\displaystyle 0.71 F = 4.12$
In other words (as you said above), you can't make the resultant zero.

Could anyone tell me the direction I should be heading in? (incidentally, the book's answer for the magnitude of F is 2.03KN)

Assuming you have the two components of the resultant, what is the magnitude?
Minimize that expression with respect to F.

 

[jonnburton]

View attachment 3086
What is the expression for the magnitude of the resultant forces (sum of the three forces)?

By the way, the answer in the book matches mine.

The magnitude is $\displaystyle F_R = \sqrt{F^2_{Rx} + F^2_{Ry}}$

In this case, I make that to lead to:

$\displaystyle \sqrt{(4.12+0.7F)^2 +(7+0.7F)^2}$

(the minus sign can be neglected as we are interested in absolute values: i.e. the magnitudes)

Which, in turn leads to:

$\displaystyle \sqrt{65.97+15.6F+0.98F^2}$

The expression inside the square root symbol can't be factorized, as the discriminant is negative. I've been trying to think where I've gone wrong here, but I think I have followed all of the steps:

1. Break the vectors down into component forces

2. Add the x components and the y components.

3. Then find the magnitude of the resultant by squaring the sums of these and finding the square root.

 

[Deleted member 4993]

$\displaystyle \sqrt{65.97+15.6F+0.98F^2}$

$\displaystyle R^2 \ = \ 65.97 \ + \ 15.6F \ + \ 0.98F^2$

Now differentiate (implicitly)

 

[Deleted member 4993]

By the way, your expression for |R| is incorrect.

Check the y-component (R_y) carefully.

 

[jonnburton]

$\displaystyle R^2 \ = \ 65.97 \ + \ 15.6F \ + \ 0.98F^2$

Now differentiate (implicitly)

$\displaystyle 2R = 15.6\frac{dF}{dR} + 2F*0.98\frac{dF}{dR}$

$\displaystyle 2R = 15.6\frac{dF}{dR}+1.96F\frac{dF}{dR}$

$\displaystyle 2R = \frac{dF}{dR}( 15.6+1.96F)$


I'm not too sure about where to go from here though

 

[Deleted member 4993]

$\displaystyle 2R = 15.6\frac{dF}{dR} + 2F*0.98\frac{dF}{dR}$

$\displaystyle 2R = 15.6\frac{dF}{dR}+1.96F\frac{dF}{dR}$

$\displaystyle 2R = \frac{dF}{dR}( 15.6+1.96F)$


I'm not too sure about where to go from here though

Your Ry is incorrect - please fix it.

You need to differentiate R with-respect-to F to calculate expression for dR/dF.

For minimum R, set dR/dF = 0

 

[jonnburton]

I'd noticed that I had missed a couple of posts on here from Subhotosh and Dr Phil.

After looking at the diagram again, I see the correct expression for the magnitude must be:

$\displaystyle R = \sqrt{ (4.12+0.7F)^2 +(7-0.7F)^2}$

$\displaystyle R = \sqrt{ 65.97-4.1F+0.98F^2}$


$\displaystyle R^2 = 65.97 - 4.1F +0.98F^2$



Now differentiating implicitly:

$\displaystyle 2R\frac{dR}{dF}= -4.1 +1.96F$

$\displaystyle \frac{dR}{dF} = \frac{1.96F-4.1}{2R}$

$\displaystyle F = 2.09$

Which is closer. I'm going to go through this again now, doing each step to more significant figures to see if that makes a difference to the final result.

 

[jonnburton]

There was a mistake in the previous working:

$\displaystyle R^2 = 65.9744-4.032F+0.98F^2$

$\displaystyle 2R\frac{dR}{dF} = -4.032+1.96F$

$\displaystyle \frac{dR}{dF} = \frac{1.96F-4.032}{2R}$

$\displaystyle F = 2.06$

 

[hassanharake]

how did you get from
dR / dF = 1.96F−4.12R

to

F=2.09

need to know the steps plz

 

[metaltaku]

how did you get from
dR / dF = 1.96F−4.12R

to

F=2.09

need to know the steps plz

man make 2R= 0

move F to the other side

dived 4.032 NOT 4.12 by -1.96