Force Vectors Problem - finding the smallest resultant force

J

jonnburton

Junior Member
Joined
Dec 16, 2012
Messages
155
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.

img011.jpg

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)
 
jonnburton said:
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)
Click to expand...
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.
 
Last edited by a moderator:
jonnburton said:
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)
Click to expand...
Assuming you have the two components of the resultant, what is the magnitude?
Minimize that expression with respect to F.
 
Last edited:
Subhotosh Khan said:
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.
Click to expand...


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

Check the y-component (Ry) carefully.
 
Subhotosh Khan said:
\(\displaystyle R^2 \ = \ 65.97 \ + \ 15.6F \ + \ 0.98F^2\)

Now differentiate (implicitly)
Click to expand...

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

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

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


I'm not too sure about where to go from here though
 
jonnburton said:
\(\displaystyle 2R = 15.6\frac{dF}{dR} + 2F*0.98\frac{dF}{dR}\)

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

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


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

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
 
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.
 
Last edited:
There was a mistake in the previous working:

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

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

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

\(\displaystyle F = 2.06\)
 
You must log in or register to reply here.
Top