To find the resultant of forces on a lamina

[gnits]

Homework Statement

To find the resultant of forces on a lamina

Relevant Equations

Equating of forces and torques

Could I please ask for help with the following question:

A lamina is in the shape of an equilateral triangle ABC, and D, E, F are the midpoints of BC, CA, AB respectively. Forces of magnitude 4N, 8N, 4N, 3N, 3N act along AB, BC, CA, BE, CF respectively, the direction of each force being indicated by the order of the letters. Find the magnitude of the resultant force on the lamina, and show that its line of action cuts AD produced at G, where DG = AD.

The lamina is kept in equilibrium by three forces acting along FE, DF, ED. Find the magnitudes of theses forces.

It is the last part where I disagree with the books answer.

The first part has the answer of 5N - with which I agree, and I have been able to show that the line of action behaves as suggested.

I define then length of each side of the equilateral triangle ABC to be a.

If I call the resultant of the five forces in the first part R, then during my answering the first part of the question I know that:

The x-component of R = Rx = -( (3*sqrt(3) + 4) / 2 )
The y-component of R = Ry = ( 4*sqrt(3) - 3 ) / 2
The point of interception of the line of action of R with the x-axis occurs at x = 8*sqrt(3)*a / (4*sqrt(3) -3)

so here is a diagram for the last part:

The book gives the answers as: 6, 10 + sqrt(3), 10 - sqrt(3)

So, from my diagram, F1, F2 and F3 have to counteract R.

Resolving horizontally I get:
F3 - F1 / 2 - F2 / 2 = (3*sqrt(3) + 4) / 2

Resolving vertically I get:
sqrt(3)*F1 - sqrt(3) * F2 = -4*sqrt(3) + 3

(A check shows that the book answers satisfy these equations, so that is encouraging )

Finally, I generate a third equation by taking anti-clockwise moments about F to give:

-F3 * a * sqrt(3)/4 = Ry * ( (1/2) + ( (4*sqrt(3) + 3) / (4 * sqrt(3) - 3) ) ) * a

I have Ry from earlier but solving this equation gives F3 = -12 - sqrt(3)

Can anyone point out my error?

Thanks.
Mitch.

 

[Steve4Physics]

Finally, I generate a third equation by taking anti-clockwise moments about F to give:

-F3 * a * sqrt(3)/4 = Ry * ( (1/2) + ( (4*sqrt(3) + 3) / (4 * sqrt(3) - 3) ) ) * a

EDIT - my original comment was wrong. Removed.]

You could try using the original individual forces (not the resultant) when balancing the moment of $F_3$ about F. See if that gives the same answer as you’ve already got.

 

[haruspex]

Not sure how you are defining Rx. You've drawn it as positive left, but I think you are using it as positive right.
Also, did you forget that F₁, F₂ and F₃ act to balance the other forces, so the moment they generate must be opposite?
I ask because I get F₃ as yours but with opposite sign.

Fwiw, I found it easiest to set AE etc. to have length 1, to take axes along AD (X) and BC, and write c, s for cos, sin of 30 degrees.
For the last part, I wouldn't bother finding the line of action of the resultant. Just write the moment balance about D using the original given forces.

 

[gnits]

Thanks both for your suggestion of using the original forces for the moment balancing. It is not agreeing with the book at the moment. Would you agree with the following:

Taking moments anti-clockwise about D:

8N force along BC has moment:
0

4N force along AB has moment:
4*sqrt(3)a/4

3N force along CF has moment:
3 * a/4

4N force along CA has moments (horizontal and vertical):
4*sqrt(3)/2*a/4

and

4*1/2*sqrt(3)a/4

3N force along BE has moments (horizontal and vertical):
-3*sqrt(3)/2 * sqrt(3)a/4

and

3*1/2*a/4

Out of F1, F2 and F3 only F1 has a non-zero moment about D

If I now equate the sum of these anticlockwise moments about D to the clockwise moment of F1 about D I get:

(in the image I write x for F1)

And this gives F1 = x = 8

whereas I was expecting 6 as per book answer.

Thanks for any help,
Mitch.

 

[kuruman]

I did this problem calculating torques about A. I did it formally by writing all force vectors and position vectors relative to the origin at A, demanded that $\sum \vec F_i=0$ and $\sum \vec r_i\times \vec F_i=0$ where the summations are over all 8 forces. I then solved the system of 3 equations and 3 unknowns.

Doing it this way allowed me to write the unknown forces as an unknown magnitude times a unit vector:$$\vec F_1=F_1\left\{-\frac{1}{2},\frac{\sqrt{3}}{2},0\right\}~;~~ \vec F_2=F_2\left\{-\frac{1}{2},-\frac{\sqrt{3}}{2},0\right\}~;~~ \vec F_3=F_3\left\{1,0,0\right\}.$$The unit vectors are consistent with your drawing in post #1. The advantage of this method is that, if the direction of an unknown vector is actually opposite to what it ought to be, the magnitude of the vector will be negative when the solution is completed.

I got $F_1=8~\text{N}$, $F_2=(12-\sqrt{3})~\text{N}$ and $F_3=(12+\sqrt{3})~\text{N}$. A check reveals that the sum of these three vectors added to the resultant from part (a) is zero. The same cannot be said when this check is applied to the given answer. If one uses the magnitudes given in the answer and the unit vectors as shown above, the sum of all the forces is $$\sum \vec F_i=\left\{-2(5+\sqrt{3}),10\sqrt{3},0\right\}.$$Something is not right. Please check that the vectors in the given solution match the labels and directions as the ones in your drawing. If they do match, then there must be some mistake or typo in the given solution. The fact that I too got F₁ = 8 N using a different method cannot be a coincidence.

 

[gnits]

Thanks to everyone for the help.

I too get:

$$F_1=8$$
$$F_2=(12-\sqrt{3})$$
$$F_3=(12+\sqrt{3})$$

I will assume here that the answer in the book is wrong.

Thanks again,
Mitch.