- #1

gnits

- 137

- 46

- 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:

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 = -

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.

*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.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.