- #1

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- Homework Statement:
- How to calculate the magnitudes of three forces around a hexagon

- Relevant Equations:
- balance of forces and torques

Can anyone please help me with the following?

Three forces which act along the sides AB, BC and CD of a regular hexagon ABCDEF of side 2a, have a resultant which acts along DF. When a couple of 4Pa in the sense CBA is added in the plane of the hexagon, the resultant acts along CA. Find the magnitudes of the three froces in terms of P.

I call the three forces F1, F2 and F3 and here is my diagram:

R is the resulatant of the three forces WITHOUT the torque.

F is the resultant of the three forces plus the torque.

answers given in book are:

F1 = 4P/sqrt(3)

F2 = -8P/sqrt(3)

F3 = 2*sqrt(3)P

I can find F1 by taking anitclockwise moments about C to give:

F1*sqrt(3)*a - 4*P*a = 0 and so F1 = 4*P / sqrt(3)

But I can't see how to find F2 and F3.

I can equate horizontal and vertical forces to get:

2*F1 + F2 - F3 = -sqrt(3)*F

and

sqrt(3)*F2 + sqrt(3)*F3 = -F

respectively.

I had hoped to substitute in my value for F1 to find F2 and F3 but that leads to these latter two equations reducing the same one.

I know that I somehow have to use the given fact that, after the torque is applied the resultant in parallel to the resultant prior to the torque being applied, but I can't see how.

Thanks for any help,

Mitch.

Three forces which act along the sides AB, BC and CD of a regular hexagon ABCDEF of side 2a, have a resultant which acts along DF. When a couple of 4Pa in the sense CBA is added in the plane of the hexagon, the resultant acts along CA. Find the magnitudes of the three froces in terms of P.

I call the three forces F1, F2 and F3 and here is my diagram:

R is the resulatant of the three forces WITHOUT the torque.

F is the resultant of the three forces plus the torque.

answers given in book are:

F1 = 4P/sqrt(3)

F2 = -8P/sqrt(3)

F3 = 2*sqrt(3)P

I can find F1 by taking anitclockwise moments about C to give:

F1*sqrt(3)*a - 4*P*a = 0 and so F1 = 4*P / sqrt(3)

But I can't see how to find F2 and F3.

I can equate horizontal and vertical forces to get:

2*F1 + F2 - F3 = -sqrt(3)*F

and

sqrt(3)*F2 + sqrt(3)*F3 = -F

respectively.

I had hoped to substitute in my value for F1 to find F2 and F3 but that leads to these latter two equations reducing the same one.

I know that I somehow have to use the given fact that, after the torque is applied the resultant in parallel to the resultant prior to the torque being applied, but I can't see how.

Thanks for any help,

Mitch.