Free body diagrams and equilibrium

[jonnburton]

Hi,

There is a worked example in the book I am using which I don't quite understand. Can anybody tell me how the following is the case?

The working comes from the following free body diagram:

The boom is used to support a 75lb flower pot. Determine the tension in the wires AB and AC.



The first step is to get the unit vector for the lines AB and AC, which is fine.

\(\displaystyle F_{AB} = F_{AB}(\frac{(2i - 6j +3k)ft}{(2ft)^2+(-6ft)^2 + (3ft)^2})\)

\(\displaystyle F_{AB} = \frac{2}{7}F_{AB}i-\frac{6}{7}F_{AB}j+\frac{3}{7}F_{AB}k\)

And, likewise \(\displaystyle F_{AC} =\frac{-2}{7}F_{AC}i -\frac{6}{7}F_{AC}j +\frac{3}{7}F_{AC}k\)



The next step is to eliminate the force reaction at O by writing the moment equation of equilibrium about O.

\(\displaystyle \sum M_o = 0\)

\(\displaystyle r_A * (F_{AB} + F_{AC} + W) = 0\)

This is the part I don't see. It is clear that the perpendicular distance from the weight force is \(\displaystyle r_A\), but \(\displaystyle r_A \) is not the perpendicular distance to point O for either \(\displaystyle F_{AB}\) or \(\displaystyle F_{AC}\)

I know I must be forgetting something I should have learned earlier here but I can't see how it can be said the moment of \(\displaystyle F_{AB}\) and \(\displaystyle F_{AC}\) about point O is obtained by multiplying these forces by something that is not the perpendicular distance. Can anyone tell me what I am missing?

 

[Deleted member 4993]

Hi,

There is a worked example in the book I am using which I don't quite understand. Can anybody tell me how the following is the case?

The working comes from the following free body diagram:

The boom is used to support a 75lb flower pot. Determine the tension in the wires AB and AC.

View attachment 3158

The first step is to get the unit vector for the lines AB and AC, which is fine.

\(\displaystyle F_{AB} = F_{AB}(\frac{(2i - 6j +3k)ft}{(2ft)^2+(-6ft)^2 + (3ft)^2})\)

\(\displaystyle F_{AB} = \frac{2}{7}F_{AB}i-\frac{6}{7}F_{AB}j+\frac{3}{7}F_{AB}k\)

And, likewise \(\displaystyle F_{AC} =\frac{-2}{7}F_{AC}i -\frac{6}{7}F_{AC}j +\frac{3}{7}F_{AC}k\)



The next step is to eliminate the force reaction at O by writing the moment equation of equilibrium about O.

\(\displaystyle \sum M_o = 0\)

\(\displaystyle r_A * (F_{AB} + F_{AC} + W) = 0\)

This is the part I don't see. It is clear that the perpendicular distance from the weight force is \(\displaystyle r_A\), but \(\displaystyle r_A \) is not the perpendicular distance to point O for either \(\displaystyle F_{AB}\) or \(\displaystyle F_{AC}\)

You are dealing with vectors here.

The moment of a force F around a point O is

the cross product of

the arm vector obtained from O to any point on F

with vector F

So you need to take cross product of Vector OA (arm vector) with all those forces (since all the forces pass through A)

However, since you are only interested to find F_AB and F_AC you need not take moments - you just consider the equilibrium of point A.


I know I must be forgetting something I should have learned earlier here but I can't see how it can be said the moment of \(\displaystyle F_{AB}\) and \(\displaystyle F_{AC}\) about point O is obtained by multiplying these forces by something that is not the perpendicular distance. Can anyone tell me what I am missing?

.

 

[jonnburton]

.

Thanks Subhotosh. Concepts that I learned earlier always seem to slip my mind later down the line. Having not done any work on this over the past week or so probably hasn't helped, either.