How do I use Bow's notation to solve vector sums with only one force?

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To solve vector sums using Bow's notation with only one force, start by identifying the forces acting at the point where the load is suspended. Graphically represent these forces in a triangle, ensuring to label each force clearly for clarity in your Bow's diagram. As you progress, draw additional triangles corresponding to adjacent joints while marking each edge with directional arrows. It's essential to maintain the proportions in your diagram, as Bow's method relies on accurate graphical representation rather than equations. Ultimately, focus on creating a clear and organized diagram to determine the forces effectively.
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I have this pass paper question I am trying to do and I can't even get started, Have been doing vector sums and just started bows notation.

I don't want a solution just where do I start. what's my first move?

Question is attached

the fact I have only one force is throwing me out.
 

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There's only one force illustrated, but clearly there are forces at X and Y.
 
haruspex said:
There's only one force illustrated, but clearly there are forces at X and Y.
So I need to fin those forces before first?
 
anthonyk2013 said:
So I need to fin those forces before first?
No, you are supposed to find them graphically.
Start at the point where the load is suspended. There are forces in the two frame members at that point. Between them, these must balance the load. You know which direction each acts in. In geometry terms, you have the length of one side (10N) and two adjacent angles. Draw the triangle.
 
haruspex said:
No, you are supposed to find them graphically.
Start at the point where the load is suspended. There are forces in the two frame members at that point. Between them, these must balance the load. You know which direction each acts in. In geometry terms, you have the length of one side (10N) and two adjacent angles. Draw the triangle.
Draw a closed triangle representing the 3 forces?
I have 10kn going vertically down but 1 and 2 go horizontal.
 
anthonyk2013 said:
Draw a closed triangle representing the 3 forces?
I have 10kn going vertically down but 1 and 2 go horizontal.
No, the three forces acting at the point where the load is suspended. There is the load itself, and the forces in the two frame members that meet at that point. This triangle is just the start of the diagram. More triangles will be added as you extend it out to incorporate all the joints.
 
I have made an attempt at solving the question, I would appreciate if someone could tell me if I'm heading in the right direction or am I a million miles away?
 

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anthonyk2013 said:
I have made an attempt at solving the question, I would appreciate if someone could tell me if I'm heading in the right direction or am I a million miles away?
That's the right start. Your labelling is a bit confusing. I don't know what you've been taught, but I would label each force (one for each strut, plus one for each of the load and the two support forces) in the original diagram and use the same labels in your Bow's diagram. (You could also label joints in the original, but in Bow's each of those becomes a triangle.)
Next step is to add the triangles corresponding to the adjacent joints.
 
haruspex said:
That's the right start. Your labelling is a bit confusing. I don't know what you've been taught, but I would label each force (one for each strut, plus one for each of the load and the two support forces) in the original diagram and use the same labels in your Bow's diagram. (You could also label joints in the original, but in Bow's each of those becomes a triangle.)
Next step is to add the triangles corresponding to the adjacent joints.

Being honest we spent an age doing vector sums and digrams and spent no time on bows notation so I am trying to teach myself, the labelling is something I not 100% sure how to do.

is my next step the joint at the top, if so do I treat it as a 60deg angle or 120deg angle?

do I use the sine rule a/sina=b/sinb=c/sinc

or cos rule a*2=b*2+c*2-2bccosA
 
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  • #10
anthonyk2013 said:
Being honest we spent an age doing vector sums and digrams and spent no time on bows notation so I am trying to teach myself, the labelling is something I not 100% sure how to do.

is my next step the joint at the top, if so do I treat it as a 60deg angle or 120deg angle?

do I use the sine rule a/sina=b/sinb=c/sinc

or cos rule a*2=b*2+c*2-2bccosA
There's one little thing to do before stepping onto another node. You should mark each edge of your triangle with an arrow to denote direction. The three arrows should run in a loop around the triangle. So for the right hand angled truss (call it A) you will get an arrow going up to the left. This tells you A is pulling up and to the left on that joint, as it should.
So let's step up to the joint at the top of A. A will still be represented by the line you have already drawn, but now the arrow is running the other way. Call the other two trusses at the top joint B and C. You now need to draw a line parallel to B from one end of A in Bow's, and a line parallel to C from the other end, and see where they cross. Which you put at which end of A does not matter in principle, but one choice will make the diagram easier to read than the other.
 
  • #11
is this what you mean?
 

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  • #12
anthonyk2013 said:
is this what you mean?
That's right. (I would build it all up in one diagram as long as it doesn't become too messy.) On to the next joint...
 
  • #13
haruspex said:
That's right. (I would build it all up in one diagram as long as it doesn't become too messy.) On to the next joint...

To work out the forces in that triangle do I use the force 11.54 and angle of 60deg?
Or do I use cos and sin rule
 
  • #14
anthonyk2013 said:
To work out the forces in that triangle do I use the force 11.54 and angle of 60deg?
Or do I use cos and sin rule
I thought the point of Bow's method is that you don't use equations at all. You just draw the diagram very accurately. All the lengths will be in proportion to the forces, and you just measure them. Of course, you could use geometry to work out the lengths, and not worry about the diagram being perfect.
 

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