# Polygon of Forces

• January 2nd 2009, 01:45 PM
David Green
Polygon of Forces
Hello all;

Just got a little query with the final understanding of a problem I have been working on. I will try to make it as brief as possible. I have Four Coplanar Forces, 4N, 8N, 7N and 5N, I have completed a Force diagram and the forces are NOT in equilibrium, their Resultant has been measured at 8.5N, the direction is in the opposite sense to the other forces.

Question.
If the resultant force is that force which would replace those other forces, does this mean that the resultant force is the average of these other forces?

Question 2.
How would a person understand which way the resultant force is applied after drawing a force diagram?

Any help appreciated

Thanks
David(Wondering)
• January 2nd 2009, 02:57 PM
Mush
Quote:

Originally Posted by David Green
Hello all;

Just got a little query with the final understanding of a problem I have been working on. I will try to make it as brief as possible. I have Four Coplanar Forces, 4N, 8N, 7N and 5N, I have completed a Force diagram and the forces are NOT in equilibrium, their Resultant has been measured at 8.5N, the direction is in the opposite sense to the other forces.

Question.
If the resultant force is that force which would replace those other forces, does this mean that the resultant force is the average of these other forces?

Question 2.
How would a person understand which way the resultant force is applied after drawing a force diagram?

Any help appreciated

Thanks
David(Wondering)

Question 1:

Well if you have n, forces, then I would define the average of these forces to be $|F_{ave}| = \frac{\sum_{i=1}^{n}|F_{i}|}{n}$, $\theta_{ave} = \frac{\sum_{i=1}^{n}\theta_{i}}{n}$. And in this sense, no, the resultant force isn't an average. The resultant force is quite simply a force whose magnitude, direction and position reproduce identical statical or dynamical conditions as the single forces that were considered. That means that if 4 forces accelerate an object with an acceleration of 50m/s^2 in the negative x direction, with 0 moments about all points, then the resultant force of these 4 must also produce an acceleration of 50m/s^2 in the negative x direction, with 0 moments about all points. However if you are talking about the direction of a resultant force it is SLIGHTLY like the average direction in some cases. If you have two forces in equal magnitude. On at -45 degrees, and the other at 45 degrees, then indeed the resultant direction is the average of this... 0!. Or if they were at 50 degrees and -40 degrees, the resultant direction would be 5 degrees

Indeed, if you have two forces acting on a body, and they are equal in magnitude, opposite in direction, and colinear, then the average of these forces is 0, and this just happens to be the resultant force also. But in general, the resultant force is not an average.

Question 2:

To replace 4 forces with a single resultant force you need consider a few things.

1) Split your 4 forces into x components and y components, and then sum the x components and sum the y components. The y component of the resultant force is equal to the sum of the y components of the other forces. Same for x.
If you draw these x and y summations, you will get a triangle triangle, and you may use pythagorus and trigonometry to calculate the magnitude and angle of the resultant force.

2) Consider the moments of your force. If your 4 forces create moments about a certain point, then your resultant force must also create the same moment about the same point. You need to choose your point carefully... try to choose a point that most of the forces pass through, because if a force passes through a point, it creates no moment about that point... the more forces that pass through the point you choose, the less work you have to do! Given the magnitude and direction of the resultant force, you should be able to find it's position by calculating it's distance from that point, if the moment it creates is to be equal to the moments created by the other 4 forces.

PS: You said your resultant is in a direction in the opposite sense to the other forces :S... That doesn't make sense! If you have 4 forces all in the same direction, then the resultant will be in the same direction with a larger magnitude...
• January 2nd 2009, 03:45 PM
David Green
Polygon of Forces
Hi Mush

I am new to this forum, the results i gave where abreviated, the forces given were not explained, but i used the descriptive title Polygon which i thought would indicate that the forces were not in the same direction.

The forces are; 4N due south, then 8N due west, then 60 degrees North west, 90 degrees North east. After drawing the Force Diagram to scale I thought i understood it until the resultant was shown in the opposite direction (sense) to the rest of the forces.

I suppose looking at the space diagram and the Force diagram side by side that I don't understand how or where the resultant force of 8.5N has been created from?

My limited understanding is asking the question, if these forces are not in equilibrium, then they must be moving, but in what direction? I looked at the resultant of each pair of adjacent forces and asked which direction the forces would move, finally i asked myelf, the resultant force of 8.5N bridging the forces 5N and 4N in the opposite sense, why would this force replace these other forces?

This is a learning curve for me, please be patient I may sound as though i don't fully understand the idea yet

David(Happy)
• January 2nd 2009, 04:10 PM
Mush
Quote:

Originally Posted by David Green
Hi Mush

I am new to this forum, the results i gave where abreviated, the forces given were not explained, but i used the descriptive title Polygon which i thought would indicate that the forces were not in the same direction.

The forces are; 4N due south, then 8N due west, then 60 degrees North west, 90 degrees North east. After drawing the Force Diagram to scale I thought i understood it until the resultant was shown in the opposite direction (sense) to the rest of the forces.

I suppose looking at the space diagram and the Force diagram side by side that I don't understand how or where the resultant force of 8.5N has been created from?

My limited understanding is asking the question, if these forces are not in equilibrium, then they must be moving, but in what direction? I looked at the resultant of each pair of adjacent forces and asked which direction the forces would move, finally i asked myelf, the resultant force of 8.5N bridging the forces 5N and 4N in the opposite sense, why would this force replace these other forces?

This is a learning curve for me, please be patient I may sound as though i don't fully understand the idea yet

David(Happy)

What are the magnitudes of the forces due NW and E?

Btw, when you say "60 degress" do you mean "60 degrees clockwise from North?"
• January 3rd 2009, 05:45 AM
David Green
Hi Mush

I have been trying to find a way to post the space diagram onto the forum, but am struggling, so I will try to explain it.

Draw a horizontal line, place an arrow pointing to the left on it at the end, write 8N. This vector is due west. Using a protractor,mark a point 60 degrees clockwise from the 8N North West, draw the second vector this is 7N, next using the protractor and moving clockwise mark a point 90 degrees from the 7N vector, draw another vector this is the 5N. The remining angle should be 30 degrees from the 5N vector to the horizontal line due West. Draw the final vector 4N due south. Label the space diagram using Bow's notation in the four quadrants moving clockwise, eg, A B C D. If you can understand this information, please draw the Force diagram and then note the resultant at 58 degrees in the opposite sense at 8.5N. Please advise when completed or if any problems

Thanks

David(Wait)
• January 3rd 2009, 07:30 AM
earboth
Quote:

Originally Posted by David Green
...

Draw a horizontal line, place an arrow pointing to the left on it at the end, write 8N. This vector is due west. Using a protractor,mark a point 60 degrees clockwise from the 8N North West, draw the second vector this is 7N, next using the protractor and moving clockwise mark a point 90 degrees from the 7N vector, draw another vector this is the 5N. The remining angle should be 30 degrees from the 5N vector to the horizontal line due West. Draw the final vector 4N due south.
...

The forces are labeled $F_1, ... F_4$

The sum $F_1 + F_2 = F_{12}$ . Since this is the addition of vectors I've drawn the corresponding parallelograms too. $F_{34}$ is the vector sum of the remaining forces.

The sum $F_{12} + F_{34} = F_{1234}$. (Painted in red)

I added the length of the resultant force and the value of the angle with respect to the North direction.

If and only if you want to stay this system of forces in equilibrium then you have to add a fifth force pointing into the opposite direction of the force $F_{1234}$ with the same magnitude.
• January 3rd 2009, 10:20 AM
David Green
Quote:

Originally Posted by earboth

The forces are labeled $F_1, ... F_4$

The sum $F_1 + F_2 = F_{12}$ . Since this is the addition of vectors I've drawn the corresponding parallelograms too. $F_{34}$ is the vector sum of the remaining forces.

The sum $F_{12} + F_{34} = F_{1234}$. (Painted in red)

I added the length of the resultant force and the value of the angle with respect to the North direction.

If and only if you want to stay this system of forces in equilibrium then you have to add a fifth force pointing into the opposite direction of the force $F_{1234}$ with the same magnitude.

Thanks for the graph data.Looking at the attachment completed on a computer program, your F2 appears to be my Resultant of 8.5N at 58 degrees, your simulation package shows it at 57.5315 degrees. Also your simulation package agrees with me that the resultant is in the opposite direction (sense), I understand the polygon of forces is not in equilibrium, but i don't understand why the Resultant is shown in the opposite sense (direction) which all seem to disagree on at the moment.
• January 3rd 2009, 10:24 AM
David Green
Quote:

Originally Posted by kakajisepucho
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Emilly

Hi Emilly

Thanks for the info, had an online chat with Jimmy who disagrees with the resultant in the opposite direction (sense), however, I know it is correct, earboth has a computer maths program which is on the forum, the results show F2 in the opposite direction (sense) which agrees with the info I have, I just don't understand why it is right?
• January 3rd 2009, 11:30 AM
earboth
Quote:

Originally Posted by David Green
...

Draw a horizontal line, place an arrow = F1 pointing to the left on it at the end, write 8N. This vector is due west. Using a protractor,mark a point 60 degrees clockwise from the 8N North West, draw the second vector = F2 this is 7N, next using the protractor and moving clockwise mark a point 90 degrees from the 7N vector, draw another vector = F3 this is the 5N. The remining angle should be 30 degrees from the 5N vector to the horizontal line due West. Draw the final vector = F4 4N due south.

Quote:

Originally Posted by David Green
Thanks for the graph data.Looking at the attachment completed on a computer program, your F2 appears to be my Resultant of 8.5N at 58 degrees No , your simulation package shows it at 57.5315 degrees. Also your simulation package agrees with me that the resultant is in the opposite direction (Thinking), ...

• January 3rd 2009, 01:52 PM
David Green
Quote:

Originally Posted by earboth

Sorry I am confused, is your F12 the resultant force, if so that would then be the 8.5N.

All this information is very much appreciated, but I have still not got to the bottom of my original purpose, and that was to find out why the resultant force is in the opposite direction?(Thinking)