Calculating equilibrant in regular hexagon

In summary, the conversation was about finding the equilibrant of given forces in a regular hexagon and verifying that it is equal and opposite to their resultant. The conversation included discussions about vector addition and drawing diagrams to better visualize the problem. The correct solution involved using components and trigonometry.
  • #1
sareba
8
0

Homework Statement


In a regular hexagon, ABCDEF, forces of magnitude 2N, 4N, 3N and 2N act along the lines AB, AC, AD and AF respectively. Find the equilbrant of the given forces and verify that is equal and opposite to their resultant.



The Attempt at a Solution


I realized that AB + BC is equal to AC, which is 4N, so BC should be 2N and AC + CD is equal to AD, which is equal to 3N, so CD should be equal to -1N? but also since AC is parallel to and twice as much as BC it should be 4N but it is 3N so which of my assumption is wrong? Also resolving didnt get me the right answer maybe i didnt do it right.

The answer is 8.7 N at 46.7° to BA.
P.S I am new to this forum so go easy if i didnt post it right or anything. :D
 
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  • #2
I realized that AB + BC is equal to AC, which is 4N, so BC should be 2N

No. AB and BC are at different angles so it's not simply a matter of AB + BC = AC. The only way 2 + 2 = 4 in vector addition is if the two vectors are parallel and they aren't.

I make the drawing something like this..
 

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  • #3
hi sareba! :smile:
sareba said:
I realized that AB + BC is equal to AC, which is 4N, so BC should be 2N and AC + CD is equal to AD, which is equal to 3N, so CD should be equal to -1N? but also since AC is parallel to and twice as much as BC it should be 4N but it is 3N so which of my assumption is wrong? Also resolving didnt get me the right answer maybe i didnt do it right.

to be honest, i don't understand any of this …

you seem to picturing it the wrong way :confused:

look at CWatters' :smile: excellent diagram, and start again

(you'll probably have to add the components, using sin and cos :wink:)
 
  • #4
I think I see his problem. If you mistakenly draw the vector acting in the direction AB the full length of the side AB (and others likewise) then easy to confuse the hexagon with a vector head-to-tail diagram.

For example this diagram is totally incorrect but it might fool you into thinking that AB+BC=AC. However it's not to scale. If the sides were 2 units long then AC would not be 4 units long nor would AD be 3 units long.
 

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  • #5
Oh... I understand now! *facepalm* I must be so dumb. Anyway thanks a lot guys! Really impressed. :approve:
 

Related to Calculating equilibrant in regular hexagon

1. What is an equilibrant in a regular hexagon?

An equilibrant in a regular hexagon is a force that is equal in magnitude but opposite in direction to the resultant force acting on the object. It is the force that is needed to balance out the other forces acting on the object and bring it into a state of equilibrium.

2. How do you calculate the equilibrant in a regular hexagon?

To calculate the equilibrant in a regular hexagon, you can use the method of vector addition. First, draw a diagram of the hexagon and label the sides and angles. Then, use the Pythagorean theorem to find the length of the resultant force. Finally, use basic trigonometry to find the magnitude and direction of the equilibrant force.

3. Why is calculating the equilibrant important in a regular hexagon?

Calculating the equilibrant in a regular hexagon is important because it helps us understand the forces acting on an object and how they are balanced. This is crucial in engineering and physics, as it allows us to predict the behavior of objects and ensure their stability.

4. Can the equilibrant ever be greater than the resultant force in a regular hexagon?

No, the equilibrant force can never be greater than the resultant force in a regular hexagon. This is because the equilibrant is meant to balance out the other forces, so it must have an equal but opposite magnitude to the resultant force.

5. Are there any real-life applications of calculating the equilibrant in a regular hexagon?

Yes, there are many real-life applications of calculating the equilibrant in a regular hexagon. This concept is used in structural engineering to design stable and balanced structures, as well as in physics to understand the forces acting on objects in motion. It is also used in sports, such as in gymnastics or diving, to maintain balance and achieve precision in movements.

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