# Three equal forces each of 1 Newton acts along side BC,CD,DE of a regular hexagon ABCDEF. calculate the magnitude of their resultant.

NASRUDHEEN N
44 Points
7 years ago
Draw the hexagon and name it as said. We consider side BC as vector BC with magnitude 1 Newton and head C as it’s direction is the direction in which C lies as seen from B.Similarly D will be the head of CD and directing it as seen from C. E will be the head directing through the the point E as seen from D, here the vector DE also have 1 N as magnitude.
The length of the resultant vector we get when we add these 3 vectors is the magnitude we need.
i.e      BC+CD+DE
Since vectors obey associative law,
BC+CD+DE=(BC+CD)+DE
From the hexagon,using triangle law of addition we get
BC+CD=BD as resultant vector with direction through point D as seen from B (means head is at D).
now BC+CD+DE=BD+DE=BE
Vector BE is the resultant.
We know a regular hexagon is made of 6 equilateral triangles and diogonals connecting centre is double the length of side. Means diogonal BE have length 2 units if each side is 1 unit. Clearly,  magnitude of the resultant= 2 Newton