Physics, asked by singhsaksham208, 10 months ago

Three equal forces each of the value 1n acts along the side bc, cd, and de of a regular hexagon abcdef . calculate the magnitude of their resultant .

Answers

Answered by vasantha2582
6

Answer:

The magnitude of force is: F=265lbF=265lb

The distance b is: b=15in

 

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Three equal forces each of 1 Newton acts...

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

2 years ago

Answers : (1)

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

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