Physics, asked by Tomboyish44, 10 months ago

In the given figure what is the value of \overrightarrow{ \sf F \ } in terms of the other vectors, if all the vectors lie along the sides of a regular hexagon?

Grade 11, Physics.

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Tomboyish44: There are options given, but I'm not sure if they're correct. Here are the options;

(a) F = A + B + C + D + E
(b) F = (A + B + C + D + E) (It's not modulus, it's a bracket)
(c) F = A + B - C + D + E
(d) F = (A + B - C + D + E)

Answers

Answered by Rohit18Bhadauria
56

To Find:

Value of vector F in terms of other vectors

Diagram:

(See the first attachment)

Solution:

With the help of diagram,

On applying triangle law of vector addition on vectors E, D and M, we get

\longrightarrow\rm{\vec{M}=\vec{D}+\vec{E}}------(1)

On applying triangle law of vector addition on vectors A, B and O, we get

\longrightarrow\rm{\vec{O}=\vec{A}+\vec{B}}------(2)

\rule{190}{1}

Now, on applying triangle law of vector addition on vectors N, O and C, we get

\longrightarrow\rm{\vec{C}=\vec{N}+\vec{O}}

\longrightarrow\rm{\vec{C}-\vec{O}=\vec{N}}

\longrightarrow\rm{\vec{N}=\vec{C}-\vec{O}}------(3)

\rule{190}{1}

Now, on applying triangle law of vector addition on vectors N, F and M, we get

\longrightarrow\rm{\vec{M}=\vec{N}+\vec{F}}

\longrightarrow\rm{\vec{M}-\vec{N}=\vec{F}}

\longrightarrow\rm{\vec{F}=\vec{M}-\vec{N}}

From (1) and (3), we get

\longrightarrow\rm{\vec{F}=\vec{D}+\vec{E}-(\vec{C}-\vec{O})}

\longrightarrow\rm{\vec{F}=\vec{D}+\vec{E}-\vec{C}+\vec{O}}

From (2), we get

\longrightarrow\rm{\vec{F}=\vec{D}+\vec{E}-\vec{C}+\vec{A}+\vec{B}}

or

\longrightarrow\rm\green{\vec{F}=\vec{A}+\vec{B}-\vec{C}+\vec{D}+\vec{E}}

\rule{190}{1}

Triangle Law of Vector Addition

If two vectors A and B lies along two consecutive sides of triangle with Head to tail position, then the third vector R taken opposite in order represents the third side of triangle and is equal to the sum of other two vectors.

i.e R=A+B

(See the diagram provided in second attachment)

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Tomboyish44: Thanks a lot! :)
Answered by Anonymous
31
 see in attachment
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