Question

There are two force vectors, one of 5 N and other of 12 N at what angle two vectors be added to get resultant vector of 17 N, 7 N and 13 N respectively

Answer 1

AnswEr :

Two force vectors 5N and 12N are resolved at an angle ∅.

Law of Cosines

$\sf R = \sqrt{A^2 + B^2 + 2AB cos \phi}$

When the resultant is 17 N,the angle is :

$\sf \: 17 = \sqrt{ {5}^{2} + {12}^{2} + 2(5)(12)cos \phi } \\ \\ \longrightarrow \: \sf \: 289 = 169 + 120cos \phi \\ \\ \longrightarrow \: \sf \: 120cos \phi = 120 \\ \\ \longrightarrow \: \sf \: cos \phi =1 (cos0)\\ \\ \longrightarrow \boxed{\boxed{ \sf \phi = 0 {}^{ \circ}} }$

The two vectors are parallel to eachother

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When the resultant is 7 N,the ange is :

$\sf7 = \sqrt{ {5}^{2} + {12}^{2} + 2(5)(12)cos \phi } \\ \\ \longrightarrow \: \sf49 = 169 + 120 cos \phi \\ \\ \longrightarrow \: \sf \: 120cos \phi = - 120 \\ \\ \longrightarrow \: \sf \:cos \phi = - 1(cos180) \\ \\ \longrightarrow \: \boxed{ \boxed{ \sf \phi = {180}^{ \circ}}}$

The vectors are antiparallel to eachother

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When the resultant is 13 N,the angle is :

$\sf13 = \sqrt{ {5}^{2} + {12}^{2} + 2(5)(12)cos \phi } \\ \\ \longrightarrow \: \sf169 = 169 + 120 cos \phi \\ \\ \longrightarrow \: \sf \: 120cos \phi = 0 \\ \\ \longrightarrow \: \sf \:cos \phi = cos90 \\ \\ \longrightarrow \: \boxed{ \boxed{ \sf \phi = {90}^{ \circ}}}$

The vectors are orthogonally inclined to eachother

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