Question

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

Answer 1

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Answer 2

Answer:

The angle for resultant vector 17 N , 7 N and 13 N will be 0°,180° and 90°.

Explanation:

Given that,

Two forces vector is

Vector A = 5 N

Vector B = 12 N

Resultant = 17 N

The angle will be

Using analytical method of vector addition

$\vec{R}= \sqrt{A^2+B^2+2ABcos\theta}$

$17 N= \sqrt{(5)^2+(12)^2+2\times5\times 12cos\theta}$

$(17)^2=(25+144+120cos\theta)$

$289=169+120cos\theta$

$289-169=120cos\theta$

$120=120cos\theta$

$cos\theta=\dfrac{120}{120}$

$cos\theta=1$

$cos\theta = cos0^{\circ}$

$\theta = 0^{\circ}$

Now, The resultant vector is 7 N.

The angle will be

$\vec{R}= \sqrt{A^2+B^2+2ABcos\theta}$

$7 N= \sqrt{(5)^2+(12)^2+2\times5\times 12cos\theta}$

$(7)^2=(25+144+120cos\theta)$

$49=169+120cos\theta$

$49-169=120cos\theta$

$-120=120cos\theta$

$cos\theta=-\dfrac{120}{120}$

$cos\theta=-1$

$cos\theta = cos180^{\circ}$

$\theta = 180^{\circ}$

Now, The resultant vector is 13 N.

The angle will be

$\vec{R}= \sqrt{A^2+B^2+2ABcos\theta}$

$13 N= \sqrt{(5)^2+(12)^2+2\times5\times 12cos\theta}$

$(13)^2=(25+144+120cos\theta)$

$169=169+120cos\theta$

$169-169=120cos\theta$

$0=120cos\theta$

$cos\theta=0$

$cos\theta = cos90^{\circ}$

$\theta = 90^{\circ}$

Hence, The angle for resultant vector 17 N , 7 N and 13 N will be 0°,180° and 90°.