# Thread: one more vector problem

1. ## one more vector problem

The quiz is as under:

two forces whose magnitudes are in the ratio 3:7 give a resultant of 3 sq.root of 79 N. If the angle of inclination be 60 degrees, find the magnitude of each force.

F1 be 3x and F2 be 7x. Resultant F is sq.root of 79. Since the inclination is 60 degree, I take the angle as 30 degrees. One side of the triangle is 3x, second one is 7x and third one is 3 sq.root 79 and one of the angles of the triangle is 30 degrees.

Whether my assumptions are correct and if so, how to go about?

Kindly guide me.

2. ## Re: one more vector problem

Hello, arangu1508!

$\text{Two forces whose magnitudes are in the ratio }3:7\text{ give a resultant of }3\sqrt{79}\text{ N.}$
$\text{If the angle of inclination is }60^o\text{, find the magnitude of each force.}$

$F_1 = 3x,\;F_2 = 7x,\;\text{ resultant }F = 3\sqrt{79}.$
$\text{Since the inclination is }60^o\text{, I take the angle as }30^o.$
$\text{One side of the triangle is }3x\text{, second is }7x\text{, and third is }3\sqrt{79}$
$\text{And one of the angles of the triangle is }30^o.$ . No!

You assumed that the diagonals of a parallelogram bisect the vertex angles.
. . This is not true . . .

We have this parallelogram:
Code:
                    7x
*  *  *  *  *  *  *
*                 *
*                 *
3x *                 * 3x
*                 *
* 60          120 *
*  *  *  *  *  *  *
7x
which gives us this triangle:
Code:
                              *
*  *
__     *     *
3√79 *        * 3x
*           *
*          120 *
*  *  *  *  *  *  *
7x
Now we can use the Law of Cosines:
. . $\big(3\sqrt{79}\big)^2 \:=\:(3x)^2 + (7x)^2 - 2(3x)(7x)\cos120^o$

Got it?

3. ## Re: one more vector problem

Mr. Soroban,

Thank you. I got it.

with regards,

Aranga