Question

two forces 3N and 4N are acting at an angle 120* between them.find the resultant force in magnitude and direction.

Answer 1

The addition of vectors $a$ and $b$ may be represented graphically by placing the start of the arrow $b$ at the tip of the arrow $a$ , and then drawing an arrow from the start of $a$ to the tip of $b$ . The new arrow drawn represents the vector $a + b$ , as illustrated below:

This addition method is sometimes called the parallelogram rule because a and b form the sides of a parallelogram and $a + b$ is one of the diagonals. If a and b are bound vectors that have the same base point, it will also be the base point of $a + b$ .

Two forces 3N and 4N are acting at an angle $120{}^{\circ}$ between them. Find the resultant force in magnitude and direction.

Let's use the parallelogram rule:

Angle between forces equals 120. Therefore angle $\alpha = 60$ .

Law of cosines:

$$x ^ {2} = 3 ^ {2} + 4 ^ {2} - 2 * 3 * 4 * \cos 6 0$$ $$x ^ {2} = 2 5 - \frac {2 4}{2} = 1 3$$ $$x = \sqrt {1 3}$$

Law of sines:

$$x / \sin (60) = 4 / \sin \alpha$$ $$\sin \alpha = \frac{4}{\sqrt{13}} \frac{\sqrt{3}}{2} = 2 \sqrt{\frac{3}{13}}$$ $$\alpha = \arcsin 2 \sqrt{\frac{3}{13}} = 74$$

Answer: magnitude equals $\sqrt{13}$ and direction is 74 degrees to force 3N.