Question

TWO FORCES 50 KILO NEWTON AND 10 KILO NEWTON ACT AT A POINT 'O'. THE INCLUDED ANGLE BETWEEN THEM IS 60 DEGREE. FIND THE MAGNITUDE AND DIRECTION OF THE RESULTANT?

Answer 1

Task. Two forces 50 kilo newton and 10 kilo newton act at a point '0'. The included angle between them is 60 degree. Find the magnitude and direction of the resultant?

Solution. Choose coordinates $(x,y)$ so that the first force $F_{1}$ is parallel to $x$-axis, so $\vec{F}_{1} = (50,0)$. Then the second force constitute the angle 60 degree, and has length 10. Therefore

$$\vec{F}_{2} = (10 \cos 60{}^{\circ}, 10 \sin 60{}^{\circ}) = \left(10 \cdot \frac{1}{2}, 10 \cdot \frac{\sqrt{3}}{2}\right) = (5, 5\sqrt{3}).$$

Therefore the resultant

$$\vec{F} = \vec{F}_{1} + \vec{F}_{2} = (50 + 5, 0 + 5\sqrt{3}) = (55, 5\sqrt{3}).$$

The length of this vector is

$$F = \sqrt{55^{2} + (5\sqrt{3})^{2}} = \sqrt{55^{2} + 25 \cdot 3} = \sqrt{3100} = 10\sqrt{31} \approx 55.68.$$

Let $\alpha$ be the angle between vector $\vec{F}$ and $x$-axis. Then

$$\cos \alpha = \frac{55}{|F|} = \frac{55}{10\sqrt{31}} = \frac{5.5}{\sqrt{31}} \approx 0.98783$$

whence

$$\alpha = \arccos(0.98783) \approx 8.96{}^{\circ}.$$

Answer. Magnitude of the resultant: $|F| = 10\sqrt{31} \approx 55.68$.

Direction of the resultant: the angle between $F$ and $F_{1}$ is $8.96{}^{\circ}$.