Answer to Question #128995 in Physics for ay

Question

Find the sum of these four vector forces: 12.5 N to the right at 35° above the horizontal, 31.8 N to
the left at 55° above the horizontal, 81.4 N to the left at 30° below the horizontal and 24.0 N to
the right at 50° below the horizontal.

Answer 1

The following simple rule will help us to find the vector sum. Projections pointing upward and to the right are positive. Projections looking downward and to the left are negative. Thus, find the components of each vector.

1) 12.5 N to the right at 35° above the horizontal:

$$F_x=12.5\text{ cos}35°=10.2\text{ N},\\ F_y=12.5\text{ sin}35°=7.17\text{ N}.$$

2) 31.8 N to the left at 55° above the horizontal

$$F_x=-31.8\text{ cos}55°=-18.2\text{ N},\\ F_y=31.8\text{ sin}55°=26\text{ N}$$

3) 81.4 N to the left at 30° below the horizontal

$$F_x=-81.4\text{ cos}30°=-70.5\text{ N},\\ F_y=-81.4\text{ sin}30°=-40.7\text{ N}$$

4) 24.0 N to the right at 50° below the horizontal.

$$F_x=12.5\text{ cos}35°=10.2\text{ N},\\ F_y=-12.5\text{ sin}35°=-7.2\text{ N}.$$

To find the x- and y-components of the resultant vector, simply add all x- and y-components algebraically:

$$X=10.2-18.2-70.5+10.2=-68.3\text{ N},\\ Y=7.17+26-40.7-7.2=-14.4\text{ N}.\\$$

The magnitude (length) of the resultant vector (that looks, as follows from X and Y, to the bottom left corner) is

$$F_\Sigma=\sqrt{X^2+Y^2}=\\=\sqrt{(-68.3)^2+(-14.4)^2}=69.8\text{ N}.$$

The angle below the horizontal is

$$\theta=\text{atan}\frac{Y}{X}=\text{atan}\frac{14.4}{68.3}=11.9°.$$