Question

Find the resultant of the following forces: 3.0N east, 4.0N west, and 5.0N in a direction north 60° west.

Answer 1

Answer:

5.5 N at 50.8° north of west.

Explanation:

To find the resultant of these forces, we have to resolve each force along the x- and y-direction, then find the components of the resultant force, and then calculate the resultant force.

The three forces are:

$F_1=3.0 N$ (east)

$F_2=4.0 N$ (west)

$F_3=5.0 N$ (at 60° north of west)

Taking east as positive x-direction and north as positive y-direction, the components of the forces along the 2 directions are:

$F_{1x}=3.0 N\\F_{1y}=0$

$F_{2x}=-4.0 N\\F_{2y}=0$

$F_{3x}=-(5.0)(cos 60^{\circ})=-2.5 N\\F_{3y}=(5.0)(sin 60^{\circ})=4.3 N$

Threfore, the components of the resultant force are:

$F_x=F_{1x}+F_{2x}+F_{3x}=3.0+(-4.0)+(-2.5)=-3.5 N\\F_y=F_{1y}+F_{2y}+F_{3y}=0+0+4.3=4.3 N$

Therefore, the magnitude of the resultant force is

$F=\sqrt{F_x^2+F_y^2}=\sqrt{(-3.5)^2+(4.3)^2}=5.5 N$

And the direction is:

$\theta=tan^{-1}(\frac{F_y}{|F_x|})=tan^{-1}(\frac{4.3}{3.5})=50.8^{\circ}$

And since the x-component is negative, it means that this angle is measured as north of west.