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

5.5 N at 50.8° north of west.

Step-by-step 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=√(F_x^2+F_y^2)=√((-3.5)^2+(4.3)^2)=5.5 N`

And the direction is:

`\theta=tan^(-1)((F_y)/(|F_x|))=tan^(-1)((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.