Question

F1= 4 N 45O North of east F2= 8 N 30o North of west F3= 3 N. 20o South of West F4= 6.5 N North

F5= 2.5 N. East FR=

Answer 1

Let's first find $x$- and $y$-components of the resultant force:

$F_{res,x}=4\ N\times cos45^{\circ}+8\ N\times cos(180^{\circ}-30^{\circ})+3\ N\times cos(180^{\circ}+20^{\circ})+6.5\ N\times cos90^{\circ}+2.5\ N\times cos0^{\circ}=-4.42\ N,$$F_{res,y}=4\ N\times sin45^{\circ}+8\ N\times sin(180^{\circ}-30^{\circ})+3\ N\times sin(180^{\circ}+20^{\circ})+6.5\ N\times sin90^{\circ}+2.5\ N\times sin0^{\circ}=12.3\ N.$

We can find the magnitude of the resultant force from the Pythagorean theorem:

$$F_{res}=\sqrt{F_{res,x}^2+F_{res,y}^2}=\sqrt{(-4.42\ N)^2+(12.3\ N)^2}=13.1\ N.$$

We can find the direction of the resultant force from the geometry:

$$\theta=tan^{-1}(\dfrac{F_{res,y}}{F_{res,x}}),$$ $$\theta=tan^{-1}({\dfrac{12.3\ N}{-4.42\ N}})=70.2^{\circ}\ N\ of\ W.$$