Question

deternine the resultant of the following coplanar forces. 15N at 30 degress from the horizaontal, 55N at 80 degrees, 90N at 210 degrees, and 130N at 260 degrees

Answer 1

$\vec F_1= 15N \ \angle(\vec F_1) =30$

$\vec F_2= 55N \ \angle(\vec F_2) =80$

$\vec F_3= 90N \ \angle(\vec F_3) =210$

$\vec F_4= 130N \ \angle(\vec F_4) =260$

$\vec F = \vec F_1+\vec F_2+\vec F_3+\vec F_4$

$\vec{F_5}=\vec{F_1}+\vec{F_3}\ \angle{(\vec{F_3},\vec{F_1})}=210-30=180$

$\vec F_5= 90-15 =75N;\angle(\vec F_5) =210$

$\vec{F_6}=\vec{F_2}+\vec{F_4}\ \angle{(\vec{F_3},\vec{F_1})}=260-80=180$

$\vec F_6= 130-55 =75N;\angle(\vec F_6) =260$

$\vec F = \vec F_5+\vec F_6$

$|\vec F_5|=|\vec F_6|$

$\angle (\vec F_6,\vec F_5)= 260-210 =50$

$\angle(\vec F) =\angle(\vec F_5)+\frac{\angle(\vec F_6)-\angle(\vec F_5)}{2}=235$

$|\vec F|^2= |\vec F_5|^2+ |\vec F_6|^2-2 |\vec F_5| |\vec F_6|\cos\angle (\vec F_6,\vec F_5)$

$|\vec F|^2= 75^2+75^2-2*75^2*\cos(50\degree)=4018.64$

$|\vec F|=\sqrt{4018.64}=63.39$

$\text{Answer: }\vec F=63.39N;\angle(\vec F) =235\degree$