Answer to Question #244900 in Mechanics | Relativity for jaen

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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