Answer to Question #100332 in Mechanics | Relativity for EIRRA EIRWANI

Question

Q5: Determine, by resolution of forces, the resultant of
the following three coplanar forces acting at a point:
12 kN acting at 32° to the horizontal; 25 kN acting at
170° to the horizontal; 40 kN acting at 240° to the
horizontal
Q6: Determine by resolution of forces the resultant of
the following three coplanar forces acting at a point:
300N acting at 20° to the horizontal; 500N acting at
165° to the horizontal; 600N acting at 250° to the
horizontal.

Answer 1

Since we deal with forces having both x- and y-components, we can find the resultant by both axles first and then find the force in "polar" form, i.e. its magnitude and angle above horizontal.

Q5:

$Ox: F_x=12\text{ cos}32^\circ+25\text{ cos}170^\circ+40\text{ cos}240^\circ=\\ =-34\text{ kN}.\\ Oy: F_y=12\text{ sin}32^\circ+25\text{ sin}170^\circ+40\text{ sin}240^\circ=\\ =-24\text{ kN}.$

The magnitude:

$F=\sqrt{F_x^2+F_y^2}=42\text{ kN},$

acting at

$\theta=180^\circ+\text{atan}\frac{F_y}{F_x}=180^\circ+55^\circ=235^\circ.$

Q6 can be solved by analogy:

$Ox: F_x=\\ =300\text{ cos}20^\circ+500\text{ cos}165^\circ+600\text{ cos}250^\circ=\\ =-406\text{ kN}.\\ Oy: F_y=\\=300\text{ sin}20^\circ+500\text{ sin}165^\circ+600\text{ sin}250^\circ=\\\\ =-332\text{ kN}.$

$F=\sqrt{F_x^2+F_y^2}=\\=\sqrt{(-406)^2+(-332)^2}=524\text{ kN},$

$\theta=-141^\circ.$