Question

ABCDEF is a regular hexagon.forces 3KN,P KN,Q KN,5KN,and 4KN act along AB,AC,AD,AE & AF respectively and are in equilibrium, determine the value of p& q

Answer 1

Given are five forces acting in a regular hexagon, Find the value of P and Q for them to be in equilibrium.

Explanation:

• As we know that in a regular polygon all the sides and angles are equal.
• In a regular hexagon all the angles are equal to $\frac{2 \pi}{3}$.
• If the angle made by a Force vector with the horizontal axis is $\theta$ then the force can be obtained along each axis by, $\vec F=|\vec F|cos\theta\hat i+|\vec F|sin\theta\hat j$  
• Here let the hexagon ABCDEF be oriented with AB along the horizontal axis. Then, the angles made by each force are,    $|\vec {AB}|=F_1=3\ kN\ \ \ \ \ \ \ \ \theta_1=0\\|\vec {AC}|=F_2=P\ kN\ \ \ \ \ \ \ \ \theta_2=\frac{\pi}{6}\\|\vec {AD}|=F_3=Q\ kN\ \ \ \ \ \ \ \ \theta_3=\frac{\pi}{3}\\ |\vec {AE}|=F_4=5\ kN\ \ \ \ \ \ \ \ \theta_4=\frac{\pi}{2}\\|\vec {AF}|=F_5=4\ kN\ \ \ \ \ \ \ \ \theta_5=\frac{2\pi}{3}$  
• Hence, for the forces to be in equilibrium the sum of both horizontal and vertical components should be zero, $H= F_1cos\theta_1+F_2cos\theta_2+F_3cos\theta_3+F_4cos\theta_4+F_5cos\theta_5 \\V=F_1sin\theta_1+F_2sin\theta_2+F_3sin\theta_3+F_4sin\theta_4+F_5sin\theta_5\\->4cos0+Pcos\frac{\pi}{6}+ Qcos\frac{\pi}{3}+5cos\frac{\pi}{2}+4cos\frac{2 \pi}{3}=0\ \ \ \ \ \ \ \ \ \ \ \\->4sin0+Psin\frac{\pi}{6}+ Qsin\frac{\pi}{3}+5sin\frac{\pi}{2}+4sin\frac{2 \pi}{3}=0\\->\frac{P\sqrt 3}{2}+\frac{Q}{2}+2=0\ \ \ \ \ \ \ \ \ \ \ \ \------(a)\\-> \frac{P}{2}+\frac{Q\sqrt 3}{2}+5+2\sqrt 3=0 \ \ \ \ \ \ \ \ -----(b)$    
• Solving equations (a) and (b) for P and Q we get,  $P= 5\ kN\ \ \ \ \ \ Q=-(5\sqrt 3+ 4)\ kN \ \ (->Q\ is\ along\ DA )$