Question

ABCDEF is a regular hexagon. Forces of 4,P,10,Q and 6 N acting along AB,CA, AD,AE and PA respectively are in equilibrium. Find the magnitude of P and Q:

Answer 1

Answer:

R

=

AB

+

AC

+

AC

+

AE

+

AF

=

ED

+

AC

+

AD

+

AE

+

CD

(∵

AB

=

ED

and

AF

=

CD

)

=(

AC

+

CD

)+(

AE

+

ED

)+

AD

=

AD

+

AD

+

AD

=3

AD

=

Answer 2

Given are the forces $4, P,10,Q,6\ N$ along the sides AB, CA, AD, AE, FA of a regular hexagon. Find P and Q if they are in equilibrium.

Explanation:

• Let the Hexagon be considered to rest on the side AB along the horizontal x-axis.
• Now the forces along the sides CA is in the oppositive direction to other forces.
• But, FA is negative in the vertical direction only.
• Now since the hexagon is regular we get the angles made by each of the forces with x-axis as,  
• $F_{AB}= 4\ N\ \ \ \ \ \ ->\theta_1=0\\F_{CA}=P\ N\ \ \ \ \ \ ->\theta_2=\frac{7\pi}{6} \\F_{AD}=10\ N\ \ \ \ \ ->\theta_3=\frac{\pi}{3} \\F_{AE}=Q\ N\ \ \ \ \ \ ->\theta_4=\frac{\pi}{2} \\F_{FA}=6\ N\ \ \ \ \ \ ->\theta_5=\frac{5\pi}{3}$    
• Now the forces to be in equilibrium the sum of their vertical components and horizontal components should be zero.
• Hence, $F_v=F_h=0$  
• $->F_v=4sin0+Psin\frac{7\pi}{6} +10sin\frac{\pi}{3}+Qsin\frac{\pi}{2} +6sin\frac{5\pi}{3} =0\\->F_h=4cos0+Pcos\frac{7\pi}{6} +10cos\frac{\pi}{3}+Qcos\frac{\pi}{2} +6cos\frac{5\pi}{3}=0\\->2\sqrt{3}-\frac{P}{2}+Q=0, \ \ \ 12-\frac{\sqrt{3} }{2}P=0$  ----(a)
• Solving the equations in (a) for P and Q we get, $P=8\sqrt{3} \ N,\ \ Q=2\sqrt{3}\ N$