Question

Six forces of equal magnllude F act on a particle such that particle remalns in equilibrium
FD) Angle between all adjacent pau of forces is 60%
P"
Column-li
Column-
(A Magnitude is equal to 3F
Bi Magnitude is equal to 27
ich Magnitude is greater than
(D) Magnitude is less than ve
(9) 5, +Fz+iz + F
(
RF + F + Fs
(s) 7, + Fz+is+1
mhi+F3 + F1+Fs​

Answer 1

vector addition can be given as

$r = \sqrt{ {a}^{2} + {b}^{2} + 2ab \cos( \alpha ) }$

so for f1 +f3, angle between them is 120

the resultant force will be

$r = \sqrt{ {f}^{2} + {f}^{2} + 2 {f}^{2} \cos(120) } \\ r = f$

because all forces are of equal magnitude and cos120 =-0.5

for f1 + f2 + f3 +f4, f1 and f4 are in opposite directions so they cancel eachother. the resultant will be f2+f3

$r = \sqrt{ {f}^{2} + {f}^{2} + 2 {f}^{2} \cos(60 ) } \\ r = \sqrt{3} f$

f3+f4+f5 will be greater then

$\sqrt{3} f$

f1+f2+f3+f6 will be

$\sqrt{3} f$

and last one will be 0