Question

six forces of magnitude 1,2,3,4,5,6 act along the sides of a regular hexagon taken in order . resultant magnitude is 1)2 (2)3 (3)4 (4)6
pls solve it out

Answer 1

Let us consider a Hexagon ABCDEF with centre O.
Let us consider AB and AE as the axes.If K is the proportionality constant .
Then the forces along the sides AB,BC,CD,DE,,EF and FA are of magnitudes K,2K,3K,4K, 5K and 6K respectively.
Let R be the resultant  vector making an angle θ with AX and at a distance d from the center of hexagon.
Resolving the forces along Ax
we get,
R Cosθ=K+2K Cos 60° +3K cos120° -4K-5K Cos 60° -6K cos 120°
=K+2k.1/2 +3k(-1/2)-4k-5k.1/2 -6K(-1/2)
=K+K-3K/2-4K-5K/2+3K
=-3KCOSθ
=-3K-----------(1)
Resolving along AY ,
we get
R sinθ=2K Sin60° +3K sin120°-5Ksin60° -6Ksin120°
=2Ksin60°+3K sin60° -5ksin60°-6ksin60°
=-6Ksin60°
=-6K.√3/2
=-3√3 .K---------------(2)
R sinθ=-3√3.K
Squaring equation 1 and 2
we get :
R²=9K² +27K²
=36K²
R=6K
So correct option (4)-- 6

Answer 2

Answer:

Hey mate this will surely help you