ABCDEF is a regular hexagon. Find the resultant of the forces P, 2P, 3P, 4P, 5P acting along AB, AC, AD, AE, AF
Answers
Answer:
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
Read more on Brainly.in - https://brainly.in/question/1220160#readmore