A force 1 of magnitude 3 kN is acting in a direction 80o from a force 2 of magnitude 8 kN. resulting force will be?

Question

anonymous · Answer

The resulting vector of two coplanar vector can be calculated by trigonometry using "the cosine rule" for a non-right-angled triangle.

$$F _{R}  = [ F1^2 + F2^2 − 2 F_1 F_2 \cos(180^o  - (α + β)) ]^ \frac {1  }{2  } $$

where

F = the vector quantity - force, velocity etc.

α + β = angle between vector 1 and 2

The angle between the vector and the resulting vector can be calculated using "the sine rule" for a non-right-angled triangle.

$$α = \sin^-1 [ F1  \sin(180^o - (α + β)) / F_R ]    $$      (2)

where

α + β = the angle between vector 1 and 2 is known

anonymous · Answer

@rishab.mission @mathlover
Do you satisfy with the answer 

FR = [ (3 kN)^2 + (8 kN)^2 - 2 (5 kN)(8 kN) cos(180o - (80o)) ]^1/2

    = 9 kN

The angle between vector 1 and the resulting vector can be calculated as

$$α = \sin-1[ (3 kN) \sin(180o - (80o)) / (9 kN) ] $$

    $$= 19.1^o $$
The angle between vector 2 and the resulting vector can be calculated as

$$α = \sin^-1[ (8 kN) \sin(180o - (80o)) / (9 kN) ] $$

    $$= 60.9^o $$