Question

Three forces of magnitude 30kN,10KN and 15KN are acting at a point O.The angles made by 30KN,10KN and 15KN
force with x-axis are 60,120,240 respectively.
Determine the magnitude and direction of the resultant.​

Answer 1

magnitude of resultant is 21.8 kN and direction is tan-¹(5√3) with x-axis.

Three forces of magnitude 30kN , 10kN and 15kN are acting at a point O.The angles made by 30kN , 10kN and 15kN force with x-axis are 60°,120°, 240° respectively.

write all forces in vector form,

F1 = 30(cos60° i + sin60° j) kN = (15 i + 15√3 j) kN

F2 = 10(cos120° i + sin120° j) kN = (-5i + 5√3 j) kN

F3 = 15(cos240° i + sin240° j) kN = (-7.5 i - 7.5√3 j) kN

now resultant force, F = F1 + F2 + F3

= (15 i + 15√3 j) + (-5i + 5√3 j) + (-7.5 i - 7.5√3 j)

= 2.5 i + 12.5√3 j

hence, magnitude of resultant = √(6.25 + 156.25 × 3) = √(6.25 + 468.75) = √(475) kN

= 21.8 kN

direction of resultant, θ = tan-¹(12.5√3/2.5) = tan-¹(5√3) with x-axis.

Answer 2

$\huge\bold\purple{Answer:-}$

Three forces of magnitude 30kN , 10kN and 15kN are acting at a point O.The angles made by 30kN , 10kN and 15kN force with x-axis are 60°,120°, 240° respectively.

write all forces in vector form,

F1 = 30(cos60° i + sin60° j) kN = (15 i + 15√3 j) kN

F2 = 10(cos120° i + sin120° j) kN = (-5i + 5√3 j) kN

F3 = 15(cos240° i + sin240° j) kN = (-7.5 i - 7.5√3 j) kN

now resultant force, F = F1 + F2 + F3

= (15 i + 15√3 j) + (-5i + 5√3 j) + (-7.5 i - 7.5√3 j)

= 2.5 i + 12.5√3 j

hence, magnitude of resultant = √(6.25 + 156.25 × 3) = √(6.25 + 468.75) = √(475) kN

= 21.8 kN

direction of resultant, θ = tan-¹(12.5√3/2.5) = tan-¹(5√3) with x-axis.