Find Magnitude coordinate direction angles of resultant for

Find Magnitude & coordinate direction angles of resultant force F_R?

Solution

>> Considering Force F1,

Component along Z Direction = 300*sin60 = 259.81 N

Component in X-Y Plane = 300*cos60 = 150 N

Component along Y direction = 150*cos45 = 106.07 N

and, Component along X Direction = - 106.07 N

=> F1 = - 106.07 i + 106.07 j + 259.81 k

>> Considering F2

Component along X-Direction = 500*cos60 = 250 N

Component along Y-Direction = 500*cos45 = 212.13 N

Component along Z-Direction = 500*cos120 = - 150 N

=> F2 = 250 i + 212.13 j - 150 k

So, Fr = Resultant Force = F1 + F2 = ( - 106.07 i + 106.07 j + 259.81 k ) + (250 i + 212.13 j - 150 k)

=> Fr = 143.93 i + 318.20 j + 109.81 k

Magnitude = [ 143.93² + 318.20² + 109.81² ]^1/2 = 366.10 N ....ANSWER...

So, Coordinate Direction Angles are {cos^-1(Frx/|Fr|) , cos^-1(Fry/|Fr|), cos^-1(Frz/|Fr|) }

= { cos^-1(143.93/366.10) , cos^-1(318.20/366.10), cos^-1(109.81/366.10) }

= { 66.85, 29.64, 72.55 } ........ANSWER.........