Question

Specify the coordinate direction angles of F1 and F2 and express each force as a Cartesian vector. State the magnitude and coordinate direction angles of the resultant vector. 

Answer 1

F2 force is in - z direction

So F2 = - 130 k

Component of force F1

One component is in z direction and one in X-y plane.

Component In z direction = F1Sin30 = 80×Sin30 = 40 lb

Component In x-y plane = F1 Cos30 = 80×Cos30

= 69.28 lb k

Now component of this component in X and Y direction

In X direction = F1Cos30×Cos40 =53.07 lb i

In Y direction = F2Cos30×Sin40 = 44.53 lb (-j)

So cartesian vector of F1 = 53.07 i - 44.53 j + 69.28 k

So resultant vector of force

= F1 + F2

= 53.07 i - 44.53 j + 69.28 k - 130 k

= 53.07 i - 44.53 j - 60.72 k

So magnitude of resultant = (53.07^2 + 44.53^2 + 60.72^2)^(1/2)

= 92.12 lb

Angle from x axis Cosx = 53.07/92.12

x = 54.82 degree

Angle from y axis Cosy = - 44.53/92.12

y = 118.88 degree

Angle from z axis Cosz = - 60.72/92.12

z = 48.77 degree

These are the angles from all axis.