1. Problem 2.133 y A Cable AB is 65 ft long, and 56 ft
the tension in that cable is
3900 lb. Determine (a) the
x, y, and z components of
the force exerted by the
cable on the anchor B, (b)
the angles qx, qy, and qz
defining the direction of
that force. D a O B
20o
50o z C x

2. Solving Problems on Your Ownx y z
56 ft D B C A
50o
20o a
Problem
Solving Problems on Your Own
Cable AB is 65 ft long, and
the tension in that cable is
3900 lb. Determine (a) the
x, y, and z components of
the force exerted by the
cable on the anchor B, (b)
the angles qx, qy, and qz
defining the direction of
that force.
1. Determine the rectangular components of a force defined by
its magnitude and direction. If the direction of the force F is
defined by the angles qy and f, projections of F through these
angles or their components will yield the components of F.

3. Solving Problems on Your Ownx y z
56 ft D B C A
50o
20o a
Problem
Solving Problems on Your Own
Cable AB is 65 ft long, and
the tension in that cable is
3900 lb. Determine (a) the
x, y, and z components of
the force exerted by the
cable on the anchor B, (b)
the angles qx, qy, and qz
defining the direction of
that force.
2. Determine the direction cosines of the line of action of a force.
The direction cosines of the line of action of a force F are
determined by dividing the components of the force by F.
cos qx= Fx F
cos qy= Fy F
cos qz= Fz F

4. Fy = + F cos qy = (3900 lb)(0.86154) Fy = + 3360 lb
Problem Solution A
Determine the direction cosines of
the line of action of a force.
65 ft qy
From triangle AOB:
cos qy =
56 ft
65 ft =
qy = 30.51o F Fy
56 ft Fx O B
(a) Fx = _ F sin qy cos 20o
= _ (3900 lb) sin 30.51o cos 20o
Fx = _1861 lb
20o Fz z x
Fy = + F cos qy = (3900 lb)( ) Fy = lb
Fz = + (3900 lb) sin 30.51o sin 20o Fz = lb

5. Fx F = _1861 lb 3900 lb cos qx = _ 0.4771 qx = 118.5o cos qz = Fz Fy
Problem Solution O x B A Fy Fz F Fx
20o qy
65 ft
Determine the direction cosines
of the line of action of a force.
(b) cos qx = Fx F =
_1861 lb
3900 lb
56 ft
cos qx = _
qx = 118.5o
From above (a): qy = 30.5o
cos qz = Fz F
= +
677 lb
3900 lb
= qz = 80.0o