the problem is 2126 with the picture below ot axes OA of the

the problem is 2-126 with the picture below ot

axes OA of the fla each cable. 26 Determine the magnitude of the projected I 26 component the pipe. 1-127, Determine the angle between pipe segments BA t of the 100-lb force acting along the axis BC of 2-127 and BC 6 3 ft 8 ft 6 ft r4 ft F= 10 lb 0 Probs. 2-126/127

Solution

force F is along CD , so

CD = D - C = (0i + 12j + 0k) - (6i +4j -2k) = -6i + 8j + 2k

direction vector CDcap = -6i + 8j + 2k / (sqrt(6^2+8^2+2^2) = (-6i + 8j + 2k) / 10.20

vector F = 100(-6i + 8j + 2k) / 10.20

BC = C - B = (6i +4j -2k) - (0i + 0j + 0k) = (6i +4j -2k)

|BC| = sqrt(6^2 + 4^2 + 2^2) = 7.48

magnitude of force along BC = F.BC / |BC| = [ (100(-6i + 8j + 2k).(6i +4j -2k) / 10.20 )]/(7.48)

= |100 x (-36 + 32 -4)/ (10.20 x 7.48)| = 10.48

component of F along BC = 10.48((6i +4j -2k) / 7.48) = 8.4i + 5.6j - 2.8k lb ......Ans

127.

BC = (6i +4j -2k)

BA = A - B = -3i + 0j + 0k

cos@ = (BC.BA) / (|BC|)(|BA|)

cos@ = (-18 / (3 x 7.48)) = - 0.802

@ = 143.33 deg