Problem 3.149

The 15-ft boom AB has a fixedend A. A steel cable is stretchedfrom the free end B of the boomto a point C located on thevertical wall. If the tension in thecable is 570 lb, determine themoment about A of the forceexerted by the cable at B.

C

B

A

x

y

z

15 ft

6 ft10 ft

1. Determine the rectangular components of a force defined byits magnitude and direction. If the direction of the force isdefined by two points located on its line of action, the force canbe expressed by: F = F = (dx i + dy j + dz k)

Fd

C

B

A

x

y

z

15 ft

6 ft10 ft Solving Problems on Your Own

The 15-ft boom AB has a fixedend A. A steel cable is stretchedfrom the free end B of the boomto a point C located on thevertical wall. If the tension in thecable is 570 lb, determine themoment about A of the forceexerted by the cable at B.

Problem 3.149

2. Compute the moment of a force in three dimensions. If r is aposition vector and F is the force the moment M is given by:

M = r x F

C

B

A

x

y

z

15 ft

6 ft10 ft Solving Problems on Your Own

The 15-ft boom AB has a fixedend A. A steel cable is stretchedfrom the free end B of the boomto a point C located on thevertical wall. If the tension in thecable is 570 lb, determine themoment about A of the forceexerted by the cable at B.

Problem 3.149

Problem 3.149 Solution

Determine the rectangularcomponents of a force definedby its magnitude and direction.

First note:

dBC = (_15)2 + (6) 2 + (_10) 2

dBC = 19 ft

Then:

TBC = (_15 i + 6 j _ 10 k) = _ (450 lb) i + (180 lb) j _ (300 lb)k570 lb19

570 N

C

B

A

x

y

z

15 ft

6 ft10 ft

Problem 3.149 Solution

Compute the moment of aforce in three dimensions.

Have:

MA = rB/A x TBC

Where: rB/A = (15 ft) i

Then:

MA = 15 i x (_ 450 i + 180 j _ 300 k)

MA = (4500 lb.ft) j + (2700 lb.ft) k

570 N

C

B

A

x

y

z

15 ft

6 ft10 ft