FE Mechanical Statics Practice Problems

The current FE Mechanical specification assigns 9-14 of the 110 exam questions to Statics. Form A contains 10 original problems in this area. The sample below is published in full, with the same worked-solution format used throughout the book.

A vertical mast is fixed at the origin with its top at point A, 12 m above the base. Two guy cables run from A to ground anchors: cable AB to anchor B, located 9 m from the base along the positive $x$ axis, and cable AC to anchor C, located 5 m from the base along the negative $y$ axis. The tension is 3.0 kN in AB and 2.6 kN in AC. The magnitude of the resultant of the two cable forces acting at A is most nearly:

(A)2.06 kN
(B)3.97 kN
(C)4.80 kN
(D)5.22 kN

Show worked solution

Answer: (D)

The two tensions combine to a resultant of 5.22 kN at A, of which 4.8 kN acts vertically downward along the mast.

Each cable force is its tension scaled by the unit vector along its own leg, so the anchor geometry alone fixes the direction cosines.

F_{A B} = 3.0 \frac{( 9 , 0 , - 12 )}{15} = (1.8, 0, - 2.4) kN

F_{A C} = 2.6 \frac{( 0 , - 5 , - 12 )}{13} = (0, - 1.0, - 2.4) kN

Concurrent forces must be added component by component before any magnitude is extracted.

R = (1.8, - 1.0, - 4.8) kN

R = 1. 8^{2} + 1. 0^{2} + 4. 8^{2} = 5.22 kN

FE Reference Handbook — Statics: Force (Vector Components and Resultant)

Why the other choices appear

(A)Magnitude of the horizontal components only, $1. 8^{2} + 1. 0^{2} = 2.06$ kN, discarding the vertical resultant.
(B)Root-sum-square of the two tension magnitudes, $3. 0^{2} + 2. 6^{2} = 3.97$ kN, as if the cables were mutually perpendicular.
(C)Sum of the vertical components alone, $2.4 + 2.4 = 4.80$ kN.