FE Mechanical Statics

Knowledge area 6 of the FE Mechanical exam: 9–14 of the 110 questions. The subtopics NCEES lists, how they are tested and worked practice problems with solutions.

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110 questions on the real 5 hours 20 minutes clock, weighted to the NCEES specifications, with a score by knowledge area and a worked solution for every answer.

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What the specification lists

  • Resultants of force systems
  • Concurrent force systems
  • Equilibrium of rigid bodies
  • Frames and trusses
  • Centroids and moments of inertia
  • Static friction

From the NCEES FE Mechanical CBT exam specifications, knowledge area 6.

Line drawing of a steel truss bridge resting on a pin and a roller support

How Statics is tested

Statics questions ask for support reactions, truss member forces, centroids, moments of inertia and friction limits. Draw the free-body diagram first, take moments about the point that removes the most unknowns, and check units. The Reference Handbook section on statics lists centroids and moments of inertia of common shapes.

On exam day the relationships you need are in the Reference Handbook section on Statics. The handbook is the only reference allowed, so practice finding each formula in it rather than memorizing a separate sheet.

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FE Mechanical Statics practice problems

Work each problem with your calculator and the handbook, then open the solution.

Problem 1 · Resultants of force systems

A horizontal beam carries three downward loads: 10 kN at 1 m, 20 kN at 3 m, and 15 kN at 6 m, all measured from the left end. The single resultant force acts at a distance from the left end of most nearly:

  1. A.2.44 m
  2. B.3.00 m
  3. C.3.33 m
  4. D.3.56 m
Show the answer and solution

Answer: D: 3.56 m

R = 10 + 20 + 15 = 45 kN. x̄ = Σ(Fx)/R = (10·1 + 20·3 + 15·6)/45 = 160/45 = 3.56 m. The positions must be weighted by the loads.

Resultants of force systems · Reference Handbook: Statics

Problem 2 · Equivalent force systems

An 800 N force acts at point P, located 1.2 m to the right of and 0.5 m above point O. The force is directed downward and to the right at 60° below the horizontal. The force is replaced by an equivalent force-couple system at O. The magnitude of the couple is most nearly:

  1. A.631 N·m
  2. B.831 N·m
  3. C.960 N·m
  4. D.1,031 N·m
Show the answer and solution

Answer: D: 1,031 N·m

Fx = 800 cos 60° = 400 N (to the right) and Fy = −800 sin 60° = −692.8 N. M_O = xFy − yFx = 1.2(−692.8) − 0.5(400) = −831.4 − 200 = −1,031 N·m (clockwise). Both components turn the body clockwise about O, so their moments add.

Equivalent force systems · Reference Handbook: Statics

Problem 3 · Force couple systems

Select all that apply.

Which statements about a couple acting on a rigid body are true? Select all that apply.

  1. A.The net force of a couple is zero
  2. B.The moment of a couple is the same about every point
  3. C.A couple can be moved anywhere in its plane without changing its external effect on the body
  4. D.The moment of a couple depends on the point about which it is computed
  5. E.A couple can be balanced by a single force
Show the answer and solution

Answer: A: The net force of a couple is zero; B: The moment of a couple is the same about every point; C: A couple can be moved anywhere in its plane without changing its external effect on the body

A couple is two equal, opposite, noncollinear forces. Its net force is zero and its moment equals one force times the perpendicular distance between them, regardless of the reference point. It is therefore a free vector that can be moved within its plane, and only another couple can balance it.

Force couple systems · Reference Handbook: Statics

Problem 4 · Equilibrium of rigid bodies

A 10 ft beam is supported by a pin at its left end A and a roller at B, 7 ft from A. The beam overhangs B by 3 ft and carries a 1,200 lbf downward load at its free right end. Neglect beam weight. The reaction at roller B is most nearly:

  1. A.514 lbf
  2. B.840 lbf
  3. C.1,200 lbf
  4. D.1,714 lbf
Show the answer and solution

Answer: D: 1,714 lbf

ΣM_A = 0: R_B(7) = 1,200(10), so R_B = 1,714 lbf upward. Vertical equilibrium then gives R_A = 1,200 − 1,714 = −514 lbf, meaning the pin pulls the beam down. An overhanging load makes the nearer support carry more than the load itself.

Equilibrium of rigid bodies · Reference Handbook: Statics

More FE Mechanical practice problems from every knowledge area

Sit the full exam under the real clock

Take a full timed FE Mechanical practice exam

110 questions on the real 5 hours 20 minutes clock, weighted to the NCEES specifications, with a score by knowledge area and a worked solution for every answer.

Or start the free Statics drill

Common questions

How many Statics questions are on the FE Mechanical exam?

The NCEES FE Mechanical specification lists 9–14 questions on Statics, out of 110. Every examinee gets the same number per knowledge area.

What does Statics cover on the FE Mechanical exam?

Resultants of force systems; Concurrent force systems; Equilibrium of rigid bodies; Frames and trusses; Centroids and moments of inertia; Static friction.

Where are the Statics formulas on exam day?

In the NCEES FE Reference Handbook, which appears on screen during the exam: the Statics section. Practice with your free copy from MyNCEES so you know where each relationship sits.

Original independent practice questions written to the published NCEES exam specifications. Not affiliated with or endorsed by NCEES (National Council of Examiners for Engineering and Surveying).