Statics is 9–14 questions on the exam and the base of the next two areas: a wrong reaction here becomes a wrong moment diagram in mechanics of materials and a wrong bolt load in machine design. FE Civil has its own statics area (8–12 questions); this site practices it with structural scenarios on the civil pages, and the general mechanics lives here.
Updated Oct 29, 2026 · Not affiliated with NCEES.
ExamFE Mechanical
questions on the exam9–14
worked problems here16
of 14 knowledge areas6
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What the area covers
The July 2020 specification lists six topics. In plain terms:
Resultants of force systems: replacing several forces and couples by one force and one couple, in 2D and 3D, with vector cross products for moments.
Concurrent force systems: forces through one point, solved by components or by the force polygon.
Equilibrium of rigid bodies: support reactions in 2D and 3D, two-force and three-force members, statically determinate checks.
Frames and trusses: method of joints, method of sections, and frames with multi-force members.
Centroids and moments of inertia: composite areas, the parallel-axis theorem, radius of gyration.
Static friction: impending slip and tipping on flat and inclined surfaces, wedges, belts.
16 statics problems, every option explained
0 / 16 answered · 0 correct
Question 1 of 16 · Statics · 9–14 items
A force F = (100î + 200ĵ − 50k̂) N is applied at point A (0, 0, 5) m on a rigid structure. The structure is designed to rotate about an axis passing through the origin (0, 0, 0) and point B (3, 4, 0) m. Determine the magnitude of the moment tending to rotate the structure about axis OB.
Given
Force F = 100i + 200j − 50k N
Point of application A = (0, 0, 5) m
Axis through O = (0, 0, 0) and B = (3, 4, 0) m
Find
Moment about axis OB (N·m)
Solution
MO = r × F = (5k) × (100i + 200j − 50k) = −1000i + 500j N·m
FE Reference Handbook, Statics: moment of a force about an axis.
Projecting MO = (−1000, 500, 0) N·m on the unit vector of OB gives −600 + 400 = −200 N·m, so the magnitude is 200 N·m.
This is only the j-component of MO; the moment about the axis needs the dot product with the axis unit vector.
This keeps only the x-term of the dot product (−1000 × 0.6) and drops the +400 contributed by the y-component.
This is the size of the i-component of r × F alone, not its projection on line OB.
This is |MO|, the full moment about point O; the moment about an axis is only the component of MO along that axis.
Question 2 of 16 · Statics · 9–14 items
A rectangular structural member has a cross-sectional area of 25 cm2 and a centroidal radius of gyration of 4 cm about its horizontal neutral axis. The member is to be modified by shifting the reference axis parallel to the neutral axis by a distance of 3 cm. What is most nearly the moment of inertia of the cross-section about this new parallel axis?
Given
Area A = 25 cm2
Centroidal radius of gyration r = 4 cm
Axis offset d = 3 cm
Find
Moment of inertia about the new axis (cm4)
Solution
Ic = A·r2 = 25 × 42 = 400 cm4
I = Ic + A·d2 = 400 + 25 × 32 = 625 cm4
FE Reference Handbook, Statics: parallel-axis theorem and radius of gyration.
k2 + d2 = 16 + 9 = 25 leaves out the area; each term of the parallel-axis theorem carries A.
Subtracting A·d2 from the centroidal value is backwards; the transfer term always adds when moving away from the centroidal axis.
This is only the transfer term A·d2; the centroidal part A·r2 = 400 cm4 is missing.
This is only the centroidal moment of inertia A·r2; the parallel-axis transfer A·d2 = 225 cm4 is missing.
I = A(r2 + d2) = 25 × (16 + 9) = 625 cm4: the centroidal inertia A·r2 plus the transfer term A·d2.
Question 3 of 16 · Statics · 9–14 items
A simply supported beam carries the loads detailed in the table below. Calculate the vertical reaction force at Support B.
Feature
Position (x)
Type
Magnitude
Support A
0 m
Pin
N/A
Support B
8 m
Roller
N/A
Load 1
4 m
Point force
50 kN (downward)
Load 2
6 m
Applied moment
120 kN·m (clockwise)
Given
Span = 8 m (pin A at 0, roller B at 8 m)
Point load = 50 kN down at 4 m
Applied moment = 120 kN·m clockwise at 6 m
Find
Reaction at B (kN)
Solution
ΣMA = 0: RB(8 m) − (50 kN)(4 m) − 120 kN·m = 0
RB = (200 + 120)/8 = 40 kN
The position of a pure couple along the span does not matter, only its sense and size.
This treats the 120 kN·m couple as counterclockwise, subtracting it from the point-load moment.
This accounts for the applied couple alone (120/8) and leaves out the 50 kN point load.
Splitting the 50 kN midspan load equally ignores the couple, which shifts reaction from A to B.
With moments about A, 8·RB = 50 × 4 + 120, so RB = 40 kN; the clockwise couple adds to the point load's moment.
This is the whole point load assigned to one support; the reactions must share it and also balance the couple.
Question 4 of 16 · Statics · 9–14 items
What is the moment of a 100 N force about point O if the force is applied at a point with position vector r = (3i + 4j) m and the force vector is F = (60i - 80j) N?
A 50 kg block rests on a plane inclined at 20° to the horizontal. If the coefficient of static friction between the block and plane is 0.4, what is the minimum force parallel to the plane required to move the block up the plane?
Given
Mass m = 50 kg
Incline angle θ = 20°
Static friction coefficient μs = 0.4
Find
Minimum force parallel to the plane F (N)
Solution
P = mg(sin θ + μs cos θ)
P = (50 kg)(9.81 m/s2)(sin 20° + 0.4 cos 20°) = 490.5 × 0.718 ≈ 352 N
FE Reference Handbook, Statics: friction.
This is the full weight mg, which is not the force needed along a 20° incline.
P = mg(sin θ + μs·cos θ) = 490.5 × (0.342 + 0.376) ≈ 352 N: the push must overcome both the weight component and friction acting down the slope.
This is only the friction force μs·mg·cos θ; the down-slope weight component is missing.
This is only the weight component mg·sin θ; friction opposes motion up the plane and must be added.
Question 6 of 16 · Statics · 9–14 items
A force system consists of three forces: 50 N at 0°, 70 N at 120°, and 90 N at 240° (all measured counterclockwise from the positive x-axis). What is the magnitude of the resultant force?
Given
Force F1 = 50 N at 0°
Force F2 = 70 N at 120°
Force F3 = 90 N at 240°
Find
Resultant magnitude R (N)
Solution
ΣFx = 50 + 70 cos 120° + 90 cos 240° = −30.0 N
ΣFy = 70 sin 120° + 90 sin 240° = −17.3 N
R = √(30.02 + 17.32) = 34.6 N
Three forces 120° apart cancel only when they are equal; here they are 50, 70 and 90 N.
This is only |ΣFy| = 17.3 N; the x-components add another −30.0 N.
This is only |ΣFx|; the y-components do not cancel and add 17.3 N in quadrature.
ΣFx = 50 − 35 − 45 = −30 N and ΣFy = 60.6 − 77.9 = −17.3 N, so R = √(302 + 17.32) ≈ 34.6 N.
Question 7 of 16 · Statics · 9–14 items
At a joint in a rigid body, two forces act concurrently: a 50 N force directed 15 degrees above the horizontal and an 80 N force directed 105 degrees from the horizontal. To maintain equilibrium, a third force must cancel their resultant. What is the magnitude and direction (from the positive x-axis) of this third force?
Given
Force F1 = 50 N at 15° from the +x axis
Force F2 = 80 N at 105° from the +x axis
Find
Magnitude and direction of the equilibrant E
Solution
Rx = 50 cos 15° + 80 cos 105° = 27.6 N, Ry = 50 sin 15° + 80 sin 105° = 90.2 N
R = √(27.62 + 90.22) = 94.3 N at tan−1(90.2/27.6) = 73°
E = −R → 94.3 N at 253°
The resultant is 94.3 N at 73° (the two forces are 90° apart); the equilibrant has the same size and points the opposite way, 73° + 180° = 253°.
This is the resultant itself; the force that restores equilibrium must point opposite to it.
Adding 50 + 80 as scalars ignores the 90° between the forces.
Subtracting the magnitudes (80 − 50) treats the forces as opposite; they are perpendicular and add as √(502 + 802).
Question 8 of 16 · Statics · 9–14 items
A T-beam is composed of a flange that is 200 mm wide and 30 mm thick, and a web that is 20 mm wide and 170 mm high (excluding the flange thickness). Determine the location of the overall centroid measured from the bottom of the web.
Given
Flange width = 200 mm
Flange thickness = 30 mm
Web width = 20 mm
Web height = 170 mm
Find
Centroid height above web bottom ȳ (mm)
Solution
Af = 200 × 30 = 6000 mm2, yf = 170 + 15 = 185 mm
Aw = 20 × 170 = 3400 mm2, yw = 85 mm
ȳ = (6000 × 185 + 3400 × 85)/9400 = 148.8 mm ≈ 149 mm
FE Reference Handbook, Statics: centroids of composite areas.
This is the centroid measured from the top of the flange, not from the bottom of the web.
Averaging the two part centroids (185 + 85)/2 ignores that the flange has nearly twice the web's area.
ȳ = (6000 × 185 + 3400 × 85)/9400 ≈ 149 mm: area-weighted centroids, with the flange centroid at 170 + 15 = 185 mm.
This uses a 200 mm web, but the 170 mm web height already excludes the flange thickness.
Question 9 of 16 · Statics · 9–14 items
A rigid L-shaped bracket is supported by a frictionless pin at point A and a smooth roller at point B. A single concentrated force, not parallel to the roller reaction, is applied at point C. For the bracket to be in static equilibrium under these three forces, which statement about their lines of action must be true?
Solution
Three-force member: with ΣM = 0 about the intersection of two lines of action, the third force must pass through that point as well (or all three are parallel, which the stem excludes).
A couple has zero resultant force, so it could never balance a single applied force.
Equilibrium requires the moments to sum to zero, not the moment arms; arm lengths alone say nothing about balance.
The meeting point of the lines of action depends on geometry and loading; nothing ties it to the centroid.
Three coplanar forces cannot all be mutually perpendicular, and perpendicularity is not an equilibrium requirement.
A three-force body whose forces are not parallel can be in equilibrium only if the three lines of action meet at one point; otherwise one force has a moment about the meeting point of the other two.
Question 10 of 16 · Statics · 9–14 items
An industrial gantry system uses a structural column that is embedded deeply into a massive concrete foundation, preventing any translation or rotation. For a rigid-body 3D static equilibrium analysis, which combination of reaction components strictly defines this boundary condition?
Solution
A support supplies one reaction for each blocked degree of freedom: 3 translations → Fx, Fy, Fz; 3 rotations → Mx, My, Mz.
Leaving out one moment would allow rotation about one axis, like a journal bearing or hinge, not an embedded column.
Without horizontal force components the support could not resist sideways translation.
Missing one force component would let the column slide along one axis.
Three forces alone describe a ball-and-socket support, which still allows rotation.
A fixed (built-in) support in 3D blocks all three translations and all three rotations, so it supplies three forces and three couples.
Question 11 of 16 · Statics · 9–14 items
What is the polar moment of inertia of a solid circular shaft with diameter 60 mm?
Given
Shaft diameter d = 60 mm
Find
Polar moment of inertia J (mm4)
Solution
J = πd4/32 = π(60 mm)4/32 ≈ 1.27 × 106 mm4
FE Reference Handbook, Statics: properties of a circle.
This is πd4/64, the area moment of inertia about one diameter; the polar moment is twice that.
J = πd4/32 = π × 604/32 ≈ 1,272,345 mm4.
Using πd4/16 doubles the correct value, as when πr4/2 is written as πr4.
This puts the diameter into πr4/2; the formula needs the radius, 30 mm.
Question 12 of 16 · Statics · 9–14 items
What is the area moment of inertia (I) for a rectangular section with width b and height h about its centroidal axis parallel to the width?
Solution
Ix′ = bh3/12 (centroidal axis parallel to b)
FE Reference Handbook, Statics: properties of a rectangle.
This is the centroidal moment about the axis parallel to the height; the cubed dimension is the one perpendicular to the axis.
For an axis through the centroid parallel to the width b, the depth h is cubed: I = bh3/12.
This is the moment of inertia about the base edge, not the centroidal axis; it adds A(h/2)2.
This is the elastic section modulus S = I/c, not the moment of inertia.
Question 13 of 16 · Statics · 9–14 items
In a truss, what are zero-force members?
Solution
Zero-force member: at an unloaded joint with two non-collinear members both are zero; with three members, two collinear, the third is zero.
A zero-force member carries no force for the particular load case; it is identified by joint equilibrium, such as two non-collinear members at an unloaded joint.
Ideal truss members are two-force members and carry only axial force; zero-force members carry none at all.
Members at supports usually carry the largest forces; location alone does not make a member zero-force.
They are zero only for the given loads; they brace other members against buckling and carry load when the loading changes.
Question 14 of 16 · Statics · 9–14 items
What is the centroid location of a semicircular area with radius r from its straight edge?
Solution
ȳ = 4r/(3π) ≈ 0.424r
FE Reference Handbook, Statics: centroids of areas (semicircle).
Halfway up would hold only for a rectangle; the semicircle has more area near its diameter, so the centroid is lower.
The centroid of a semicircular area lies 4r/(3π) ≈ 0.424r from the diameter.
A distance r is the top of the arc itself; the centroid must sit inside the area, closer to the wide base.
2r/3 is the sector formula 2r sin α/(3α) with sin α ≈ α, valid only for a very narrow sector; for a semicircle α = π/2 gives 4r/(3π).
Question 15 of 16 · Statics · 9–14 items
A simply supported truss carries loads at its top-chord joints, and no joint has only two unknown forces. Using the method of joints, what must be done before the first joint can be solved?
Solution
With no joint having only two unknowns, solve ΣFx = 0, ΣFy = 0, ΣM = 0 for the whole truss first, then start at a support joint.
Truss member forces under joint loads do not depend on the centroid of the truss.
The method of sections is an alternative technique, not a step of the method of joints, and it also needs the reactions first.
Every joint has more than two unknowns until the reactions are known, so the whole-truss equilibrium equations come first.
Member forces in a statically determinate truss come from equilibrium alone; areas matter only for stress afterward.
Question 16 of 16 · Statics · 9–14 items
Which of the following is NOT a condition for static equilibrium of a rigid body?
Solution
Static equilibrium of a rigid body: ΣF = 0 and ΣM = 0, both at once.
This is a required condition: without it the body would accelerate in translation.
This is a required condition: without it the body would start to rotate.
This is required too; a couple has ΣF = 0 yet still causes rotation.
Equilibrium is about forces and moments; a body moving at constant velocity is also in equilibrium, so no velocity condition exists.
Answer every question to see your score
Common traps
Statics errors are rarely about the formula. They come from the diagram and from sign bookkeeping. The usual suspects, as treated in standard texts such as Hibbeler's Engineering Mechanics: Statics and Beer and Johnston's Vector Mechanics for Engineers:
A pin is two unknowns, a roller is one. Drawing one arrow at a pin leaves an equation short, and the answer still looks reasonable.
Mixed moment signs. Pick counterclockwise positive (or the right-hand rule in 3D) before writing the first moment sum, and keep it for the whole body.
A distributed load goes on the diagram as its resultant, at the centroid of the load shape. For a triangular load that point is one third of the length from the tall end, not the middle.
Two non-centroidal axes. The parallel-axis theorem always runs through the centroid, so moving between two other axes takes two steps.
Static friction is a limit, not a value. μs·N is the most friction the surface can supply; the actual friction force comes from equilibrium until slip is impending. Kinetic friction belongs to dynamics.
Slip or tip. A block on a rough surface can tip before it slides. Check both and report the smaller load.
Area moment, not mass moment
The second moment of area (m4 or in4) belongs here and in bending. The mass moment of inertia (kg·m2) belongs to rotating bodies in dynamics. The two sit in different parts of the FE Reference Handbook, and a units check catches the mix-up in seconds.
Where statics feeds the other areas
Every support reaction you solve here is the starting point for shear and moment diagrams in mechanics of materials, and the same equilibrium sums size bolts, keys and brackets in mechanical design and analysis. Centroids and area moments come back in beam bending.
FE Civil has a statics area of its own; on this site it is practiced with bridge and building loads on the civil structural engineering page, while the generic mechanics lives here. Vector algebra for 3D moments is part of the shared mathematics section.
Is statics on the FE the same as in a college statics course?
The topics match a first statics course. The difference is speed: 5 h 20 min for the whole exam leaves under three minutes per question, with formulas looked up in the on-screen Handbook rather than recalled.
How many statics questions are on the FE Mechanical exam?
The specification gives a range of 9–14 questions out of 110. Some of those are unscored pretest items, so the scored count is a little smaller.
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