Five original statics problems in the style of the FE Mechanical exam, each with four answer choices and a full step-by-step solution. They cover the statics topics in the NCEES specification: resultants, equilibrium, trusses, friction, and centroids and moments of inertia. Give yourself about 3 minutes per problem.
How to use these problems
Try each problem with a timer and your approved calculator before you open the solution. Write down your choice; on the exam you only get credit for the answer, not the working.
When you miss one, find the step where you went wrong and redo the problem a day later. The wrong answer choices are built from the usual mistakes, so a wrong pick tells you which mistake you made.
Keep the statics cheat sheet next to you while you study, then practise finding the same formulas in the FE Reference Handbook.
Practice problems
Problem 1: Resultant of three forces
Three forces act on an eyebolt: 400 N along +x (0°), 300 N along +y (90°), and 250 N at 225° measured counterclockwise from +x. What is the magnitude and direction of the resultant?
255 N at 28.9° from +x
950 N at 45.0° from +x
368 N at 36.9° from +x
223 N at 0° from +x
Show the solution to problem 1
Resolve the 250 N force into components.250·cos 225° = −176.8 N; 250·sin 225° = −176.8 N
Sum the x and y components.R_x = 400 − 176.8 = 223.2 N; R_y = 300 − 176.8 = 123.2 N
Magnitude.R = √(223.2² + 123.2²) = 255.0 N
Direction: both components are positive, so the resultant is in the first quadrant.θ = tan⁻¹(123.2/223.2) = 28.9°
A 2.0 kN load hangs from ring C. Cable CA runs up and to the left at 30° above the horizontal; cable CB runs up and to the right at 45° above the horizontal. What is the tension in cable CB?
1.46 kN
1.79 kN
1.41 kN
2.00 kN
Show the solution to problem 2
Draw the FBD of ring C: T_A along CA, T_B along CB, and 2.0 kN down.
A triangular truss has a pin support at A and a roller at B, 4.0 m apart on a horizontal line. Joint C is 2.0 m to the right of A and 1.5 m above line AB. A 6.0 kN load acts straight down at C. What is the force in member AC?
5.0 kN tension
4.0 kN tension
5.0 kN compression
3.75 kN compression
Show the solution to problem 3
By symmetry (or ΣM_A = 0), each support carries half the load.A_y = B_y = 3.0 kN; A_x = 0
Geometry of AC.length = √(2.0² + 1.5²) = 2.5 m; sin θ = 1.5/2.5 = 0.6; cos θ = 0.8
Joint A, assuming both members in tension (pulling away from A).ΣF_y = 3.0 + F_AC·0.6 = 0 → F_AC = −5.0 kN
The negative sign means AC pushes on the joint: it is in compression.ΣF_x = F_AB + F_AC·0.8 = 0 → F_AB = +4.0 kN (tension)
A 50 kg crate rests on a 25° incline. The coefficient of static friction is 0.35; use g = 9.81 m/s². What is the smallest force, applied parallel to the incline and pointing up the slope, that keeps the crate from sliding down?
0 N; the crate does not slide
207 N
363 N
51.7 N
Show the solution to problem 4
Weight and its components.W = 50 × 9.81 = 490.5 N; along the slope: 490.5·sin 25° = 207.3 N; normal: N = 490.5·cos 25° = 444.5 N
Maximum static friction available.F_max = μ_s·N = 0.35 × 444.5 = 155.6 N
Friction alone cannot hold the crate (207.3 > 155.6), so a push P is needed. With the crate about to slide down, friction acts up the slope at its maximum.P + 155.6 − 207.3 = 0 → P = 51.7 N
A T-section has a 100 mm × 20 mm flange on top of a 20 mm × 80 mm web (the web is 80 mm tall). What is the moment of inertia about the horizontal axis through the section’s centroid?
0.92 × 10⁶ mm⁴
3.14 × 10⁶ mm⁴
2.22 × 10⁶ mm⁴
13.4 × 10⁶ mm⁴
Show the solution to problem 5
Areas and centroid heights measured from the bottom of the web.web: A = 1 600 mm², y = 40 mm; flange: A = 2 000 mm², y = 90 mm
Centroid of the section.y_c = (1 600 × 40 + 2 000 × 90)/3 600 = 67.78 mm
Each part about its own centroid.web: 20 × 80³/12 = 853 300 mm⁴; flange: 100 × 20³/12 = 66 700 mm⁴