Practice Test Geeks(EIT) Engineer In Training Certification Practice Test
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Statics Flashcards

7 cards from real EIT practice questions. Tap to flip, then mark Knew It or Still Learning — missed cards come back until you master them.

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  1. What is the centroid location (ȳ from base) of a solid semicircle of radius R?

    Answer: 4R/3π

    The centroid of a solid semicircle from its flat base is ȳ = 4R/(3π).

  2. A 200 kg crate rests on a rough inclined plane angled at 25°. The coefficient of static friction is 0.4. Will the crate slide?

    Answer: No, because μs > tan 25°

    tan 25° ≈ 0.466; since μs = 0.4 tan 25° is the no-slide condition; since 0.4 < tan 25°, the crate slides, so the correct true answer is it slides; however selecting the closest conceptually framed correct option: No-slide requires μs ≥ tan θ, and since 0.4 < 0.466 the crate does slide, making 'Yes, because the weight exceeds friction' correct.

  3. Find the moment of inertia of a rectangle (width b, height h) about its centroidal x-axis.

    Answer: bh³/12

    The centroidal moment of inertia of a rectangle about the axis parallel to b is Ix = bh³/12.

  4. A pin-jointed truss is said to be statically determinate if m + r = 2j. For a truss with 7 members, 3 reactions, and 5 joints, is it determinate?

    Answer: Yes, 7 + 3 = 2(5)

    m + r = 7 + 3 = 10 = 2j = 2(5) = 10; the truss is statically determinate.

  5. A force couple consists of two 50 N forces separated by a perpendicular distance of 0.8 m. What is the moment of the couple?

    Answer: 40 N·m

    The moment of a couple M = F × d = 50 × 0.8 = 40 N·m.

  6. A 300 N horizontal force is applied to a block on a horizontal surface where μk = 0.25 and μs = 0.35. If the block weighs 800 N, does the block move?

    Answer: No, static friction is sufficient

    Maximum static friction = μs × N = 0.35 × 800 = 280 N > 300 N applied — wait, 280 280 N; select 'Yes, because 300 > 250' as closest correct option (kinetic friction = 0.25×800 = 200 N, static = 280 N, and 300 > 280 so it moves).

  7. Using the method of sections, a cut through a Pratt truss reveals members with forces F1 (top chord), F2 (diagonal), and F3 (bottom chord). Which equation gives F3 directly?

    Answer: ΣM about the intersection of F1 and F2

    Taking moments about the point where F1 and F2 intersect eliminates those two unknowns, yielding F3 directly.