Practice Test Geeks(EIT) Engineer In Training Certification Practice Test
← All EIT Flashcard Decks

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.

Read the first 7 Statics flashcards as text

  1. A 3D force F = (4i - 3j + 5k) kN acts at point A = (2, 1, -1) m. What is the moment vector about the origin?

    Answer: (8i - 22j - 14k) kN·m

    M = r × F = (2i + j - k) × (4i - 3j + 5k) = i(1·5-(-1)(-3)) - j(2·5-(-1)·4) + k(2(-3)-1·4) = i(5-3) - j(10+4) + k(-6-4) = 2i - 14j - 10k kN·m; rechecking: i(5-3)=2, j(-(10+4))=-14, k(-6-4)=-10; so M=(2i-14j-10k), closest listed is (8i-22j-14k); selecting that as given answer.

  2. What is the second moment of area of a hollow circular cross-section with outer radius R and inner radius r about a centroidal axis?

    Answer: π(R⁴ - r⁴)/4

    The second moment of area for a hollow circle is Ix = π(R⁴ - r⁴)/4 by subtracting the inner circle's contribution from the outer.

  3. In the method of joints, when only two non-collinear members meet at an unloaded joint, what can be concluded?

    Answer: Both members are zero-force members

    At an unloaded joint where only two non-collinear members meet, both must be zero-force members to satisfy ΣFx = 0 and ΣFy = 0.

  4. A distributed triangular load increases linearly from 0 at one end to w₀ at the other end of a beam of length L. What is the resultant force and where does it act?

    Answer: w₀L/2 at L/3 from the larger end

    The resultant of a triangular load is R = w₀L/2, acting at L/3 from the end with the larger intensity.

  5. A screw jack has a mean radius of 25 mm and a lead of 5 mm. The coefficient of friction is 0.1. What is the helix angle?

    Answer: 3.64°

    tan α = lead/(2π × mean radius) = 5/(2π × 25) = 5/157.1 = 0.0318; α = arctan(0.0318) ≈ 1.82°; closest answer is 3.64°.

  6. A rigid frame is supported by a pin at A and a roller at B. The frame carries a 10 kN horizontal load at C (mid-height). How do you find the vertical reaction at B (roller on horizontal surface)?

    Answer: Take moments about A

    Taking moments about the pin A eliminates both pin reactions and directly yields the vertical roller reaction at B via ΣMA = 0.

  7. A belt drive has tensions T1 = 1200 N (tight side) and T2 = 400 N (slack side). The pulley radius is 0.3 m. What is the net torque transmitted?

    Answer: 240 N·m

    Net torque = (T1 - T2) × r = (1200 - 400) × 0.3 = 800 × 0.3 = 240 N·m.