Statics practice questions

Answer: C) 1.27 m

The centroid of a semicircular area lies 4r/(3π) from its diameter. For r = 3 m, this gives about 1.27 m.

• AIncorrect. 2.00 m is 2r/3, two-thirds of the radius. That fraction has no role in this centroid formula, so it overstates the distance.
• BIncorrect. 1.50 m is r/2. Treating the centroid as halfway out ignores the fact that more of the semicircle's area lies near the diameter.
• CCorrect. For a semicircular area, ȳ = 4r/(3π) = 4(3)/(3π) = 12/(3π) ≈ 1.27 m from the diameter.
• DIncorrect. 0.64 m is 2r/(3π), which drops a factor of 2 from the correct 4r/(3π). It is the answer you get by mixing up this formula with a similar-looking one, such as the one for a semicircular arc (2r/π).

Answer: A) 11.4 kN

Use the parallelogram law: R = √(F₁² + F₂² + 2F₁F₂cos θ), where θ is the angle between the forces. Here that gives √129 ≈ 11.4 kN.

• ACorrect. With θ = 60° between the forces, R = √(F₁² + F₂² + 2F₁F₂cos θ) = √(25 + 64 + 40) = √129 ≈ 11.4 kN.
• BIncorrect. 13.0 kN is the plain sum 5 + 8. You can only add magnitudes directly when the forces act in the same direction (θ = 0°).
• CIncorrect. 9.4 kN is √(25 + 64), which treats the forces as perpendicular. At 60° the cosine term adds 2·5·8·cos 60° = 40 kN² under the root.
• DIncorrect. 7.0 kN is √(25 + 64 − 40) = √49. It applies the law of cosines with a minus sign, treating 60° as the interior angle of the force triangle. That interior angle is actually 180° − 60° = 120°.

Answer: D) 2 kN downward

• AIncorrect. They don't cancel.
• BIncorrect. They oppose each other.
• CIncorrect. The larger force is downward.
• DCorrect. The forces are opposite.

Answer: A) 30°

• ACorrect. tan θ = μs.
• BIncorrect. That needs μs = 1.
• CIncorrect. That needs μs ≈ 1.73.
• DIncorrect. The angle is the arctangent.

Answer: A) 2.0 m

• ACorrect. 600 × 1.5 = 450 × x.
• BIncorrect. That inverts the ratio.
• CIncorrect. Weights differ.
• DIncorrect. That doubles the distance.

Answer: B) 3: two forces and a moment

• AIncorrect. That's a roller.
• BCorrect. It resists translation and rotation.
• CIncorrect. That's a pin.
• DIncorrect. That's a fixed support in 3D.

Answer: C) 100 N

• AIncorrect. That uses cos 30°.
• BIncorrect. Use the vertical component, T sin 30°.
• CCorrect. 2T sin 30° = 100, so T = 100.
• DIncorrect. Recheck: 2T(0.5) = 100.

Answer: C) Both zero

• AIncorrect. They're zero-force members.
• BIncorrect. Equilibrium forces both to zero.
• CCorrect. They're zero-force members.
• DIncorrect. Both are zero.

Answer: D) Equal, opposite and collinear

• AIncorrect. Unequal forces wouldn't balance.
• BIncorrect. They must be collinear.
• CIncorrect. They must be opposite to balance.
• DCorrect. They act along the line joining the two points.

Answer: B) 13

A simple plane truss satisfies m = 2j − 3, so 8 joints give 2(8) − 3 = 13 members.

• AIncorrect. This uses m = j + 3. For a simple truss, members grow by two per joint, not one, so m = 2j − 3 = 13.
• BCorrect. The starting triangle has 3 joints and 3 members. The 5 added joints bring 2 members each, so 3 + 10 = 13, which matches m = 2j − 3 = 2(8) − 3.
• CIncorrect. This uses m = 2j and forgets that the starting triangle has only 3 members for its 3 joints. That is why the formula subtracts 3: m = 2j − 3 = 13.
• DIncorrect. This adds 3 instead of subtracting it (2j + 3). A truss with 19 members on 8 joints would have redundant members and would not be a simple truss.