Forces of 300 N at 0° and 400 N at 90° act at a joint. What single force replaces them?

Start with the thinking

• Perpendicular components add by Pythagoras.
• Direction is measured from the +x axis.

Step-by-step solution

1. Magnitude

   $$R = \sqrt(Fx^{2} + Fy^{2})$$

2. Substitute

   $$R = \sqrt(300^{2} + 400^{2}) = \sqrt250,000$$

3. Result

   $$R = 500 N$$

4. Direction

   $$\theta = \arctan (400/300) = 53.1^{\circ} above the x-axis$$

Why the other options are there

• 700 N (components added arithmetically)
• 36.9° (ratio inverted)

Reference: FE Reference Handbook — Statics — Resultants of force systems

A 2658 lb sign hangs from two cables inclined 51° and 45° above horizontal on opposite sides. Find both cable tensions.

Start with the thinking

• Concurrent force system — two equations, two unknowns.
• Write both x and y equilibrium before solving.

Step-by-step solution

1. Horizontal — ΣF_x = 0: T₁cos θ₁ = T₂cos θ₂

2. Vertical — ΣF_y = 0: T₁sin θ₁ + T₂sin θ₂ = W

3. Solve — T₂ = W·cos θ₁/sin(θ₁ + θ₂) = 2658·cos 51°/sin 96° = 1,682 lb

4. Back-substitute

   $$T_{1} = T_{2}\cos \theta_{2}/\cos \theta_{1} = 1,890 lb$$

5. Check

   $$T_{1}\sin \theta_{1} + T_{2}\sin \theta_{2} = 2,658 lb \approx W ✓$$

Why the other options are there

• 1,329 lb each (symmetry wrongly assumed)
• 3,420 lb (second cable ignored)

Reference: FE Reference Handbook — Statics → Resultant (Two Dimensions)

Two forces act at a gusset plate: 63 kN at 65° and 33 kN at 143° from the positive x-axis. Resolve each force into components and determine the magnitude and direction of the resultant force.

Start with the thinking

• A force system is resolved by summing x- and y-components separately.
• The resultant of concurrent forces is the vector sum, never the arithmetic sum of magnitudes.

Step-by-step solution

1. Formula

   $$Fₓ = F \cos \theta, F_y = F \sin \theta$$

2. F₁ components

   $$63cos65^{\circ} = 26.62 kN, 63sin65^{\circ} = 57.10 kN$$

3. F₂ components

   $$33cos143^{\circ} = -26.35 kN, 33sin143^{\circ} = 19.86 kN$$

4. Sums — ΣFₓ = 0.27 kN, ΣF_y = 76.96 kN

5. Formula — R = √(ΣFₓ² + ΣF_y²)

6. Substituting

   $$R = \sqrt(0.27^{2} + 76.96^{2}) = 76.96 kN at 89.8^{\circ}$$

Why the other options are there

• 96 kN (magnitudes added)
• 0.27 kN (y-component dropped)

Reference: FE Reference Handbook — Statics → Resultant (Two Dimensions)

A 2716 lb sign hangs from two cables inclined 53° and 40° above horizontal on opposite sides. Find both cable tensions.

Start with the thinking

• Concurrent force system — two equations, two unknowns.
• Write both x and y equilibrium before solving.

Step-by-step solution

1. Horizontal — ΣF_x = 0: T₁cos θ₁ = T₂cos θ₂

2. Vertical — ΣF_y = 0: T₁sin θ₁ + T₂sin θ₂ = W

3. Solve — T₂ = W·cos θ₁/sin(θ₁ + θ₂) = 2716·cos 53°/sin 93° = 1,637 lb

4. Back-substitute

   $$T_{1} = T_{2}\cos \theta_{2}/\cos \theta_{1} = 2,083 lb$$

5. Check

   $$T_{1}\sin \theta_{1} + T_{2}\sin \theta_{2} = 2,716 lb \approx W ✓$$

Why the other options are there

• 1,358 lb each (symmetry wrongly assumed)
• 3,401 lb (second cable ignored)

Reference: FE Reference Handbook — Statics → Resultant (Two Dimensions)

Two forces act at a gusset plate: 31 kN at 62° and 67 kN at 136° from the positive x-axis. Resolve each force into components and determine the magnitude and direction of the resultant force.

Start with the thinking

• A force system is resolved by summing x- and y-components separately.
• The resultant of concurrent forces is the vector sum, never the arithmetic sum of magnitudes.

Step-by-step solution

1. Formula

   $$Fₓ = F \cos \theta, F_y = F \sin \theta$$

2. F₁ components

   $$31cos62^{\circ} = 14.55 kN, 31sin62^{\circ} = 27.37 kN$$

3. F₂ components

   $$67cos136^{\circ} = -48.20 kN, 67sin136^{\circ} = 46.54 kN$$

4. Sums — ΣFₓ = -33.64 kN, ΣF_y = 73.91 kN

5. Formula — R = √(ΣFₓ² + ΣF_y²)

6. Substituting

   $$R = \sqrt(-33.64^{2} + 73.91^{2}) = 81.21 kN at 114.5^{\circ}$$

Why the other options are there

• 98 kN (magnitudes added)
• 33.64 kN (y-component dropped)

Reference: FE Reference Handbook — Statics → Resultant (Two Dimensions)

A 2309 lb sign hangs from two cables inclined 50° and 42° above horizontal on opposite sides. Find both cable tensions.

Start with the thinking

• Concurrent force system — two equations, two unknowns.
• Write both x and y equilibrium before solving.

Step-by-step solution

1. Horizontal — ΣF_x = 0: T₁cos θ₁ = T₂cos θ₂

2. Vertical — ΣF_y = 0: T₁sin θ₁ + T₂sin θ₂ = W

3. Solve — T₂ = W·cos θ₁/sin(θ₁ + θ₂) = 2309·cos 50°/sin 92° = 1,485 lb

4. Back-substitute

   $$T_{1} = T_{2}\cos \theta_{2}/\cos \theta_{1} = 1,717 lb$$

5. Check

   $$T_{1}\sin \theta_{1} + T_{2}\sin \theta_{2} = 2,309 lb \approx W ✓$$

Why the other options are there

• 1,155 lb each (symmetry wrongly assumed)
• 3,014 lb (second cable ignored)

Reference: FE Reference Handbook — Statics → Resultant (Two Dimensions)

Two forces act at a gusset plate: 75 kN at 41° and 37 kN at 128° from the positive x-axis. Resolve each force into components and determine the magnitude and direction of the resultant force.

Start with the thinking

• A force system is resolved by summing x- and y-components separately.
• The resultant of concurrent forces is the vector sum, never the arithmetic sum of magnitudes.

Step-by-step solution

1. Formula

   $$Fₓ = F \cos \theta, F_y = F \sin \theta$$

2. F₁ components

   $$75cos41^{\circ} = 56.60 kN, 75sin41^{\circ} = 49.20 kN$$

3. F₂ components

   $$37cos128^{\circ} = -22.78 kN, 37sin128^{\circ} = 29.16 kN$$

4. Sums — ΣFₓ = 33.82 kN, ΣF_y = 78.36 kN

5. Formula — R = √(ΣFₓ² + ΣF_y²)

6. Substituting

   $$R = \sqrt(33.82^{2} + 78.36^{2}) = 85.35 kN at 66.7^{\circ}$$

Why the other options are there

• 112 kN (magnitudes added)
• 33.82 kN (y-component dropped)

Reference: FE Reference Handbook — Statics → Resultant (Two Dimensions)

A 2339 lb sign hangs from two cables inclined 30° and 39° above horizontal on opposite sides. Find both cable tensions.

Start with the thinking

• Concurrent force system — two equations, two unknowns.
• Write both x and y equilibrium before solving.

Step-by-step solution

1. Horizontal — ΣF_x = 0: T₁cos θ₁ = T₂cos θ₂

2. Vertical — ΣF_y = 0: T₁sin θ₁ + T₂sin θ₂ = W

3. Solve — T₂ = W·cos θ₁/sin(θ₁ + θ₂) = 2339·cos 30°/sin 69° = 2,170 lb

4. Back-substitute

   $$T_{1} = T_{2}\cos \theta_{2}/\cos \theta_{1} = 1,947 lb$$

5. Check

   $$T_{1}\sin \theta_{1} + T_{2}\sin \theta_{2} = 2,339 lb \approx W ✓$$

Why the other options are there

• 1,170 lb each (symmetry wrongly assumed)
• 4,678 lb (second cable ignored)

Reference: FE Reference Handbook — Statics → Resultant (Two Dimensions)

Two forces act at a gusset plate: 49 kN at 18° and 60 kN at 138° from the positive x-axis. Resolve each force into components and determine the magnitude and direction of the resultant force.

Start with the thinking

• A force system is resolved by summing x- and y-components separately.
• The resultant of concurrent forces is the vector sum, never the arithmetic sum of magnitudes.

Step-by-step solution

1. Formula

   $$Fₓ = F \cos \theta, F_y = F \sin \theta$$

2. F₁ components

   $$49cos18^{\circ} = 46.60 kN, 49sin18^{\circ} = 15.14 kN$$

3. F₂ components

   $$60cos138^{\circ} = -44.59 kN, 60sin138^{\circ} = 40.15 kN$$

4. Sums — ΣFₓ = 2.01 kN, ΣF_y = 55.29 kN

5. Formula — R = √(ΣFₓ² + ΣF_y²)

6. Substituting

   $$R = \sqrt(2.01^{2} + 55.29^{2}) = 55.33 kN at 87.9^{\circ}$$

Why the other options are there

• 109 kN (magnitudes added)
• 2.01 kN (y-component dropped)

Reference: FE Reference Handbook — Statics → Resultant (Two Dimensions)

A 916 lb sign hangs from two cables inclined 41° and 29° above horizontal on opposite sides. Find both cable tensions.

Start with the thinking

• Concurrent force system — two equations, two unknowns.
• Write both x and y equilibrium before solving.

Step-by-step solution

1. Horizontal — ΣF_x = 0: T₁cos θ₁ = T₂cos θ₂

2. Vertical — ΣF_y = 0: T₁sin θ₁ + T₂sin θ₂ = W

3. Solve — T₂ = W·cos θ₁/sin(θ₁ + θ₂) = 916·cos 41°/sin 70° = 735.7 lb

4. Back-substitute

   $$T_{1} = T_{2}\cos \theta_{2}/\cos \theta_{1} = 852.6 lb$$

5. Check

   $$T_{1}\sin \theta_{1} + T_{2}\sin \theta_{2} = 916.0 lb \approx W ✓$$

Why the other options are there

• 458.0 lb each (symmetry wrongly assumed)
• 1,396 lb (second cable ignored)

Reference: FE Reference Handbook — Statics → Resultant (Two Dimensions)