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

Two forces act at a gusset plate: 47 kN at 53° and 45 kN at 145° 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

   $$47cos53^{\circ} = 28.29 kN, 47sin53^{\circ} = 37.54 kN$$

3. F₂ components

   $$45cos145^{\circ} = -36.86 kN, 45sin145^{\circ} = 25.81 kN$$

4. Sums — ΣFₓ = -8.58 kN, ΣF_y = 63.35 kN

5. Formula — R = √(ΣFₓ² + ΣF_y²)

6. Substituting

   $$R = \sqrt(-8.58^{2} + 63.35^{2}) = 63.92 kN at 97.7^{\circ}$$

Why the other options are there

• 92 kN (magnitudes added)
• 8.58 kN (y-component dropped)

Reference: FE Reference Handbook — Statics → Systems of Forces

Two forces act at a gusset plate: 24 kN at 58° and 84 kN at 127° 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

   $$24cos58^{\circ} = 12.72 kN, 24sin58^{\circ} = 20.35 kN$$

3. F₂ components

   $$84cos127^{\circ} = -50.55 kN, 84sin127^{\circ} = 67.09 kN$$

4. Sums — ΣFₓ = -37.83 kN, ΣF_y = 87.44 kN

5. Formula — R = √(ΣFₓ² + ΣF_y²)

6. Substituting

   $$R = \sqrt(-37.83^{2} + 87.44^{2}) = 95.27 kN at 113.4^{\circ}$$

Why the other options are there

• 108 kN (magnitudes added)
• 37.83 kN (y-component dropped)

Reference: FE Reference Handbook — Statics → Systems of Forces

Two forces act at a gusset plate: 71 kN at 71° and 33 kN at 102° 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

   $$71cos71^{\circ} = 23.12 kN, 71sin71^{\circ} = 67.13 kN$$

3. F₂ components

   $$33cos102^{\circ} = -6.86 kN, 33sin102^{\circ} = 32.28 kN$$

4. Sums — ΣFₓ = 16.25 kN, ΣF_y = 99.41 kN

5. Formula — R = √(ΣFₓ² + ΣF_y²)

6. Substituting

   $$R = \sqrt(16.25^{2} + 99.41^{2}) = 100.7 kN at 80.7^{\circ}$$

Why the other options are there

• 104 kN (magnitudes added)
• 16.25 kN (y-component dropped)

Reference: FE Reference Handbook — Statics → Systems of Forces

Two forces act at a gusset plate: 42 kN at 42° and 51 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

   $$42cos42^{\circ} = 31.21 kN, 42sin42^{\circ} = 28.10 kN$$

3. F₂ components

   $$51cos136^{\circ} = -36.69 kN, 51sin136^{\circ} = 35.43 kN$$

4. Sums — ΣFₓ = -5.47 kN, ΣF_y = 63.53 kN

5. Formula — R = √(ΣFₓ² + ΣF_y²)

6. Substituting

   $$R = \sqrt(-5.47^{2} + 63.53^{2}) = 63.77 kN at 94.9^{\circ}$$

Why the other options are there

• 93 kN (magnitudes added)
• 5.47 kN (y-component dropped)

Reference: FE Reference Handbook — Statics → Systems of Forces

Two forces act at a gusset plate: 67 kN at 48° and 65 kN at 100° 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

   $$67cos48^{\circ} = 44.83 kN, 67sin48^{\circ} = 49.79 kN$$

3. F₂ components

   $$65cos100^{\circ} = -11.29 kN, 65sin100^{\circ} = 64.01 kN$$

4. Sums — ΣFₓ = 33.54 kN, ΣF_y = 113.8 kN

5. Formula — R = √(ΣFₓ² + ΣF_y²)

6. Substituting

   $$R = \sqrt(33.54^{2} + 113.8^{2}) = 118.6 kN at 73.6^{\circ}$$

Why the other options are there

• 132 kN (magnitudes added)
• 33.54 kN (y-component dropped)

Reference: FE Reference Handbook — Statics → Systems of Forces

Two forces act at a gusset plate: 48 kN at 20° and 21 kN at 113° 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

   $$48cos20^{\circ} = 45.11 kN, 48sin20^{\circ} = 16.42 kN$$

3. F₂ components

   $$21cos113^{\circ} = -8.21 kN, 21sin113^{\circ} = 19.33 kN$$

4. Sums — ΣFₓ = 36.90 kN, ΣF_y = 35.75 kN

5. Formula — R = √(ΣFₓ² + ΣF_y²)

6. Substituting

   $$R = \sqrt(36.90^{2} + 35.75^{2}) = 51.38 kN at 44.1^{\circ}$$

Why the other options are there

• 69 kN (magnitudes added)
• 36.90 kN (y-component dropped)

Reference: FE Reference Handbook — Statics → Systems of Forces

Two forces act at a gusset plate: 73 kN at 61° and 23 kN at 135° 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

   $$73cos61^{\circ} = 35.39 kN, 73sin61^{\circ} = 63.85 kN$$

3. F₂ components

   $$23cos135^{\circ} = -16.26 kN, 23sin135^{\circ} = 16.26 kN$$

4. Sums — ΣFₓ = 19.13 kN, ΣF_y = 80.11 kN

5. Formula — R = √(ΣFₓ² + ΣF_y²)

6. Substituting

   $$R = \sqrt(19.13^{2} + 80.11^{2}) = 82.36 kN at 76.6^{\circ}$$

Why the other options are there

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

Reference: FE Reference Handbook — Statics → Systems of Forces

Two forces act at a gusset plate: 47 kN at 36° and 53 kN at 169° 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

   $$47cos36^{\circ} = 38.02 kN, 47sin36^{\circ} = 27.63 kN$$

3. F₂ components

   $$53cos169^{\circ} = -52.03 kN, 53sin169^{\circ} = 10.11 kN$$

4. Sums — ΣFₓ = -14.00 kN, ΣF_y = 37.74 kN

5. Formula — R = √(ΣFₓ² + ΣF_y²)

6. Substituting

   $$R = \sqrt(-14.00^{2} + 37.74^{2}) = 40.25 kN at 110.4^{\circ}$$

Why the other options are there

• 100 kN (magnitudes added)
• 14.00 kN (y-component dropped)

Reference: FE Reference Handbook — Statics → Systems of Forces

Two forces act at a gusset plate: 26 kN at 53° and 78 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

   $$26cos53^{\circ} = 15.65 kN, 26sin53^{\circ} = 20.76 kN$$

3. F₂ components

   $$78cos143^{\circ} = -62.29 kN, 78sin143^{\circ} = 46.94 kN$$

4. Sums — ΣFₓ = -46.65 kN, ΣF_y = 67.71 kN

5. Formula — R = √(ΣFₓ² + ΣF_y²)

6. Substituting

   $$R = \sqrt(-46.65^{2} + 67.71^{2}) = 82.22 kN at 124.6^{\circ}$$

Why the other options are there

• 104 kN (magnitudes added)
• 46.65 kN (y-component dropped)

Reference: FE Reference Handbook — Statics → Systems of Forces