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Scalars, Graphical Vectors and Force Components

Key Takeaways

  • Scalars have magnitude; vectors have magnitude and direction.

  • Draw vectors tip to tail to find a graphical resultant.

  • Resolve a force using the stated angular reference and check the resultant quadrant.

  • A shallow symmetric sling angle increases leg tension; ratings and the lifting plan govern actual use.

Last updated: October 2026

Scalars vs. Vector Quantities

All physical quantities in mechanics are classified as either scalars or vectors:

ClassificationDefinitionKey CharacteristicsPower Plant Examples
Scalar QuantityA quantity possessing magnitude only (a numerical value and unit of measurement).Adds algebraically (3+4=73 + 4 = 7). Direction is meaningless.Mass (500 kg500\text{ kg}), Temperature (180∘C180^\circ\text{C}), Time (60 s60\text{ s}), Energy/Enthalpy (2,780 kJ/kg2,780\text{ kJ/kg}), Volume (15 m315\text{ m}^3), Pressure (1,200 kPa1,200\text{ kPa})
Vector QuantityA quantity possessing magnitude and direction.Adds geometrically according to vector addition laws.Force (25 kN25\text{ kN} at 30∘30^\circ), Weight (gravity pulling downward), Velocity (15 m/s15\text{ m/s} North), Acceleration (m/s2\text{m/s}^2), Torque/Moment (N⋅m\text{N}\cdot\text{m} CCW)

Vector Addition & Rectangular Components

When multiple forces act simultaneously on a single body (concurrent forces), their combined effect is represented by a single resultant vector (RR).

Vector Addition Laws

  1. Parallelogram Law of Forces: If two concurrent forces acting on a point are represented in magnitude and direction by two adjacent sides of a parallelogram, their resultant is represented by the diagonal passing from that point.
  2. Triangle Law of Forces: If two forces acting on a point are drawn tip-to-tail, the vector drawn from the initial tail to the final tip represents the resultant force.

Analytical Resolution into Perpendicular Components

In engineering practice, resolving vectors into orthogonal horizontal (xx) and vertical (yy) components is much faster and more accurate than graphical scale drawings.

For a force FF acting at an angle θ\theta measured relative to the horizontal axis:

Fx=Fcos⁡θF_x = F \cos\theta Fy=Fsin⁡θF_y = F \sin\theta

To recombine multiple resolved components into a single resultant force RR:

R=(∑Fx)2+(∑Fy)2R = \sqrt{(\sum F_x)^2 + (\sum F_y)^2} θR=arctan⁡(∣∑Fy∣∣∑Fx∣)\theta_R = \arctan\left(\frac{|\sum F_y|}{|\sum F_x|}\right)

Rigging & Hoisting Mechanics

When lifting equipment with a multi-leg bridle sling, the tension in each sling leg depends heavily on the sling angle (θ\theta) measured relative to the horizontal surface of the load.

Sling Tension Formula

For a symmetrical two-leg sling lifting a total load weight WW:

T=W2sin⁡θT = \frac{W}{2 \sin\theta}

Where:

  • TT = Tension in each sling leg (N\text{N} or kN\text{kN})
  • WW = Total weight of load (m×gm \times g, in N\text{N} or kN\text{kN})
  • θ\theta = Angle between the sling leg and the horizontal load plane

Alternatively, using the ratio of sling length (LL) to vertical height (HH from load to hook):

Load Angle Factor (LAF)=LH=1sin⁡θ\text{Load Angle Factor (LAF)} = \frac{L}{H} = \frac{1}{\sin\theta} T=(W2)×(LH)T = \left(\frac{W}{2}\right) \times \left(\frac{L}{H}\right)

The Critical Danger of Shallow Sling Angles

As the sling angle becomes shallower, sin⁡θ\sin\theta approaches zero, causing the internal tension TT to increase nonlinearly as the reciprocal of sine:

Sling Angle to Horizontal (θ\theta)sin⁡θ\sin\thetaLoad Angle Factor (1/sin⁡θ1/\sin\theta)Tension per Leg for 10,000 kg10,000\text{ kg} Load (W=98.1 kNW = 98.1\text{ kN})
90∘90^\circ (Straight vertical)1.0001.00049.05 kN49.05\text{ kN} (0.50×W0.50 \times W)
60∘60^\circ (Equilateral bridle)0.8661.15556.64 kN56.64\text{ kN} (0.58×W0.58 \times W)
45∘45^\circ (Standard sling)0.7071.41469.34 kN69.34\text{ kN} (0.71×W0.71 \times W)
30∘30^\circ (Example angle)0.5002.00098.10 kN98.10\text{ kN} (1.00×W1.00 \times W)
15∘15^\circ (Very high tension)0.2593.864189.53 kN189.53\text{ kN} (1.93×W1.93 \times W)

Crucial Exam Fact: At a 30∘30^\circ horizontal sling angle, each sling leg is subjected to a tensile force equal to the entire weight of the load! Use the sling manufacturer’s ratings and the approved lifting plan; this arithmetic does not authorize a lift.

Construct a graphical resultant

Choose a scale such as 1 cm for 2 kN and mark north/east or positive axes. Draw the first force arrow to scale. From its tip draw the next arrow in its correct direction. The resultant runs from the first tail to the final tip. Its length gives magnitude and its measured angle gives direction. An equilibrant has equal magnitude and the opposite direction. Never add arrow lengths as scalars unless the forces share a line of action and their signs have been considered.

For 6 kN east and 8 kN north, draw perpendicular arrows 3 cm and 4 cm long. The closing diagonal measures 5 cm, giving a 10 kN resultant at about 53.1 degrees north of east. Analytical checking gives the same magnitude from the square root of 6 squared plus 8 squared. The inverse tangent supplies only a reference angle; use the signs of both components to select the correct quadrant. A resultant pointing southwest cannot be described by the northeast reference angle alone.

In a symmetric two-leg lifting calculation, state that both legs share the load equally, the load is stationary, and the angle is measured from the horizontal. The simple tension equation does not establish the capacity of hooks, shackles, attachment points or the load itself. Unequal leg lengths or a displaced centre of gravity invalidate equal sharing. Four sling legs do not automatically divide a load into four equal parts. The exam tests vector resolution; actual rigging additionally requires rated equipment and competent lift planning.

Source notes

Official scope and technical references checked 2026-10-17: Reference 1; Reference 2. Archived DOE handbooks are used for physical principles, not current Canadian legal requirements.

Test Your Knowledge

Which of the following physical quantities encountered in power engineering is classified as a vector quantity?

A

Weight of a boiler drum

B

Temperature of superheated steam

C

Enthalpy of saturated liquid

D

Mass of fuel oil in a storage tank

Test Your Knowledge

For perpendicular forces of 6 kN east and 8 kN north, what is the resultant magnitude?

A

2 kN

B

7 kN

C

14 kN

D

10 kN

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