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23-Ind-B6 Human Factor in Design · May 2013

Question 4 of 7: Manual Materials Handling — Task Design and a Lifting-Rate Calculation

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

National Examinations, May 2013 — 98-Ind-B6, Workplace Design (3-hour closed-book exam, Casio/Sharp approved calculators only. The front page states any 5 of the 7 questions, each worth 20 marks, constitute a complete paper; all 7 are answered below.)

Reference texts: Sanders & McCormick, Human Factors in Engineering and Design (7th ed.) — controls and displays, anthropometry and workstation design, physical work and manual materials handling, and human-machine system arrangement; Niebel & Freivalds, Methods, Standards, and Work Design — workplace layout and posture.

Question 4: Manual Materials Handling — Task Design and a Lifting-Rate Calculation (20 marks: i–7, ii–5, iii–8)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

(i) Task Characteristics for Minimizing MMH Hazard

The task characteristics an ergonomics analyst must examine when designing (or redesigning) a manual-materials-handling (MMH) job are, principally: the load itself — its weight, size/bulk, and whether it has usable handles or a stable, graspable shape; the vertical lift/lower distance and the horizontal distance the load travels away from the body (a load held far from the torso multiplies the effective spinal-compression moment even at unchanged weight); the lifting frequency and duration over a shift (repetition drives cumulative fatigue and MSD risk independently of any single lift's peak force); the asymmetry of the lift (twisting the trunk while lifting sharply increases spinal loading versus a purely sagittal-plane lift); the coupling quality between hand and load (a good handle vs. a slippery, sharp-edged, or bagged load changes the safe weight limit substantially); and the quality of the lift itself — posture at the start/end of the lift, floor surface/footing, and available recovery time between lifts.

(ii) Two Most Promising Engineering Solutions

Rather than relying on training or administrative controls (which are the least reliable layer of the hazard-control hierarchy), the two most promising engineering interventions are: (1) mechanization/automation of the handling task — conveyors, hoists, vacuum lifters, powered manipulators, or automated material-transfer equipment that removes the human from the force-generating role entirely for the heaviest or most awkward lifts; and (2) workstation and task redesign to reduce the physical demand of any lift that remains manual — reducing the vertical travel distance and horizontal reach by positioning storage/staging at a better height, providing adjustable-height lift tables/pallet positioners so the load starts and ends near waist height, and reducing unit load size/weight (e.g., splitting a large container into smaller, more frequent handling units) so no single lift approaches the population's safe-lifting capacity.

(iii) Lifting-Rate Calculation

Given. Lift height $H = 4\text{ ft}$; weight lifted $W = 55\text{ lb}$; energy consumption for the task $k = 5$ gram-calories per ft·lb of work done; desirable energy-expenditure limit $\dot{E}_{\text{limit}} = 200\text{ kcal/hr}$.

Find. The number of lifts per hour consistent with staying within the desirable energy-expenditure limit.

Approach. Compute the mechanical work done per lift, convert it to an energy cost per lift using the given energy-consumption rate, then divide the hourly energy budget by the energy cost per lift.

  1. Work done per lift. Work = weight × height lifted: $$w = W \times H = 55\text{ lb} \times 4\text{ ft} = 220\text{ ft}\cdot\text{lb per lift}$$
  2. Energy cost per lift, in gram-calories. Apply the given energy-consumption rate of 5 gram-calories per ft·lb: $$e = k \times w = 5\ \tfrac{\text{g-cal}}{\text{ft}\cdot\text{lb}} \times 220\text{ ft}\cdot\text{lb} = 1100\text{ gram-calories per lift}$$
  3. Convert to kilocalories. Since 1 kcal = 1000 gram-calories: $$e = \dfrac{1100}{1000} = 1.1\text{ kcal per lift}$$
  4. Allowable lifts per hour. Divide the desirable hourly energy limit by the energy cost per lift, then round down to a whole, practicable lift count (a fractional lift cannot be performed, and rounding down keeps the operator inside the limit): $$n = \dfrac{\dot{E}_{\text{limit}}}{e} = \dfrac{200\text{ kcal/hr}}{1.1\text{ kcal/lift}} = 181.8\text{ lifts/hr} \;\Rightarrow\; \boxed{n \approx 181\text{ lifts per hour}}$$
Final Results
QuantityValue
Work done per lift220 ft·lb
Energy cost per lift1.1 kcal
Allowable lift rate≈ 181 lifts/hr (181.8 before rounding down)
Check — the exam gives no coupling/duty-cycle detail beyond the four stated quantities, so this solution treats the 200 kcal/hr figure as a net task-energy budget already accounting for the worker's baseline metabolic rate, and rounds the theoretical rate (181.8) down to the nearest whole lift (181/hr) rather than up, since exceeding the stated limit — even by a fraction of a lift — is the outcome the limit exists to prevent.