Human and ergonomics - Speech or Presentation Example

Ergonomics Problems Q1. The worker is holding the toolbox with weight of 8 kg on the right hand and on the left hand he is carrying the bulky material of 6.5 kg. The left arm and the right arm are making 30 degree and 45 degree with torso. The length of the arm is 72cm and weight of the arm is negligible.Ans: The momentum of the right arm= Mass * Load arm = 8 kg * 9.8m/s2 * 72/100 meter * cos45 = 56.44 * 0.71 = 40.

31 The momentum of the left arm= Mass * Load arm = 6.5 * 9.8 * 72/100 * cos30 = 6.5 * 9.8 * 72/100* 0.5 =22.93 Net torque on the back = 40.31- 22.93 = 17.38Nm towards the rightQ2. Boxes weigh 10 kg box lifted from wooden pallet 17cm and lifted through 80 cm and making 90 degree and 45 cm and 38 cm in horizontal distance at original and destination position frequency of lift was 1box/30 minutes and temperature was 32 degree.

The NIOSH lifting equation for lifting weights is given by:Recommended Weight Limit (RWL) = LC x HM x VM x DM x AM x FM x CMwhere:LC = load constant (51 lbs.)HM = horizontal multiplier = 10/HVM = vertical multilpier = (1- (0.0075 [V-30])DM = distance multiplier = (0.82 + (1.8/D))AM = asymmetric multiplier = (1 - (0.0032A))FM = frequency multiplier (from table)CM = coupling multiplier (from table)A = angle of asymmetry = angular displacement of the load from the saggital plane, measured at the origin and destination of the liftwhereH = horizontal location of the load forward of the midpoint between the ankles at the origin of the lift (in inches)V = vertical location of the load at the origin of the lift (in inches)D = vertical travel distance between the origin and the destination (in inches)F = average frequency of lifts (lifts/minute)Since the frequency of lifts were less than 1 hour hence the frequency multiplier is 0.

85 as the work is over one hour.The coupling multiplier is taken as poor as there was no handles and the work was done while standing and hence according to table is 0.9.RWL at original position= 51 lbs * 10/17.71 * ( 1-( 0.0075[6.69-30]) *(0.82 +(1.8/31.49)) * ( 1-(0.0032*90) *0.85* 0.90= 51 *0.56 *1.17 * 0.87* 0.77 * 0.9= 17.11 pound= 7.76kgLifting Index at original position= Actual weight lifted/ RWL at original position= 7.76kg /10kg= 0.77RWL at destination position= 51 lbs * 10/14.9 * ( 1-( 0.0075[6.69-30]) *(0.82 +(1.8/31.49)) * ( 1-(0.0032*90) *0.85* 0.90= 51 *0.67 *1.17 * 0.87* 0.77 * 0.9= 20.48 pounds=9.

28Lifting Index at destination position= Actual weight lifted/ RWL at original position= 9.28kg/10kg= 0.92InterpretationThus it can be seen that the lifting index is more 0.92 in the destination position which is alright as it is below 1 but it approaches 1 nearly and may lead to acute or chronic musculoskeletal disorders as because the lifting index should be always less than at least 1, stress of lifting is less in the original position as the lifting index is 0.77 which was also less than 1.Q3. Alternative 1 Number of workers needed =5Wages =$10/hrDuration of operation= 3 hr/worker/weekInsurance costs= $3 per worker per weekWorkers Compensation= $2 per worker /weekTraining =$120/ workerNumber of working days = 3 days /weekEstimated life cycle = 6 yearsAlternative 2:1 robotInstallation charge of robot = $22000Operation costs =$15/hourDuration of operation 1 hour/dayMaintenance $ 12/ weekSalvage value =$ 1500Number of working days 3 days / week Estimated life cycle = 6 yearsEach month is considered 4.

33 weeks and interest rate 3% /monthPresent worth analysisPV= A * (1+i)n -1/ (1+i)n A for alternative 1( 1 worker) = 90$ *52 +52*3 + 2*52 +20(training/year)= 4960$ For 5 workers = 24800$/year A for alternative 2 = 22000/6 + 45$ *52 + 1500/6 +12$ *52=3666.6 +2340+ 250 +624=6880.6$/yearPV for alternative 1 = 24800 * (1+0.03)6 -1/ (1+0.03)6= 3959.66$PV for alternative 2= 6880.6$ * (1+0.03)6 -1/ (1+0.03)6 = 1098.

58$Hence the implementation of alternative 2 is better as it costs less at PV