W2016 Test 3 Marking Guide

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University of Waterloo *

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162

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Mechanical Engineering

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Jan 9, 2024

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BME 162 Winter 2017 Test 3 Answers

1. (5 marks, one for each term ) In each case, provide the terms that best explain the alarm problem that is occurring:

2. (5 marks, one for each term ) In each case, provide the terms that best explain the visual problem that is occurring: If your answer for (1) is contract sensitive, you can get 0.5 point.

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3. (3 marks) A biomedical engineering student drinks too much coffee before his midterm and feels after that it may have influenced his performance. Is he right? Use the Yerkes Dodson law to discuss his situation (1)Coffee would lead to high arousal. 1 mark (2)(1)People actually perform better at medium levels of arousal. 1 mark (3)(1)Particularly on complex tasks, lower levels of arousal are needed. 1 mark

4. (4 marks) One biomedical engineering student has a workterm job where her manager keeps her very busy. She works 60 hour weeks, doesn’t like what she’s doing but it has been made clear to her that if she complains she will be let go. Another biomedical engineering student is also having a busy workterm. She has been given an independent project to take in whatever direction she likes. She finds it very engaging and easily puts in 60 hours a week working on it. Who is having the more stressful workterm? Use Karasek’s demand control model to explain your answer. (1)Student 1 has high job demands and low decision latitude. This makes for higher stress. (2) Student 2 also has high job demands but high decision latitude. This can be a very healthy situation. You need to identify the two components for each student. The best answers will use the terms properly And you will be taken a mark off if you are messy with your concepts.

5. A. (5 marks) A hospital has just upgraded their CAT scanning equipment. If the equipment is better than the previous equipment, what would you be expecting to see, from a signal detection theory perspective (1) You should state that the new equipment should spread out the curves, making signal more easily differentiated from noise. 1 mark (2) This should result in more correct diagnoses (hits), more correct rejections, fewer false positive diagnoses (false alarms), and few missed positives (misses). 4 mark (3) You can draw graphs, but you should certainly explain them.

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Question 6. Biomechanics • The following slides shows step by step how to solve this question. • Due to the nature of the question, if you miss one step (e.g. fail to notice that Fdesk = Fchair), it is possible that you lose more marks as the Forces are used again when calculating moments.

27 cm 60 60 60 60 38 c m 19 cm 32 cm Mass of head = 4.7 kg Mass of torso = 21 kg Mass of upperarm = 3.1 kg Mass of each forearm = 2.8 kg Assume the mass of the head acts at the middle of the head Assume the mass of the torso acts at the middle of the torso Assume the mass of the upper arm acts at 1/3 down from the shoulder Assume the mass of the forearms acts at the middle of the segment. Assume the elbows are pushed together and act as a single point of contact. You can ignore the neck Assume the only points of contact are the stars. The heels of the feet only brush lightly against the chair for some stability but are not bearing weight. The elbows and the seat points each take 50% of the weight of the forearms, upperarms, head and torso. You can ignore the mass of the legs for this problem. Question 6. Biomechanics

27 cm 38 c m 19 cm 32 cm Mass of head = 4.7 kg Mass of torso = 21 kg Mass of upperarm = 3.1 kg Mass of forearm = 2.8 kg Assume the mass of the head acts at the middle of the head Assume the mass of the torso acts at the middle of the torso Assume the mass of the upper arm acts at 1/3 down from the shoulder Assume the mass of the forearm acts at the middle of the segment You can ignore the neck FrCHAIR FrDESK FHEAD FTORSO FUARM FFARM Build a free body diagram. Static equilibrium. FrDESK+FrCHAIR=FFARM+FUARM+FHEAD+FTO RSO MrDESK+MrCHAIR+MFARM+MUARM+MHEAD+M TORSO=0 1 mark 1 mark Question 6. Biomechanics

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27 cm 3 8 c m 19 cm 32 cm Mass of head = 4.7 kg Mass of torso = 21 kg Mass of upperarm = 3.1 kg Mass of forearm = 2.8 kg Assume the mass of the head acts at the middle of the head Assume the mass of the torso acts at the middle of the torso Assume the mass of the upper arm acts at 1/3 down from the shoulder Assume the mass of the forearm acts at the middle of the segment You can ignore the neck Calculate the forces and where the forces act. 60 60 60 60 Forearm Point of action = 16cm along the segment A=cos(60).16m A=0.08m FFARM=2.8kg*9.8N/kg*2forearms FFARM=54.88N at .08m from the elbow Upperarm Point of action= 1/3 of the way along the segment 3A=cos60*.27m A=0.045m from the shoulder FUARM=3.1kg*9.8N/kg*2upperarms FUARM=60.76N at .045m from the shoulder 1 mark 1 mark .5 mark .5 mark Question 6. Biomechanics

27 cm 3 8 c m 19 cm 32 cm Mass of head = 4.7 kg Mass of torso = 21 kg Mass of upperarm = 3.1 kg Mass of forearm = 2.8 kg Assume the mass of the head acts at the middle of the head Assume the mass of the torso acts at the middle of the torso Assume the mass of the upper arm acts at 1/3 down from the shoulder Assume the mass of the forearm acts at the middle of the segment You can ignore the neck Calculate the forces and where the forces act. 60 60 60 60 Head Point of action = 19cm+38cm= 57cm along the segment A=cos(60).57m A=0.285m from the seat point FHEAD=4.7kg*9.8N/kg FFARM=46.1N at .285m Torso Point of action= 1/2 of the way along the segment 2A=cos60*.38m A=0.095m from the seat point FTORSO=21kg*9.8N/kg FTORSO=205.8N at .095m from the shoulder (or the seat) 1 mark 1 mark Question 6. Biomechanics

You can solve the reaction forces now. FrCHAIR FrDESK FHEAD FTORSO FUARM FFARM FrDESK+FrCHAIR=FFARM+FUARM+FHEAD+FTO RSO FrDESK+FrCHAIR=54.88N+60.76N+46.1N+205. 8N FrDESK+FrCHAIR=367.54N We also know that FrDESK=FrCHAIR 2FrDESK=367.54N FrDESK=FrCHAIR=183.88N [up] 1 mark Question 6. Biomechanics

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Time to work on the moments. .08m .045m .135m .285m .095m .19m MrDESK+MrCHAIR+MFARM+MUARM+MHEAD+MTORSO=0 Remember we only need MrDESK MrCHAIR+183.88N*(.19m+.135m) MrCHAIR=59.71N*m[cc] MFARM=54.88N*.08m MFARM=4.39N*m[cc] MUARM=60.76*(.135-.045)m MUARM=5.468N*m[c] MHEAD=46.1N*(.135m-(.285-.19)m) MHEAD=46.1N*(.04m) MHEAD=1.84N*m[c] MTORSO=205.8N*(.135+.095)m MTORSO=47.334N*m[c] MrDESK=5.468+1.84+47.334-59.71-4.39 MrDESK=9.458N*m[cc] 1 mark 1 mark 1 mark 1 mark 1 mark 1 mark 1 mark – signs, units, and clarity Question 6. Biomechanics

Question 7. Ambulance Design • This was a difficult question. The trick was to identify that the cot can be adjusted below hip height (crotch height), allowing the attendants to pass at thigh level. Here is a breakdown of marks • 2.4m-(2x.76m). There are .88m of width left unused in the ambulance. – 1 MARK • 95 th pc male is 53.2cm at the shoulders, so this is the cot width. So the patient fits. The length is also no problem, #805, 190cm. – 1 MARK

• If you set the cot at a height lower than 65.2 cm (the crotch height of the 5 th pc female, measure 246), the attendants can pass by at thigh level. (if it were higher than this they might pass by at buttock level which needs more space) – 1 MARK • For a squeeze by, assume they can touch the cot and rotate sideways. This allows us to use less space to fit one attendant. (e.g. we can think an attendant as a cylinder with a radius of 0.5 thigh clearance (see below) – 1 MARK • Finally, use thigh clearance of the 95 th pc male, 19.1 cm. For two attendants to pass you need 38.2cm so it is short of the requirements by 4cm. – 1 MARK Question 7. Ambulance Design

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8. (10 marks, 7 for modifiers, and 3 for the final calculation and conclusion) A worker at the hospital manually lifts trays of instruments from a cart to a shelf. Each tray of instruments weighs 15 lbs. The worker holds the trays at 10 inches from their midpoint (taken from the ankles). The vertical distance at the start of the lift is 40 inches. The vertical distance at the top of the lift is 60 inches. The worker rotates 45 degrees during the lift and lifts 3 trays per minute for a 3 hour shift. Assume the trays have good handles. Is the lift safe at both the origin and the destination of the lift? HM=10/h=10/10=1 1 mark VM=1-0.0075|v-30|=1-.0075(10)=0925 0.5 mark DM=0.82+1.8/d=0.82+1.8/20=0.91 1 mark AM=1-0.00032a=1-0.0032*45=0.856 1 mark FM= 0.55 1 mark CM=1 1 mark RWL=51*1*.925*.91*.9856*.55*1 RWL=23.3lbs 0.5 mark For the end position, the only value that changes is VM VM=1-0.0075(60-30)=0.775 0.5 mark RWL=16.93 0.5 mark LI=0.886 < 1 Lift is still safe for the majority of the population. 1 mark

9. (4 marks) In the column on the right, put in the term or words that best fits the definition on the left.

10. (5 marks) A dental hygienist can see about 8-10 patients a day. During these appointments she inspects their teeth, gums and overall mouth health. She is responsible for conducting an oral cancer screening. On half her patients she needs to conduct scaling which is a deep cleaning where she physically scrapes hardened plaque from people’s teeth. She also takes xrays which require a careful positioning of an xray film. She mostly sits on a stool that she can adjust the height on and move around the patient. Her appointments generally start with an inspection, then xrays if needed, then scaling if needed, then cleaning. Conduct a task analysis of her job. Identify the most challenging cognitive and physical demands. (1)TA: inspection -> are xrays needed? ---yes->xrays- go to scaling if needed  Is scaling needed? - yes scaling- go to cleaning  Cleaning 2 mark (2)Cognitive demands 8-10 times a day. ( at least two point of these ) - Recognition of what oral cancer looks like - Inspection of gum/teeth health - Positioning of xray films You should note that the recognition of health problem is probably a challenging task. It is abit of a signal detection task. 1.5 mark (2) Physical demands Posture – sitting but often leaning over Scaling – heavy work on hands and wrists. Repetitive, done several times a day. 1.5 point

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