Back Exam 1A - Please review and complete the required...

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Unformatted text preview: Please review and complete the required information on this cover page before answering the questions. Biomechanics BMED-4540 Department of Biomedical Engineering Rensselaer Polytechnic Institute In-class closed book exam (Attach Crib Sheet) Date: September 22, 2003 Duration: One hour & 50 minutes (2 PM to 3:50 PM) MQfi: Total pomts: 100 Mam 3 W45, Name: ANSWER KEY This examination accounts for 15% of your final grade. Please read the instructions carefully and answer each question to the best of your abilities. Total number of pages including cover: 10 NB: Not all Questions carry equal weight. The points are indicated in parenthesis by each Question. Allocate your efforts accordingly and do not exceed the space provided. Be brief and to the point. 1. Muscle length can remain constant (static contraction), increase (eccentric contraction) or decrease (concentric contraction) during a particular movement of the body. Give one example of each of the three types of muscle contractions. In the examples, clearly indicate the type of movement and the name of the muscle involved. (12 points) Muscle involved 0th T eof oint Motion Static contm‘7ti0": Biceps Elbow — held in a flexed position Quadriceps without movement Knee — held in an extended position without movemet . . Biceps Elbow — extension Eccentric contraction: . . Hamstrmgs Knee — extenswn Quadrieeps Knee -flexion Concentric contraction: Biceps Elbow —flexion Hamstrings Knee -flexion Quadriceps Knee - extension 2. Consider the total hip joint prosthesis shown below. The geometric parameters of the prosthesis are such that [1 2 50 mm, 12 = 100 mm, 91: 45°, and 82 = 90°. Assume that when standing on both feet, a net joint reaction force of F = 500 N is acting at the femoral head. Determine the moments . generated about the points B and C due to the joint reaction force F. (10 points) F A 31 M3 = 0 because point B lies on the B 61 line of action of the joint reaction 92 force F. ‘2 MC =de=Fxlg(c0s(t91)) = 500 d N (.1 m )(cos(45 a C MC = 35.36 Nm (cow) 3. The figure (a) below illustrates a person using an exercise machine. The L-shaped beam shown in figure (b) represents the left arm of the person. Points A and B correspond to the shoulder and elbow joints, respectively. Relative to the person, the upper arm (AB) is extended towards the left (x- direction) and the lower arm (BC) is extended forward (2 direction). At this instant the person is holding a handle that is connected by a cable to a suspending weight. The weight applies an upward (in the y—direction) force with magnitude F on the arm at point C. The lengths of the upper arm and the lower arm are a:25 cm and b=30 cm, respectively, and the magnitude of the applied force is F2200N. Use vector method (cross-product) to determine the magnitudes and direction of moments developed at the lower and upper arms by force F. (16 points) (a) T .l’ x . 13 C a. fit .1 H II S: '11 a... “"- + #3“ a“ || .1 x ~11 = ai+bk)><(Fj) =aF(i><j)+bF(k>< j) sassaaiw g § a; s = (0.25)(200)k — (0.30)(200)i || JI c: an I o 9 Mx = 60Nrn (»x direction) (Lower arm) M2 = SONm (+2 direction) (Upper arm) 4. Calculate the tension in the tricep muscle and the joint reaction force at the elbow (point ‘0”) for the elbow extensor exercise shown below. Consider that the muscle attaches at point “A” such that 0A = 2.5 cm. The weight of lower arm (20 N) acts through the center of gravity of the lower arm located at the point B and OB = 15 cm. The load strap carrying a weight of 100 N acts on the arm at point C and OC 2 25 cm. (12 points) T' - ' 211M Ell/Io: 0 ~M(2.5cm) - (20 N)(156m) + (100 N)(25 cm) = 0 M: 880N 215:0 880N+R-20N+100N=0 R=-960N 5. The diagram below shows a dead weight of 100N attached to a patient’s foot by means of a pulley. The distance between the point of knee rotation and the point on the foot where cable is attached is 60 cm (measured along the long axis of the lower leg). The patient uses this set—up while sitting on a chair to exercise her quadriceps muscles. Assume that the weight of the patient’s foot and lower leg is 3ON and their combined center of mass is located 20 cm (measured along the long axis of the lower leg) below the knee axis. The tendon of the quadriceps muscles (patellar tendon) inserts in the anterior (front) part of the tibia at 10 cm (measured along the long axis of the lower leg) below the point of knee rotation and the angle of tendon insertion remains at a constant value of 25 degrees (measured clockwise from the long axis of the lower leg) throughout the exercise. What is the magnitude of force necessary for the quadriceps muscles to maintain the exercise load at 0 (position D) and 90 degrees (position A) of flexion? Note that when the knee is at 0 degree of flexion, the pulley rope forms an angle of 19 degrees with the long axis of the leg. (16 points) At 0° flexion: fillcm IDIJN ZMJ = (M sin 25)(0.1) — (30)(0.2) — (lOOsinl9)(0.6) = 0 : (30)(0.2)-+(1008in19)(0.6) 2604.2N(CCW) I (sm 25)(0. 1) kill 2 FA. = 4x —604.2cos 25 —100c0519 : 0 3 Jx = 642.17N _> ‘i A“: 2P} = Jy + My ~W — Fy = 0 = Jy + (60423111 25) — 30 — (1008in19) :> Jy =192.8N i 135% g In!“ I; gig“ ' a J =‘/(Jx)2 +(Jv)2 = (642.17N)3 +(—192.8N)2 :670.49N L ' g1 62Tan_1[—192.8N 642.17N M H R )2 —16.71° 90° Flexion: “ling lflDN EM, = (M sin 25)(0.1) — (100)(0.6) = 0 ((100)(0.6) (sin 25)(0.l) 2 Ft = Jx —100 + 1419.7 sin 25 = 0 :> Jx = SOON <— \ I1 :13 : Jy+My —W = 0 = Jy + (1419.7cos 25)—30 :> Jy =1256.8N ¢ :1 5:”er J =1/(Jfi)2 +(Jy)2 = N/(fisoof + (—1256.8N)2 =1352.61N #1256.8N —500N M 2 21419.7N(ccw) t9 = Tan'l[ ]: 68,3o 6. A11 890 N man is standing on one foot. Find the magnitude and the direction of the joint reaction force on the head of the supporting femur. The schematic below shows the location and the magnitude of the centers of gravity for the- body and the lower leg. “M” is the unknown tension in the hip abductor muscle group (attached at point 0) acting at 71° with the horizontal. IX and Jy are the unknown x- and y-components of the joint reaction force. (20 points) 50°m138N 2M0: 0 -7cm(Jy) — (138 N)(10 cm) + (890 N)(17.5 cm): 0 J), = 2027.86 N down 213 = 0 890 N +Msin 71 0- 138 N - 2027.86 N = 0 M = 1349.38 N ZFx = 0 (1349.38 N)cos71 0- L; = 0 JI = 439.32 N to the lefi J = (sz + Jfiflm’ = ((439.32 N)2 + (-2027.86 NW”) J = 2075 N 0: tan'1(Jy/Jx) = tan'1(-2027.86/-439.32 N) = 77.8 0 from the horizontal 7. State true or false giving reasons. Clearly state if you agree or disagree with a given statement and explain WHY? (16 points) (a) Initiation of human walking involves disturbing the state of equilibrium present in the resting position by the contraction of a number of muscle groups throughout the human body. True, initiation of human walking involves disturbing the state of equilibrium by contracting muscles to move the center of gravity forward. (b) For the loading condition given below in which the weight W is resisted by the tension in quadriceps, the presence of patella in the knee increases the amount of muscle tension in the quadriceps (indicated by M). W x dw = dm x M The presence of the patella increases “dm” and thus reduces M. Therefore the statement is False. (c) Knee is a more stable joint than the joint formed between the two vertebrae of spine. Joint between two vertebrae False, because the knee is a diarthrodial joint, this allows E varying degrees of movement, while the joint between the vertebrae of the spine is an amphiarthrodial joint, which only allows for slight relative motion. (d) The schematic below represents the use of crutches to increase the stability of a person. Determine if the statement below is true or false. Statement: Positions A and B are primarily secure in the anterior-posterior directions while position C is more stable in the frontal plane than in the saglttal plane" True, because the center of ,l I c m'avity in positions A and B n shift primarily in the iterior-posterior directions 'tile the center of gravity in sition C can move imarily in the frontal ene. ‘ ' "“n? ill 1". J2. 10 ...
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