125_208 Homework 4 Solutions

14:125:208: Introduction to Biomechanics Spring 2010 Homework Assignment #4

Due 2/22/10

Problem 1: (Anatomical Terms) 1) The ______ plane divides the body into right and left halves.

A) transverse B) sagittal C) coronal D) oblique E) para-sagittal

2) The armpit or axilla is __________ to the hip.

A) superficial B) deep C) superior D) inferior 3) The "6-pack of abs" is due to the rectus abdominis muscle that lies within the _______ abdominal wall.

A) anterior B) posterior C) superior D) inferior

4) The big toe or hallux (otherwise known as the piggie that went to market) is _______ to the piggie that went wee-wee-wee all the way home (the little toe).

A) medial B) intermediate C) lateral D) median

5) The arm is _________ to the hand.

A) medial B) lateral C) proximal D) distal

6) The scapula (shoulder blade) is _________ to the vertebral column.

A) anteromedial B) posterolateral C) proximal D) distal

7) The leg is _________ to the knee.

A) medial B) lateral C) proximal D) distal

8) Supination of the arm requires the use of the following muscles:

A) pronator teres B) biceps brachii C) brachioradialis D) brachialis E) triceps brachii F) Supinator

9) Arm curls work out all of the following muscles except:

A) coracobrachialis B) biceps brachii C) brachioradialis D) brachialis

10) In the anatomical position, wrist flexion moves the hand _________.

A) posteriorly B) laterally C) medially D) superiorly E) anteriorly


Due 2/21/10

Problem 2. (More Anatomical Terms) i) Label the bones and muscles in the following picture (which you have seen in class)

1 Humerus

2 Supinator

3 Pronator Teres

4 Ulna

6

5 Pronator Quadratus


Due 2/21/10

ii) Explain why the supraspinatus is required to initiate arm abduction When the arm is in the adducted position The deltoid acts primarily in a vertical direction (and can even have an x-directed force towards the body) and cannot produce a sufficient CW moment about the shoulder joint to oppose the CCW moment from the arm weight. The deltoid moment is either very small in the CW direction, zero or even small in the CCW direction. The supraspinatus on the other hand acts in a horizontal direction and can produce a CW moment on the shoulder joint. As abduction proceeds the deltoid acts in a more horizontal position to produce a CW moment and can take over arm abduction after ~15 shoulder rotation


Due 2/21/10

iii) Together the Vastus lateralis, Rectus femoris, Vastus medialis and Vastus intermedius make up the __Quadriceps______ __Femoris______.muscle. (Two words) The _Meniscus, menisci_____________ act as shock absorbers within the knee The femoral head articulates within the __Acetabulum_____ which is a landmark on the Ilium of the pelvis The biceps brachii originates on the _Scapula_______ and inserts on the __Radius_______. The major hip abductor is the ___gluteus________ __medius____________ (Two words) ii) In soccer (football) if you wish to complete a push pass you need to contact the ball with the medial side of your foot by swinging your kicking leg forward and medially as you complete the forward motion phase to kick the ball. This motion requires __Flexion________ and _adduction_______ of the hip and __extension_____ of the knee during the forward phase of the kick (Describe the kick using the anatomical definitions of motion)


Due 2/21/10

In the sarcomere, the thin filament is composed primarily of __Actin____________, ___Troponin_______________ and ___Tropomyosin______ (3 proteins). In the picture on the right you can see the Latissimus dorsi muscle. This muscle is responsible for __Extension______ and __Adduction_______ of the arm (two distinct motions) If an athlete steps down with their knee hyperextended while twisting their leg, they are likely to tear their _Anterior______ __Cruciate______ __Ligament_________ (3 words) The Infraspinatus originates on the __Posterior_____ side of the scapula while the Subscapularis originates on the __Anterior______ side Taken together the biceps femoris, semimembranosus and semitendinosus make up the __Hamstring_____ muscles The bone of upper arm is called the __Humerus_____ Dr. Zahn goes into the doctor for a chest x-ray. One of the images is shown below. This image shows the ___Coronal___________ plane of his chest You are watching a little league baseball game and the pitcher throws the baseball. In order to throw a speedball, the proper completion of the acceleration phase of the throw is critical. The acceleration phase requires the pitcher to bring their upper arm to the front of their body with the shoulder joint acting as a junction point, supporting the rotational acceleration motion of the arm, and throw the ball with their elbow straight and their palms facing downward as they move into the deceleration phase of the throw This motion requires ___Flexion__________ of the arm at the shoulder, ___Extension__ at the elbow and ___Pronation_________ of the forearm in order to throw the ball. (Describe how the pitcher moves his arm to the position shown below during the acceleration phase of the pitch using the anatomical definitions of motion by filling in the three blanks above)

SuperiorInferior


Due 2/21/10

Problem 3. (Static Analysis of a Pushup)

A paraplegic patient weighing 70 kg is performing a pushup in the prone position (below). The lower legs and hands are supported by the floor. A force plate, placed under the right hand, shows a vertical force of 200 N. The moment arm of this force relative to the center of motion of the elbow joint is 0.08 m.

The red line represents the triceps muscle. It is at a 5 deg angle below the horizontal, attaches 1.5 cm from the humeroulnar joint on the olecranon process, and 2.5 cm from the glenohumeral (Shoulder) joint on the humerus. The blue line represents the pectoralis muscle. It makes a 30 deg angle with the horizontal and attaches 5 cm from the glenohumeral joint. The length of the humerus from the radioulnar joint to the glenohumeral joint is 0.2 m. The humerus makes an 8 deg angle with the horizontal. Based on this information, the triceps has a moment arm 1.4489 cm from the elbow joint and 0.1308 cm from the shoulder joint.. The elbow joint reaction force in the x direction Rex has a moment arm 2.7835 cm from the shoulder joint, the elbow joint reaction force

70deg


Due 2/21/10

in the y direction Rey has a moment arm 19.8054 cm from the shoulder joint and the pectoralis muscle has a moment arm 3.0783 cm from the shoulder joint. If we assume that the triceps and pecs are the only muscles supporting the elbow and shoulder, we can draw FBDs of the lower arm/hand and the upper arm, and apply the equations of static equilibrium. Lower arm Upper arm We can see that we have 6 equations and six unknowns. If we write these equations as a system of linear equations we can simultaneously solve for all unknowns.

Ncm1600200*8F1.44890RlFl

0MN200RF0.087156

0200R5sinF0F

0RF0.9960R5cosF

0F

tri

groundelbowtritrielbow

zelbow

extri

eytri

y

extri

extri

x

===+

===+

===

=

R=200 N

Ftri

Rex

Rey

00783.38054.197835.21308.00

0

05.0087156.0030sin5sin

0

0866.0996.0030cos5cos

0

=++=++

==++

=++=

=++=++

=

peceyextri

pecpecshouldereyelbowxshoulderexelbowyshouldertritrishoulder

zshoulder

sypecextri

sypeceytri

y

sxpecextri

sxpecextri

x

FRRFFlRlRlFl

M

RFRFRFRF

F

RFRFRFRF

F

Rsx

Rsy Ftri

Fpec

Rey

Rex


Due 2/21/10

Write the 6 equations of static equilibrium in linear form Ax=b, where

=

sy

sx

pec

ey

ex

tri

RRFRRF

x

Using a program such as Matlab find the inverse matrix A-1 and solve the system of equations to simultaneously determine the magnitude of the triceps force, the magnitude of the pecs force, and the magnitude and direction of reactions at the elbow and shoulder. Print out your computer results clearly showing the definition of matrix A, A-1 and the solution matrix solving for all the values of x. In Matlab a matrix can be defined by the following command A=[a11 a12 a13 a21 a22 a23 a31 a32 a33] where this would define a 3X3 matrix with a space after each element in a row defines the next column and a carriage return defines each new row. To find the inverse of matrix A in matlab you would type inv(A) Notice that this segmented system (upper and lower arm) can be solved simultaneously quite quickly

=

000

16002000

000783.38054.197835.21308.0105.010087156.001866.001996.0000004489.100010087156.000001996.0

sy

sx

pec

ey

ex

tri

RRFRRF

2/21/11 2:41 PM MATLAB Command Window 1 of 2

This is a Classroom License for instructional use only. Research and commercial use is prohibited. To get started, select MATLAB Help or Demos from the Help menu. >> A=[0.996 -1 0 0 0 00.087156 0 -1 0 0 0-1.4489 0 0 0 0 0-0.996 1 0 0.866 -1 0-0.087156 0 1 -0.5 0 1-0.1308 2.7835 -19.8054 3.0783 0 0] A = 0.9960 -1.0000 0 0 0 0 0.0872 0 -1.0000 0 0 0 -1.4489 0 0 0 0 0 -0.9960 1.0000 0 0.8660 -1.0000 0 -0.0872 0 1.0000 -0.5000 0 1.0000 -0.1308 2.7835 -19.8054 3.0783 0 0 >> inv(A) ans = 0 0 -0.6902 0 0 0 -1.0000 -0.0000 -0.6874 0.0000 -0.0000 0 -0.0000 -1.0000 -0.0602 -0.0000 -0.0000 -0.0000 0.9042 -6.4339 0.2052 0.0000 -0.0000 0.3249 -0.2169 -5.5717 0.1777 -1.0000 -0.0000 0.2813 0.4521 -2.2169 0.1026 0.0000 1.0000 0.1624 >> B=[0-200-1600000] B = 0 -200 -1600 0 0 0 >> x=inv(A)*B x =

2/21/11 2:41 PM MATLAB Command Window 2 of 2

1.0e+003 * 1.1043 1.0999 0.2962 0.9584 0.8300 0.2792 >>


Due 2/21/10

Problem 4: (Hip Forces while carrying loads)

A) Find the joint reaction force, R, and the abductor muscle force, M, in the right hip when a woman stands on her right leg while carrying a 25kg suitcase in the left hand (Fig 7.7A). You can use the following data:

Body weight = 600N Weight of standing leg = 100N Moment arm of the weight of the suitcase = 0.31m Moment arm of the abductor muscle force = 0.05m lateral to the femoral head Moment arm of the body weight = 0.05m medial to the femoral head Moment arm of the body weight excluding the standing leg = 0.06m Moment arm of the standing leg = 0.01m medial to the femoral head Moment arm of the ground reaction force = 0.06m medial to the femoral head Abductor muscle force (M) acts at a 70 degree angle to the transverse plane.

(Please Note: the given moment arms are the distance from the hip joint shown by the x in the picture to the force and are always perpendicular to the direction of the force)

B) Find R and M for the right hip joint when the woman stands on the right leg while carrying a 25kg suitcase in each hand (Fig 7.7b). The moment arm for the weight of the suitcase in her right hand is 0.14m. All other weights and dimensions are as above.

Which produces a larger hip joint reaction force: carrying 25kg in one hand or 50kg evenly distributed between two hands?

600 N

600 N


Due 2/21/10

For this problem there are several ways to solve it. I will first solve it the book way. A) We can first draw a free body diagram of the body excluding the leg

We can also solve this problem considering the whole body and the ground reaction force/moment Now we consider the leg

Now we can write the equations of Static Equilibrium

NRRR

NRNFR

NF

FFlWlFl

M

WWRFF

RFF

yx

jy

mjx

m

m

loadloadbodymm

zO

loadjym

y

jxm

x

2625

277070sin215025050073570cos

215005.0

5.7730025031.050006.005.0

00

0sin0

0cos0

22

2

2

=+==++=

===+=

==

==+

==+

=

rr

Fload

NmMFlWlRlM

M

NRWWRF

xinforcesNoF

g

loadloadbodygyreactiong

zhip

gyloadgy

y

x

5.5685006.025031.060005.0

0

0

8502506000

0

0

=+==+

==+==

=

=

W Fload

Rg, Mg

NFNF

NF

FlWlRlMM

FFWRF

FFF

jy

jx

m

mmlegleggyreactiong

zhip

jymyleggy

y

jxm

x

275270sin2130100850

5.72870cos2130

213005.0

10001.085006.05.560

0

0

0

0cos0

=+===

=+==+

==+

==

=

The slight difference in numbers is due to the inexactness of the given moment arms to the body weights

Rg, Mg


Due 2/21/10

To solve Part B) I will use the rest of body Method

Comparing case A to Case B we find the joint reaction force from carrying 1 case is greater than that for carrying the 2 cases. This is because in the second case by carrying the second case the bodyweight is more balanced which shifts the COM closer to the femoral head thus decreasing the moment arm to the COM so the abductor muscles dont have to work as hard (less force) to overcome the moment due to the weights leading to a lower reaction force.

We can write the equations of Static Equilibrium

NRRR

NRNFR

NF

FFlFlWlFl

M

WWRFF

RFF

yx

jy

mjx

m

m

loadloadloadloadbodymm

zO

loadjym

y

jxm

x

2414

236370sin1450500500

49670cos

145005.0

355.7730025014.025031.050006.005.0

00

02sin

0

0cos0

22

22112

2

=+==++=

===+=

=+=+

==+

==+

=

rr

Fload Fload


Due 2/21/10

Problem 5. (Static Analysis of the Sun Salutation)

You have just begun taking a Yoga class and your instructor is teaching you the Surya Namaskara or the Sun Salutation. The Hasta Uttanasana or Raised Hands Posture is one of the poses taken during the Sun Salutation in which you raise the arms upward and slowly bend backwards stretching the arms above the head. As the instructor takes the Hasta Uttanasana pose you are suddenly enlightened and you realize that you can analyze this posture biomechanically.

Your Yoga Instructor Whole Body Lower Legs

You are enlightened with the following information:

Your Yoga Instructor weighs 700 N in total

The instructors COM is 10 cm posterior to his feet (in the z direction).

His legs makes contact with the floor an angle of 75 with respect to the horizontal and he keeps his knees extended.

The lower legs are 45 cm long to the knee joint. This results in a distance from the feet to the Tibiofemoral joint of 11.65 cm in the z-direction and 43.47 cm in the y-direction

The weight of the lower legs is 80 N and the COM of the legs is 20 cm distal to the knee along the tibia, placing the COM 5.18 cm posterior and 19.32 cm inferior to the knee joint in the z and y directions respectively.

The dashed line represents the patella tendon. It is at a 15 degrees angle with respect to the tibia, and attaches 5 cm distal to the tibiofemoral joint along the tibia bone.

The instructors two legs are idealized as a single member, noting that since the feet are placed together any forces calculated would be distributed equally between the two legs.

a) Draw a free body diagram of the entire body in this pose. Determine any reactions at the instructors feet.

b) If we assume that the quadriceps are the only muscles supporting the legs then draw FBDs of the lower legs/feet and determine the magnitude of the quadriceps force, and the magnitude and direction of any reactions at the knee. This problem may be solved by either analyzing the lower leg or by rest of body analysis. If we remove the weight of the legs the COM of the rest of the body shifts to 12.125 cm posterior where the feet would be.

75

10 cm

z

y

W=700 N Wl=80 N

11.65 cm

Tibiofemoral Joint 5.18 cm

15

125_208 Homework 4 Solutions

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