Rutgers : BIO 101 101 : 125_201+Chapter_1_Biomech

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to Introduction Engineering Section 2: Fundamentals of Engineering Rutgers University Langrana, Ph.D David Ph.D Morad Bouzit, Rachel Bareither, MS Martin MD, Ph.D -1- CHAPTER 1: INTRODUCTION TO Derived from classical the field of biomechanics is used to how biological systems respond to exerted upon them. The principles are applied across the from cellular mechanics to movement analysis.

As in classical responses to perturbations in the biological can be divided into three these are described in figure 1-1 as body mechanics, deformable mechanics, and fluid mechanics. and deformable bodies will be the focus of the following chapters in of different systems and how the two can be compared and

Figure 1-1: Description of Mechanics Rigid Body Body Fluid Mechanics Dynamics Elasticity Plasticity Liquids Gases -Kinematics 1.1 Rigid Body and Deformable Definitions When a force on a body, two things can happen: the can move, translating and/or through space, and it can also changing its shape in response to the forces. Think about a golf ball off of a tee.

The club applies a force to the and the ball flies through the translating and spinning. But if you have seen the impact of a club-on-ball high speed video, you that the golf ball significantly changes shape 1-1).

In mechanics, we can separate two responses into separate rigid body and deformable mechanics and we can do the same thing in (Ref: Figure Impact of a Golf Club on a Ball -2- In rigid body we assume that the bodies are interacting mechanically cannot shape.

Newtons Laws of which we learned in Physics, are the principles of rigid body and largely deal with forces and conserving energy and When a body is not accelerating, body analysis reduces to of forces and moments, where the sum of the must equal zero. is called a static analysis.

forces cause the bodies to we need to be able to predict trajectories, both translational and and how these positions change time this is the field of and we need to know how forces these movements this is the of dynamics. In deformable body we allow the bodies to change in and shape. The degree to which occurs depends on the specific of the body.

These properties out to be of critical importance in classical because they define materials, and therefore structures out of these materials, break. By the forces that act on a body and the that result, we can ensure designs for structures can withstand the of day-to-day loading (for understanding the deformation that the disc undergoes during daily loading will aid in the of a replacement — seen in 1.2).

These properties are of greater importance in biological Many cells and tissues in our respond to the deformations and stresses during loading by performing functions. One response could be an electrical impulse, like cells in the cochlea do when bend from sound

Another response could 1.2 Deformation of the be proliferation (to make cells and more proteins to Intervertebral Disc during of the Spine the tissue extra for example, growing more as (Ref: result of print.asp) Biomechanics takes combined principles of rigid and deformable body mechanics and them to principles of biology and thus enabling us to understand the and external forces that act on the body. Modeling the mechanical of the human body allows to simulate loading conditions and

The musculoskeletal system is the primary in our bodies that acts in a fashion (through the use of levers and to create motion), and many of the covered in the next few chapters deal specifically with muscles, tendons, and ligaments. of these biomechanical principles in biomechanics include orthopedic rehabilitation engineering, physical sports medicine, ergonomics, and forensics.

However, many tissues and organs perform functions. For example, the cardiovascular pumps blood through our and skin provides a protective barrier. We are now discovering that forces are critical regulators of at all scales, from controlling the of specific peptide sequences of matrix proteins (Call out to dictating stem cell

Other critical regulations are organ function, like the sensing when it is full, and functions, like running and -3- Call Out Box: Why does not fall off ( Skin is up of two distinct layers, namely the (outer layer) and the dermis layer). The epidermis is joined to the at the basement membrane. This is up of a layer of collagen and other of the basal cells.

Collagen is a durable, fibrous protein makes up 90 percent of the dermis and strength to the skin. How tough? collagen from cattle is otherwise known as leather. the integrity of these layers is to their stability and durability. Bullosa (EB) is a rare condition in which the skin and body linings blister at the knock or rub, causing open wounds.

The condition has a of distinct forms. At its mildest, the is confined to the hands and feet holding things and walking painful. In more severe all the body is affected and the wounds very slowly, giving to scarring, physical deformity and disability.

For many affected by the condition, the is not limited to the skin, but also the inner body linings, as the mouth and esophagus. EB can be divided three main types. division is made on the basis of the disruption (mechanical defect) in the membrane of the skin.

The location of the determines the symptoms of each of EB. EB Simplex; EB Simplex is characterized by a of adhesion of the skin right the basement membrane. The most form of EB simplex has blistering of the and feet, where rubbing to be the main cause (especially of the feet by footwear). Junctional EB: EB is characterized by lack of adhesion of the through the basement membrane.

The of junctional EB is usually rather where half of the children within the first two years as the of malnutrition and anemia which are by serious blistering in the pharynx and the There are only a few patients this severe type survive the second or third of life. The other half of the have a milder form of EB which does not impose restrictions.

Dystrophic EB: Dystrophic EB is by a lack of adhesion of the skin the basement membrane. Dystrophic EB its name from the tendency of the to heal with scarring. process can lead to contraction of the fusion of the fingers and toes, of the mouth membranes and narrowing of the In the following chapters, we will elements of classical, solid in the context of biomechanics.

We will how static analyses of the musculoskeletal parallels similar analyses classical solid mechanics. will allow us to understand how and moments are distributed throughout the We will then examine how forces and moments generate and strain in materials, including tissues and their replacements.

We learn how to identify the mechanical that govern the way a material in response to forces from tests, and how to use this information to design. Lastly, we will analysis of the dynamics of the musculoskeletal to examine how the body responds movement to forces. -4- 1.2 Forces are all around us; they hold us to the they accelerate us in our vehicles, even occur in and around the cells that make up our

The basic definition of a force is the of one body on another body. phenomena produce forces in They may occur at the sub-atomic, molecular, and macroscopic levels. For our we will mainly be dealing two types of forces.

The first is Of course, we all know that is really an acceleration, which on is equal to 9.81m/s2. Gravity on the mass of a body, which is the multiplied by the volume.

Unless we do tricky, we are always accelerating the center of the Earth, and our bodies actually adapted to this magnitude of gravitational acceleration Box effects of gravity on physiology travel). The other type of is a surface force because it over a surface area than a volume). Most the surface force is a contact that results when two are in contact with each

Gravity is often what that contact. As we stand on the there is a body force and a force acting on us. Gravity us to contact the ground, which the contact force.

Callout Effects of Microgravity on Physiology are well-adapted to the physical conditions occur at the surface of the Earth, gravity. As such, human relies on this force to certain functions within the Since humans have and inhabited orbiting stations for durations, the exposure to weightlessness, or has been demonstrated to have adverse effects to health.

effects are widespread throughout the Bone and mineral metabolism. of the evidence to date suggests calcium loss from progresses inexorably during of microgravity, and some studies that full recovery is slow, or indeed does not on return to the 1-G environment. Therefore, changes may be the most important factor for long-term human

Bone loss occurs in trabecular bone as well as bone due to substantial calcium Some studies suggest a of an average of 1-2% of bone density per month in some Bone demineralization is also during prolonged bed rest, not to the same extent as in microgravity.

to reduce bone mineralization weight-loading exercises, wearing a that provides continuous of the skeleton, and nutritional supplements. physiology. All skeletal muscles to the presence or absence of motor therefore, it is not surprising that atrophy is a feature of microgravity.

In the Gemini flights, weight of the astronauts occurred as a result of a reduced body fluid and a reduced muscle mass. In one shuttle flight, decreases in volume showed decreases in leg of 4-6%. Muscle biopsies in have shown that only 5 days of microgravity, cross-sectional areas of muscle were 11 and 24% smaller in type I and II fibers, respectively.

The reduction in mass is accompanied by reduced force and increased muscle Countermeasures for muscle atrophy vigorous exercise programs. The represents only a few of the systems the body that are affected by

Ref:, Physiology of a Environment Historical Perspectives: in microgravity John B. West, J Physiol 89: 379-384, 2000) -5- Forces -Forces are vectors The basic and important element of is to recognize that forces are Thus, they possess and direction, and when you operate with forces, you do so vectorally.

For instance, when you add two forces, you resolve each force its individual components, which act in directions. Then add those together, and express the final as a vector. So if we are dealing with a Cartesian coordinate system you add the x-components together, the y-components and the z-components together, and express the force as these three (seen in Figure 1.4).

F = Fx + Fy + Fz Figure 1.4 ( Equation 1.1 can be expressed in terms of its unit j F = ai + b + ck j Where a, b, and c are the magnitudes and i. and k are the unit of F in the x, y and z directions respectively, as seen in 1.4a. z F F c k j a i y b x Figure 1-4a -6- The of each component relates to the of the resultant which can be found simple Pythagorean geometry. ( Fx ) 2 + ( Fy ) 2 + ( Fz ) 2 = F To obtain the magnitude of each F must be resolved using the the force makes with axis from figure Therefore: Fx = a = F sin cos Fy = b = F sin sin Fz = c = F cos Alternatively, three new can be defined as the angle that the cosine between the resultant (hypotenuse) and the respective axis 1-5).

Fx = F cos x (1.4) Fy = F cos y Fz = F cos z A (1.5) B Figure 1-5: Components of In a) the components of the two-dimensional force F are The components for a three-dimensional force are in b).

The addition of several more acting through the origin can be in the same manner. By adding all the components in each axis, a vector R can be obtained. r n n n R = a 1 + a 2 +. + a n i + b1 + b2 +. + bn + c1 + c 2 +. + c n k j 1 1 1 -7- Example 1 A throws a ring buoy to a who is having problems swimming in the The lifeguard lies on his stomach and the child to safety.

The child a force of 50N at the point shown in the below. What are the angles x, y, and z. and the components of the 50N force? The first is to use trigonometry and find the diagonal of the box in the figure. z d = x2 + y2 + z2 d= (6)2 + (2)2 + ( z 6m y F d = 6.34m 2m We can use this to determine the 0.5 m y angles, x 2 x x x = cos 1 = cos 1 = 71.61 d 6.34 y 6 y = cos 1 = cos 1 = d 6.34 z 0.5 z = cos 1 = cos 1 = 94.52 d 6.34 We now may for the magnitude of the forces in the three x, y, z: 2 Fx = F cos x = 50 = 15.77 N 6.34 6 Fy = F cos y = 50 = 47.31N 0.5 Fz = F cos x = 50 = 3.94 N 6.34 We can check by equation 1.1, ( Fx ) 2 + ( Fy ) 2 + ( Fz ) 2 = F. F= (15.77 ) + ( ) + ( 3.94 ) 2 2 2 = 50 N When resolving a onto a non-conventional axis the law of sines and cosines may be utilized to for the forces.

A review of this law can be in appendix __. 1.3 Moments The result of acting at a distance. Forces do not cause bodies to translate also can cause the bodies to about some point.

describe this quantity. If a is in equilibrium, then these must be balanced, just forces. For example, when a has to walk the plank on a pirate the plank attempts to rotate (in case, clockwise) about the where it is nailed to the ship, O (Figure 1.6).

Because it is to the boat it is prevented from The plank -8- therefore bends due to the that is produced by the weight of the and is proportional to the distance away O. Intuitively, we know that as the walks further down the the plank bends more. The weight (gravitational force, downward) has not changed, but the distance of the to Point O has increased (Figure

The magnitude of the moment M is the force F by the perpendicular distance (d) between the of action of the applied force and O. Like forces, moments are and the magnitude of the moment is written as, M =dF The distance, d, is often called the arm. The line of action is an that is colinear with the vector, and the perpendicular distance of line with the point of is also the smallest distance the line and the point.

Thus, if the of action runs right the point the sailor is directly point O so that his weight intersects with O then the is zero because d is zero. O d F 1-6: The bending of the plank point O is caused by the force by the person standing at the end of it. F M r O d z y As mentioned, the used to find the x moment is the shortest perpendicular distance the force to the point or axis.

In 1-7: A schematic of a force cases in 2-D and very often in creating a moment about O. determining the perpendicular distance is not Vector analysis can be applied to the process. The moment can be determined for F in Figure 1-7, acting point O, using the cross M = rF (1.11) = r F sin Where r in equations is the position vector from O to the origin of the force.

Recall from Physics the hand rule of predicting the of rotation of a moment as shown in 1-8. Using the right point your fingers in the of r and curl into F through the between them. Figure A schematic of the right-hand rule -9- your right hand positive r to positive F will you if it is a clockwise or counterclockwise rotation and your thumb in the positive of the moment.

The moment can be calculated by 1.9-1.11, the result is the rotation or of this point. Equations can be used to solve any 2-dimensional Example 2: A 150N force is to point A at a 75 degree angle in 1-9 for a fixed object. Find the of the force about point O and the distance d that is perpendicular point B to the force line of

This can be accomplished using scalar analysis and vector Scalar analysis. First the rectangular components of the 150 N force. Fx = F cos 75 = 150 cos 75 = N Fy = F sin 75 = 150 sin 75 = 144.89 N After Fx and Fy have calculated we can find the moment point O, choosing the force the object counter clockwise is and the force rotating the object is negative.

5 cm 150 N 75 A 10 cm O 7.5 cm B y x Figure 1-9 M O = Fx (0.10) + Fy = 38.82(0.10) + 144.89(0.05) = 3.36 Nm to find the distance, d, from to B the line of action of the 150 N force, the about B needs to be found, and the perpendicular distance may be found, 10 — M B = Fx (0.175) + Fy (0.05) = + 144.89(0.05) = 0.45 Nm MB = F d d= MB F = 0.45 = = 3mm 150 Vector analysis. Again we to find the force vector F = 150(cos 75 o + sin 75 o ) = 150(0.258i + 0.966 ) = + 144.9 ) N j j r = rA O = (0.05i + 0.10 j Note, there are differences F from the equations of Fx and Fy above due to

From vector mathematics, that rA O = rA rO = ( x A xO )i + ( y A y O ) j Using the equation, M = r F. for MO, i M O = rA O F = rx Fx k j ry Fy i j k 0 = 0.05 0.10 0 = [(0.05)(144.9 ) )(38.7 )]k = 3.375 Nm(k ) 0 144.9 0 Finally, the last of the problem, we find MB in the same as MO above, We use the position vector, r = rA B = + 0.175 ) j M B = rA B i F = rx Fx j ry Fy k i j k 0 = 0.05 0.175 0 = ) (0.175)(38.7 )]k = 0.47 Nm(k ) 0 144.9 0 MB = F d d= MB F = 0.47 = 0.003m = 3mm 150 For cases, as in Figure 1-10a, the can be calculated using the Equations and 1.12. i M O = r F = rx j ry k rz = (ry Fz rz F y )i + (rz Fx rx Fz ) + (rx F y ry Fx )k j Fx Fy Fz (1.12) j = M Ox i + M Oy + M Oz k Where the of the moment represented in Figure can be found using the equation, 2 2 2 M O = M Ox + M Oy + M Oz — 11 — And the direction of component, representing the directions are related with the unit of the moment, represented in Figure can be calculated as well. As shown in 1-10, the moment can also be as: M O = M Oe (1.14) Where e = cos x i + cos y + cos z k. leading to the j x = cos 1 y = cos 1 z = cos 1 M Ox MO (1.15) M Oy (1.16) MO M Oz MO (1.17) 1-10 Example 2 You are walking dog and he sees a squirrel ahead.

He a force of (0, -180 N, 360 N) to your hand, which is located at a roughly (-0.2m, 0.2m, from your center of (Figure 1.11). Calculate the x, y, and z of the moment that is created due to the of the dog about the center of mass of the y Center of Mass (-0.2, 0.5) x F=0i 180j z Figure 1-11 — 12 First, we can find the moment the center of mass by using 1.11 knowing r r that r = + 0.2 + 0.5k (meters) and F = 0i 180 + 360k j j i j k M O = r F = 0.2 0.2 0.5 = 162i + 72 + 36k ( Nm) j 0 180 360 The magnitude of the moment can now be by Equation 1.13, 2 2 2 M O = M Ox + M Oy + M Oz = 162 2 + 72 2 + 36 2 = 180.9 Nm to solve for the directions associated the unit vector of the moment the center of mass of the dog walker, we can use 1.15-1.17.

M 162 cos x = Ox = = 0.90 MO 180.9 x = cos 1 = 25.8 1.3.1 Moments and in Biomechanics Just like moments must balance for a to be in equilibrium. If there is a net moment on a or a system of bodies, then (or more appropriately, a change in will follow.

For static r F =0 and r M = 0 In our plank example, in order for the to remain fixed in space a rigid body analysis, and ignoring the deformation that with bending), the fixed end of the must exert two reactions a force, equal and opposite to the of the sailor, assuming the plank has no and a reaction moment, equal and to moment produced by the sailor he exerts a force (his at a distance from the fixed end of the As he walks farther down the the reaction force stays the but the reaction moment increases with his distance.

What does this for biomechanics? Consider this example. Take an ordinary of free weights lets light, say 5lbs mounted on a that is 3ft long, one at either Hold the barbell straight your head with two evenly spaced on the barbell.

No right? The 10lbs is transferred your skeleton to the ground, exerts a 10lb force up on you. Each arm is holding up and the symmetry of the forces cancels out the the barbell is not forcing you to bend or to maintain the upright position. bones and muscles dont to work too hard to maintain posture. Now take the same and fix them both to one end of the barbell.

the other end of barbell, with hands, so the end of the barbell touches abdomen and points straight out in of you, parallel to the ground. You likely need to lock wrists. Assuming you — 13 can maintain this position therefore remain in static what reactions must you The mass of the weights is the same, you must exert the same force as when the weights over your head.

Why is it so to maintain this position? The are acting at a distance (

3ft) the point of application of the balancing your hands/wrists. You can actually the weights try to rotate your down this is what the produced by the weights at the end of the barbell like! To keep the barbell you must provide the reaction and generating that reaction requires a significant amount of

This problem is akin to the walking the plank the weights are the the barbell is the plank, and your including the muscles, tendons, and ligaments, are the nails that the plank to the ship. In classic we would analyze the forces and in the nails and on the plank, and then how they would deform and they would break in to those forces and moments.

Honda Xaxis

In we can do the same analysis on muscles, and bones. But to do so, we must first a technique that allows us to the forces and moments internal to the of bodies we must section the and draw a free body Example 3 H X Figure 1-13 1-12: Cyclist for example 3 a cyclist standing on a bike one foot and holding the handle bar one hand as in Figure 1-12.

A on the handle bar measures a force acting vertically as shown in 1-13. The weight of the whole is 600N while the weight of the leg is 100N. Suppose that F is the of the resultant force exerted by the hip muscle and R the magnitude of the joint force applied by the pelvis on The angle between the line of of the resultant muscle force and the is =70 as shown FBD of the leg alone (Figure

Following are the moment arms to the hip joint marked by X: — 14 F X Figure 1-14 Moment arm of the on the hand (force H) = 0.16m arm of the flexor muscle force F) = 0.05m Moment arm of the body = 0.08m Moment arm of the body excluding the standing leg = 0.08m arm of the standing leg = 0.04m Moment arm of the reaction force = 0.08m a free-body diagram of: (i) the bike the cyclist, (ii) the bike (iii) the cyclist alone, the standing leg alone, and (v) the body the leg. In each drawing all external forces.

Complete analysis on (iv) the standing leg to solve for the resultant forces Rx and Ry at the hip 1-15). (i) (ii) X Wc H Wb Wb N N (iii) P N N (v) Ry Rx F X Rx H Wl Wc Ry F Wc-l P P Figure 1-15 15 — H Now to solve for the standing leg, we first can easily the force P by using free diagram (iii) and the information in the question, P = WC H = 600 150 = 450 N Now we can move to free diagram (iv) and use static to solve the equations. Fx = 0 R x F cos = 0 F = 0 M =0 y R y + F sin Wl + P = 0 + P. f Wl .e F .b = 0 O z From O Mz = 0 P. f Wl .e b 450 100 0.04 F= = 640 N 0.05 F= From F x =0 R x = F cos R x = cos( 70 / 180 ) = 218 .9 N From F y =0 R y = Wl P F sin R y = 100 450 640 sin( 70 / = 951 .40 N 1.4 Free-Body Diagram (FBD) diagrams are an essential part of any body analysis, and mastery of the almost ensures a correct

Free body diagrams the proverbial A picture is worth a words. The diagrams include or drawings of the essential elements and all forces, moments both the and direction and dimensions associated those elements. Perhaps importantly, individual bodies, or or bodies, must be sectioned to separated or free from the of the elements to complete the analysis.

By separating a body, you create two new on which forces can act. new forces are always equal in and opposite in direction. — 16 Table 1: Drawing free diagrams 1 Examine the problem and which body or bodies to be drawn into the free-body and which need to be isolated the rest of the environment 2 Draw or a simple diagram, with the geometries of the body or bodies 3 identify all of the forces that are upon the body or bodies on the diagram 4 Pick a coordinate that will be used you are solving your free-body and sketch it.

Also remember to add any to your free-body diagram to it complete. Example 4 Consider a of manual lifting. We are interested in reaction forces created at the lumbar spine (L5/S1) These forces will be due to body weight, posture and with or without an external Figure 1-16A shows a two schematic of a person in a straight posture with a box in his hands.

The and external weight (from the create clockwise moments needs to be balanced. In a simplified way of one of the back muscles called the Spinae muscle will the counter clockwise moment to this moment. To maintain the rather large compressive and joint forces are created as in Figure 1-16B.

A B Figure Schematic diagram of sagittal lift — 17 — 1.5 Analysis Now that we can draw body diagrams we are ready to the forces and moments that things from moving. equilibrium is defined as a condition in the bodies in question are at rest, or in a motion.

As mentioned, for an object to be in equilibrium, both the net force and the net must be zero, r F =0 r M =0 and for this to be true, the individual components of the vectors of force and moments also be equal to zero: Fx = 0 Mx = 0 F F y z M M =0 =0 y =0 z =0 for a static equilibrium analysis, are 6 equations that can be solved. In cases, an analysis reduces to two for instance, in the co-planer system in 1-12.

In these cases, the of each body reduces to 3 Fx = 0 F M =0 y z =0 Since there are no forces components in the z-direction, and z=zero for all in the xand y- directions, there can be no about the x- or y-axes, and any moment be about the z-axis. LM LT LH LA For example 4, above in Figure 1-16, the step is to draw the upper and all the relevant forces, moments, and (Figure 1-17).

Clearly, we can force vector acting from body segment and the box, and up from L5/S1 reaction force. But are these to understand what is happening at the Note that the body is vertical; therefore, orientation of force and compressive joint are in the vertical direction too.

But will not always be the situation. this lifting is only in the plane, we can reduce this to a 2D with the x-axis aligned the axis of the shear joint and y direction pointed upward. in our equilibrium analysis, we will the same 3 equations as in the plank shown in Figure 1-6.

Fx = 0, F y = 0, M z = 0 LB 1-17 — 18 — The information can be calculated from data provided in Chapter 2: Head Trunk Arms Box Weight (N) 50 280 65 100 Moment Arm (cm) LH = 22 L T = 12 LA = 25 L B = 42 LM = 6 the body is in a static position, the sum of the about any point is zero. at the L5-S1 position of the lower the muscle force is calculated as M L5-S1 = 0 Therefore: FM x 6cm [50N x + 280N x 12cm + 65N x 25cm + x 42cm] = 0 6cm FM = 10,285 FM = N Now lets look at another A policeman is holding his arm out to his side extended (Figure 1-12).

In his is an apple, which weighs one-third of a pound, or 150g. How force do his muscles need to to keep his arm from rotating i.e. maintain static Figure 1-12. (Ref: — 19 — The first is to draw the policemans arm and all the relevant moments, and distances (Figure

In this case, the joint in is the shoulder, so lets generate a FBD of the from the hand to the shoulder. that we can reduce this to a 2D with the xaxis aligned the axis of the arm, and y direction upward. Thus, in our equilibrium we will have the same 3 as in the plank example.

Lapple (known direction, unknown Fshoulder (unknown direction, Wapple Warm Figure (Ref: Langrana lecture We know our free body the arm, has a weight (Warm is generally known from studies), and we want that acting at the center of gravity of the arm measuring from the shoulder the of our sectioning cut). The apple has a (Wapple) acting at a distance from the shoulder.

Both of forces are acting down, and will create moments make the arm rotate downwards. If the arm a plank, then nails hold it upright. The nails the degrees of freedom for the plank, or the directions of translation and rotation, to

The shoulder, however, has 3 degrees of It cannot translate your arm not separate from your but it can rotate about all 3 axis. The (or at least part of the shoulder) is an of a ball-and-socket joint.

In the next we will review different joints and their mechanical and which muscles contribute to the balances for those joints. To the shoulder from rotating, the must exert a force his deltoid muscles, which the humerus to the back. We have sectioned right through the when we generated our FBD.

The connect to the upper arm at a point close to the shoulder joint, at a Ldelt, and at an angle. When the contracts, it pulls on the arm at the point of at this angle. Generally, anatomical features, like the points of muscle tendons and angles of muscles, are known.

We can now draw this force in our FBD diagram). Examining the FBD, we see there are now 2 forces acting (Wapple and Warm) and one force up (the vertical component of It is possible that these 3 balance each other.

But now is also a horizontal component Fdelt that will the arm into the rest of the body balances this? The bones of the joint support this and the bones can actually support (but not moments) in all three We can draw the relevant forces for analysis in our diagram now, as (Fshoulder, which has a horizontal and 20 — vertical component).

forces act at the shoulder joint, or in our coordinate system. Now we are ready to our analysis! Fx = 0, F y = 0, M z = 0 First, look at the

What forces act in the x-direction? Fx = 0 Fdelt x = 0 r Fshoulderx Fdelt cos = 0 In equation, there are 2 unknown and the magnitude of Fdelt. Now look at the What forces act in the y-direction? FY = 0 + Fdelty Wapple Warm = 0 r + Fdelt sin Wapple Warm = 0 We added one more unknown and one more equation, so we now have 2 and 3 unknowns.

Our final equation is the balance in the z-direction. Before we this equation, however, we choose a location from we take the moments. It turns out this point can be anywhere in the but that choosing some makes the problem easier choosing others. We will the shoulder: Mz =0 Fdelt y Ldelt Larm Wapple Lapple = 0 which forces contribute to the (Fdelty, Warm, Wapple) and do not (Fdeltx, Fshoulderx, Fshouldery).

The forces do no contribute to the moment they act at the point about we are taking the moment they no moment arm. The horizontal of the deltoid force also has a moment arm because its line of also passes through the joint. The remaining forces force the arm to rotate down and negative) or up (counterclockwise and positive).

is our 3rd equation, and we have not introduced any unknowns, so we can solve this of equations. Notice that we left our analysis in general leaving variable names in the instead of replacing them their numerical values. is exceptionally advantageous in solving problems for two reasons. First, it us to see the relationship among the forces and how changing one variable makes the increase or decrease.

Second, it us to swap different values for the and use a computer to solve the equations. The are the same if the person is holding an or a brick all we have to do is insert the value for W at the appropriate location (or any other variable for that

Lets put some real into this problem: = 5kg x 9.8m/s2 = 49N Wapple = 0.15kg x 9.8 = 1.47N Larm = 26.5cm shoulder Lapple = 60cm shoulder Ldelt = 10cm shoulder = 20 — 21 — in the x-direction: r Fshoulderx Fdelt cos = 0 r Fdelt cos 20 o = 0 Forces in the y-direction: r + Fdelt sin Wapple Warm = 0 r + Fdelt sin 20 1.47 49 = 0 And the moments the z-axis: Fdelt y Ldelt Larm Wapple Lapple = 0 y 10 49 26.5 1.47 60 = 0 r Fdelt sin 20 10 49 1.47 60 = 0 Solving this yields r Fdelt = 405 N Plugging value into the 2 above r FshoulderY + Fdelt sin Wapple = 0 FshoulderY = 88 N And r Fshoulderx Fdelt cos = 0 = 381N So holding an apple in outstretched arm, which weigh a little over requires you to exert over of force or over 8 times the This also generates force at the shoulder (a magnitude of In the next few chapters, we will about musculoskeletal anatomy and and solve similar biomechanical for the shoulder and other joints. 22 —

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Honda Xaxis
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