Statics of the Body

can be resolved into components in the x, y, and z directions ($$F_{x},F_{y},F_{z}$$). In a static condition the sum of the forces $$\mathbf {F}$$ in each the x, y, and z directions is zero:



$$\begin{aligned} \sum F_{x}=0,\qquad \sum F_{y}=0,\qquad \sum F_{z}=0. \end{aligned}$$

(2.1)
The speed of the center of mass of the object in each direction is then constant, and will usually be assumed to be zero here. These forces can be in balance either for the entire body or for any part of the body.

Similarly, each torque $${\tau }$$ can be resolved into components in the x, y, and z directions ($$\tau _{x},\tau _{y},\tau _{z}$$). In a static condition, the torques $$\varvec{\tau }$$ about the x-, y-, and z-axes also each sum to zero for the entire body and for any body part:


$$\begin{aligned} \sum \tau _{x}=0,\qquad \sum \tau _{y}=0,\qquad \sum \tau _{z}=0. \end{aligned}$$

(2.2)
The speed of angular rotation of the object about each axis is then constant, and will usually be assumed to be zero here.

What actually is a torque? Forces describe changes in linear motion—which means changes in velocities, while torques describe how these same forces can change angular motion—which means changes in angular velocities. The diagram in Fig. 2.1b shows that a force $$\mathbf {F}$$ applied in the positive y direction (with component F) a distance $$+\,r$$ from the z-axis, leads to a torque about the z-axis $$\tau _{y}$$ of magnitude rF. This leads to motion in the counterclockwise direction, caused by an angular acceleration that increases the angle $$\theta $$. This is defined as a positive torque about this axis. A negative torque would occur, for example, if the force were applied in the negative y direction. This would lead to motion in the clockwise direction, caused by an angular acceleration that decreases the (signed) angle $$\theta $$.

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Fig. 2.1
Torques and relevant axes

In general, the torque (vector $${\tau }$$) about any axis is defined as the vector cross product between the distance vector from that axis to the point where the force is applied $$\mathbf {r}$$ and the force vector $$\mathbf {F}$$ (Fig. 2.1e)


$$\begin{aligned} {\tau }=\mathbf {r}\times \mathbf {F}. \end{aligned}$$

(2.3)
(You do not need to understand or use this vector cross product, just the results that are given below.)

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Fig. 2.2
Direction of torques, showing a positive and b negative torques

Because vectors can be translated anywhere, things may be clearer if we move both $$\mathbf {r}$$ and $$\mathbf {F}$$, as in Fig. 2.1b, d, so they originate from where the force is applied, as in Fig. 2.1c, e, respectively. We will call the angle from the $$\mathbf {r}$$ vector to the $$\mathbf {F}$$ vector $$\theta $$. The torque $$\tau _{z}$$ about the upward axis is


$$\begin{aligned} \tau _{z}=rF \sin \theta , \end{aligned}$$

(2.4)
where r is the magnitude of vector $$\mathbf {r}$$ (the distance from the axis to the point where the force is applied) and F is the magnitude of vector $$\mathbf {F}$$. For $$0^{\circ }<\theta <180^{\circ }$$ (or $$0<\theta <\pi $$ in radians), $$\sin \theta $$ is positive and the torque is positive (Fig. 2.2a), while for $$180^{\circ }<\theta <360^{\circ }$$ ($$\pi <\theta <2\pi $$), $$\sin \theta $$ is negative and the torque is negative (Fig. 2.2b). When $$\theta $$ = 90$$^{\circ } ($$=$$\pi /2) $$, $$\sin \theta =1$$ and the torque is rF, as above. When $$\mathbf {r}$$ and $$\mathbf {F}$$ are either parallel ($$\theta =0^{\circ } (=0)$$) or antiparallel ($$\theta =180^{\circ } (=\pi ))$$, the torque is zero.

Clearly, only the component of the $$\mathbf {r}$$ normal to the $$\mathbf {F}$$, which we will call $$r^{\prime }$$, contributes to the torque action. In fact, as Fig. 2.3 proves, $$\tau _{z}= r^{\prime }F$$. Equivalently, only the component of $$\mathbf {F}$$ normal to $$\mathbf {r}$$, i.e., $$F^{\prime }$$, contributes to the torque action, and $$\tau _{z}= rF^{\prime }$$. As we will see, sometimes information is provided where these normal components of displacement or force are provided, and the torques can be calculated without explicitly determining the angle between the displacement and force vectors. Consequently,


$$\begin{aligned} \tau _{z}=rF \sin \theta =r^{\prime }F=rF^{\prime }. \end{aligned}$$

(2.5)
This is true for any axis. The axis can be chosen cleverly for a particular problem to simplify analysis.

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Fig. 2.3
Determining torques from using components of the displacement and force vectors that are normal to the force and displacement vectors, respectively

In linear motion , a force leads to an acceleration $$\mathbf {a} =\mathrm{d}^{2}\mathbf {r}/\mathrm{d}t^{2}$$, which is equivalent to a change in velocity $$ \mathbf {v}=\mathrm{d}\mathbf {r}/\mathrm{d}t$$ (the magnitude of which is the speed v) or momentum $$\mathbf {p}=m\,\mathrm{d}\mathbf {v}/\mathrm{d}t$$ , by


$$\begin{aligned} \mathbf {F}=m\mathbf {a}=m\frac{\mathrm{d}\mathbf {v}}{\mathrm{d}t}=\frac{\mathrm{d}\mathbf {p}}{\mathrm{d}t}, \end{aligned}$$

(2.6)
where m is the mass of the object. Similarly, a torque leads to an analogous change in the angle $$\theta $$ and angular frequency $$\varOmega =\mathrm{d}\theta {/}\mathrm{d}t$$ , and orbital angular momentum L


$$\begin{aligned} \tau =I \frac{\mathrm{d}\varOmega }{\mathrm{d}t}=\frac{\mathrm{d}L}{\mathrm{d}t}, \end{aligned}$$

(2.7)
where I is the moment of inertia and $$L=I\varOmega $$. In static situations the sum of the forces and torques on the object is zero, so the right-hand side of (2.7) equals zero.

By the way, what we are defining as torques are indeed commonly called “torques” in connection to the rotational and twisting motions of objects, as in this chapter and Chap. 3, but are instead called “moments” in connection to the bending of objects, as in Chap. 4.



2.2 Statics: Motion in One Plane and Levers


Many problems involve motion in one plane, say the xy plane—for which z is a constant. For example, the motion of knees and elbows is in one plane. Some problems involving motion of the leg about the hip can be treated in these two dimensions. The six equations in (2.1) and (2.2) then reduce to three equations:


$$\begin{aligned} \sum F_{x}=0,\qquad \sum F_{y}=0,\qquad \sum \tau _{z}=0. \end{aligned}$$

(2.8)
(We will adopt this xyz coordinate system because it is conventionally used in two-dimensional problems, even though it differs from the coordinate system convention we adopted for the body in Fig. 1.​1.)

These types of problems can be classified as one of the three types of levers (Fig. 2.4). There are examples in the body of each. They can be described by how a weight W and a force M, provided by a muscle, act on a solid object, say a bone resting on a fulcrum; this represents an articular joint . The weight can include that of parts of the body as well as external weights. The weight and muscle act at distances $$d_{\mathrm {W}}$$ and $$d_{\mathrm {M}}$$ from the joint. For each type of lever the total torque is zero when


$$\begin{aligned} Md_{\mathrm {M}}=Wd_{\mathrm {W}}, \end{aligned}$$

(2.9)
so


$$\begin{aligned} M=\frac{d_{\mathrm {W}}}{d_{\mathrm {M}}} \; W. \end{aligned}$$

(2.10)
The relative directions of the forces and the relative distances of the weight and muscle forces from the joint are different for each type of lever .

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Fig. 2.4
Three types of levers, a first, b second, and c third class levers. The large increase in the distance the weight moves over the change in muscle length in the third class lever is also seen in part (c)

In a first class lever, the weight and muscle act on opposite sides of the fulcrum and are in the same direction (Fig. 2.4a). This is the least common type of lever in the body. Using the x, y coordinate system shown, there are clearly no forces in the x direction so the first equation in (2.8) is automatically satisfied. Since the weight and muscle both act in the same direction—downward—force balance in the y direction requires that the fulcrum provides an upward force of $$W+M$$. Balancing torques in the z direction requires a choice of a z-axis. Any axis normal to the xy plane can be chosen. The simplest one is an axis at the fulcrum. The weight provides a torque of $$Wd_{\mathrm {W}}$$, while the muscle provides a torque of $$-Md_{\mathrm {M}}$$. The signs are consistent with the above discussion. The fulcrum provides no torque about this axis because the distance from the fulcrum to the axis is zero. So


$$\begin{aligned} \sum \tau _{z}=Wd_{\mathrm {W}}-Md_{\mathrm {M}}=0. \end{aligned}$$

(2.11)
This leads to (2.9), which tells how large the muscle force must be to maintain equilibrium. If the muscle cannot provide this large of a force, there can be no static condition. (Example: The lead ball is too heavy to hold up.) If the muscle provides more than this force, there is purposeful motion. (Example: The baseball is being thrown, as we will see later.) In these two cases, $$\sum \tau _{z}$$ in (2.11) is not zero.

It may seem that we cheated by choosing the axis at the fulcrum. Actually, we could have chosen the axis anyway in the xy plane. To prove this let us choose the axis anywhere along the bone, say a distance x to the right of the weight (Fig. 2.5). The torques provided by the weight, fulcrum, and muscle are now $$Wx,(W+M) (d_{\mathrm {W}}-x),$$ and $$-M(d_{\mathrm {W}}+d_{\mathrm {M} }-x),$$ respectively. Balance requires


$$\begin{aligned} \sum \tau _{z}=Wx+(W+M) (d_{\mathrm {W}}-x)-M(d_{\mathrm {W}}+d_{\mathrm {M}}-x)=0, \end{aligned}$$

(2.12)
which reduces to (2.11) again.

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Fig. 2.5
Displacing the axis for calculating torques to the right of the weight by a distance x, as shown for a first class lever. For the axis chosen at the fulcrum $$x=d_{\mathrm {W}}$$. The axis can be laterally displaced anywhere, to the left or right (as shown) of the lever, above or below it, or in it


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Fig. 2.6
Examples of first (a, b), second (c), and third (d) class levers in the body

One type of the first class lever is a seesaw or teeter totter. A second type is the head atop the spinal cord , where the weight of the head is balanced by the downward effective force of the muscles (Fig. 2.6a). In a third example, the triceps brachii pull on the ulna about the elbow pivot balanced by the forces on the forearm . With the upper arm down, the triceps brachii can balance an upward force pushing the hand up (Fig. 2.6b). (With the upper arm pointed up, the triceps brachii can also balance the hand holding a weight.) Because $$d_{\mathrm {M}} \ll $$ $$d_{\mathrm {W}}$$ here, the force that the triceps brachii needs to exert is much greater than the forces exerted at the hand .

In a second class lever, the muscle and weight act on the same side of the fulcrum, and the weight is nearer to the fulcrum, so $$M<W$$ (Fig. 2.4b). The second class lever has a mechanical advantage , which means that the mechanical advantage, defined as $$W/M = d_{\mathrm {M}}/ d_{\mathrm {W}}$$, is $$ > 1$$” src=”http://basicmedicalkey.com/wp-content/uploads/2017/06/A114622_2_En_2_Chapter_IEq64.gif”></SPAN>. This type of lever is the second most common in the body. One example is standing on tiptoes: the rotation of the foot about the toes (the fulcrum), which would be caused by the weight of the foot, is balanced by the muscle force transmitted by the Achilles tendon (Fig. <SPAN class=InternalRef><A href=2.6c). Another example is pushing down with the triceps brachii.

In third class levers, the muscle and weight are again on the same side of the fulcrum, but now the muscle is nearer to the fulcrum than the weight (Fig. 2.4c). This is the most common example in the body. Because often $$d_{\mathrm {M}} \ll d_{\mathrm {W}}$$, we see that $$M=(d_{\mathrm {W}}/d_{\mathrm {M}})W\gg W$$. This arrangement means that very large forces must be exerted by the muscles because of this $$d_{\mathrm {W}}/d_{\mathrm {M}}$$ amplification, which seems to be a big disadvantage (and is literally a mechanical disadvantage). However, something else is gained in this tradeoff in design. As seen in the Fig. 2.4c, when the bone rotates a given angle, causing a vertical displacement $$y_{\mathrm {M}}$$ at the muscle, there is an amplification of the distance traveled at the position of the weight by $$d_{\mathrm {W}}/d_{\mathrm {M}}$$. As we will see in Chap. 5, muscles are able to contract only a small fraction of their length—which amounts to at most several cm in many muscles. The length of the biceps is about 25 cm, and the maximum contraction is by $$\sim $$7–8 cm. With this amplification, the weight can now move much more than this. One example of a third class lever is the balancing of the lower arm by the biceps brachii inserted on the radius (Fig. 2.6d). Another is holding a weight with an outstretched arm.


2.3 Statics in the Body


We will examine the planar forces in the static equilibrium of the lower arm, at the hip, and in the back. In analyzing the lower arm, we will choose successively more complex and realistic models. We will see that the forces in the hip and back are quite large, much more than one would expect, and explains why people often have problems in these parts of the body, problems that can lead to hip replacements and life-long lower back pain [1, 8, 36]. The approach for these problems is the same. We consider all elements in one plane and examine the forces in the (as defined) x and y directions and the torque in the z direction. Some of the approaches of [3, 28] are followed.


2.3.1 The Lower Arm


We will examine the equilibrium of the forearm balanced by the contraction of the biceps brachii inserted on the forearm long bone called the radius; this is a continuation of the discussion of third class levers. The relevant bones are shown in Figs. 2.7 and 2.8. In equilibrium, the biceps brachii force counters the potential rotation about the elbow joint by the weight held in the hand (Fig. 2.9a). We will examine this example for different models, using Fig. 1.​15 and Tables 1.​6 and 1.​7 to provide anthropometric information. The forearm is 0.146 H long and the hand length is 0.108 H, where H is the body height, so the in the hand is about $$(0.146 + 0.108/2)H =0.2H$$ from the pivot. (The ball is in the middle of the hand.)

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Fig. 2.7
Bones of the arm, anterior view. (From [20]. Used with permission)


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Fig. 2.8
Bones of the arm, posterior view. (From [20]. Used with permission)


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Fig. 2.9
a Weight held in the hand, showing the biceps brachii muscles. (b–d) Forces for the equilibrium of a weight held in the hand for Cases 1–3

Case 1

The biceps brachii insert about 4 cm from the pivot axis. Say there is a weight $$W_{\mathrm {W}}$$ held in the hand, which is $$d_{\mathrm {W}}= 36$$ cm from the pivot. (With $$H = $$ 180 cm, 0.2$$H = $$ 36 cm.) Therefore $$M=(d_{\mathrm {W}}/d_{\mathrm {M}})W_{\mathrm {W}} = $$ (36 cm/4 cm )$$W_{\mathrm {W}} = $$ $$9W_{\mathrm {W}}$$. So for a weight of 100 N the muscle must provide a force of 900 N for balance. Here N stands for the MKS/SI unit of newtons. Since 1 N $$\simeq 0.225$$ lb, equivalently, a 22 lb weight is balanced by 200 lb of force exerted by the biceps brachii (Fig. 2.9b).

We have made several assumptions and approximations in this example without explicitly stating them. It is always good to start with simple models. It is equally important to understand exactly what assumptions and approximations are being made. Then, the model can be made more realistic. Here, we have assumed that the forearm and upper arm make a 90$$^{\circ }$$ angle. We have also neglected the mass of the forearm.

Case 2

Now let us improve the model by including the weight of the forearm $$W_{\mathrm {F}}$$ (Fig. 2.9c). This is $${\approx }0.022W_{\mathrm {b}}$$, where $$W_{\mathrm {b}}$$ is the body weight (Table 1.​7). For a 70 kg (700 N, 160 lb) person, this is $${\approx }15\,\mathrm{N} (3.4\,\mathrm{lb})$$. We can treat the effect of the weight of the forearm as if it were acting at its center of mass, which is approximately in the middle of the forearm, $$d_{\mathrm {F}} = 0.146H/2 = $$ 13 cm from the pivot:


$$\begin{aligned} \sum \tau _{z}= & {} Md_{\mathrm {M}}-W_{\mathrm {W}}d_{\mathrm {W}}-W_{\mathrm {F}}d_{\mathrm {F}}=0 \end{aligned}$$

(2.13)



$$\begin{aligned} Md_{\mathrm {M}}= & {} W_{\mathrm {W}}d_{\mathrm {W}}+W_{\mathrm {F}}d_{\mathrm {F}}\end{aligned}$$

(2.14)



$$\begin{aligned} M= & {} \frac{d_{\mathrm {W}}}{d_{\mathrm {M}}}W_{\mathrm {W}}+\frac{d_{\mathrm {F}}}{d_{\mathrm { M}}}W_{\mathrm {F}}. \end{aligned}$$

(2.15)
The ratio $$d_{\mathrm {F}}/d_{\mathrm {M}}$$ ($$=$$13 cm/4 cm), so now $$M=9\,W_{ \mathrm {W}}+3.25\,W_{\mathrm {F}}$$ and the muscle force required to maintain equilibrium has increased to 900 N + (3.25) (15 N) $$=$$950 N (210 lb).

Case 3

What happens if we no longer assume that the forearm and upper arm make a 90$$^{\circ }$$ angle? Let us keep the upper arm vertical and let the forearm make an angle $$\theta $$, which can range over 142$$^{\circ }$$ (Table 1.​10). The force due to the muscle is then still vertical, and those due to the weights of the forearm and ball are, of course, downward. From Fig. 2.9d we see that the torque caused by each of these three forces is multiplied by $$\sin \theta $$. Now


$$\begin{aligned} \sum \tau _{z}=Md_{\mathrm {M}}\sin \theta -W_{\mathrm {W}}d_{\mathrm {W}}\sin \theta -W_{\mathrm {F}}d_{\mathrm {F}}\sin \theta =0 \end{aligned}$$

(2.16)
and we arrive at the same result that


$$\begin{aligned} Md_{\mathrm {M}}=W_{\mathrm {W}}d_{\mathrm {W}}+W_{\mathrm {F}}d_{\mathrm {F}}. \end{aligned}$$

(2.17)
Actually, we made additional assumptions in this example that we will re-examine later. The distance from the pivot where the biceps brachii insert on the radius really changes with $$\theta $$ (Fig. 3.​46). Also, while this analysis suggests that the muscle force M required for equilibrium is the same for all angles, there is a subtlety in this result. Equation (2.16) gives the muscle force needed to maintain equilibrium. As we will see in Chap. 5, muscles can exert forces up to a maximum value. If the M from (2.17) can be achieved, then there can be equilibrium; if it cannot, then the static condition cannot be achieved. The maximum force that a muscle can exert depends on its length, which, from Fig. 2.9, is clearly a function of $$\theta $$. So the M in (2.17) may be achievable at some angles (nearer $$90^{\circ }$$, where the maximum force turns out to be greatest) and not at others.

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Fig. 2.10
Sketch of the elbow joint for analyzing the statics of the lower arm for Case 4, with the three muscles, the biceps (brachii), brachialis, and brachioradialis, shown in (b–d). (From [18])

Case 4

The biceps brachii are not the only muscles used to flex the elbow. What happens if we also include the contributions of these other muscles? Fig. 2.10 shows that the biceps brachii, the brachialis, and the brachioradialis all contribute to this flexing. Assuming that $$\theta =90^{\circ }$$ (which may not be a good assumption for each muscle), (2.13) is modified to


$$\begin{aligned} \sum \tau _{z}=M_{1}d_{\mathrm {M}_{1}}+M_{2}d_{\mathrm {M}_{2}}+M_{3}d_{\mathrm {M} _{3}}-W_{\mathrm {W}}d_{\mathrm {W}}-W_{\mathrm {F}}d_{\mathrm {F}}=0 \end{aligned}$$

(2.18)



$$\begin{aligned} M_{1}d_{\mathrm {M}_{1}}+M_{2}d_{\mathrm {M}_{2}}+M_{3}d_{\mathrm {M}_{3}}=W_{\mathrm {W} }d_{\mathrm {W}}+W_{\mathrm {F}}d_{\mathrm {F}} \end{aligned}$$

(2.19)
where $$M_{1}$$, $$M_{2}$$, and $$M_{3}$$ represent the forces exerted by the three muscles $$M_{i}$$, respectively. If the physiological cross-sectional areas of the three muscles are $$A_{1}$$, $$A_{2}$$, and $$A_{3}$$, respectively (which we usually call PCAs), and the muscle force for each can be assumed to be proportional to this area (which is a pretty good assumption), then $$M_{i}=kA_{i}$$, for $$i= 1,2,3$$. (We will see that this is a good assumption with k reaching a maximum of $$\sim $$40 N/cm$$^{2}$$ or so.) So,


$$\begin{aligned} kA_{1}d_{\mathrm {M}_{1}}+kA_{2}d_{\mathrm {M}_{2}}+kA_{3}d_{\mathrm {M}_{3}}=W_{ \mathrm {W}}d_{\mathrm {W}}+W_{\mathrm {F}}d_{\mathrm {F}} \end{aligned}$$

(2.20)



$$\begin{aligned} k=\frac{W_{\mathrm {W}}d_{\mathrm {W}}+W_{\mathrm {F}}d_{\mathrm {F}}}{A_{1}d_{\mathrm {M} _{1}}+A_{2}d_{\mathrm {M}_{2}}+A_{3}d_{\mathrm {M}_{3}}} \end{aligned}$$

(2.21)



$$\begin{aligned} M_{1}=kA_{1}=A_{1}\frac{W_{\mathrm {W}}d_{\mathrm {W}}+W_{\mathrm {F}}d_{\mathrm {F}}}{ A_{1}d_{\mathrm {M}_{1}}+A_{2}d_{\mathrm {M}_{2}}+A_{3}d_{\mathrm {M}_{3}}} \end{aligned}$$

(2.22)



$$\begin{aligned} M_{2}=kA_{2}=A_{2}\frac{W_{\mathrm {W}}d_{\mathrm {W}}+W_{\mathrm {F}}d_{\mathrm {F}}}{ A_{1}d_{\mathrm {M}_{1}}+A_{2}d_{\mathrm {M}_{2}}+A_{3}d_{\mathrm {M}_{3}}} \end{aligned}$$

(2.23)



$$\begin{aligned} M_{3}=kA_{3}=A_{3}\frac{W_{\mathrm {W}}d_{\mathrm {W}}+W_{\mathrm {F}}d_{\mathrm {F}}}{ A_{1}d_{\mathrm {M}_{1}}+A_{2}d_{\mathrm {M}_{2}}+A_{3}d_{\mathrm {M}_{3}}}. \end{aligned}$$

(2.24)
Using the parameters from Table 2.1, we get $$M_{1} = $$ 262 N (biceps), $$M_{2} = $$ 399 N (brachialis), and $$M_{3} = $$ 85 N (brachioradialis) when we generalize Case 2. This compares to the $$M_{1} = $$ 696 N that we would obtain for Case 2 with the biceps alone, using $$d_{1} = $$ 4.6 cm (instead of the 4 cm used before, which led to 800 N). The total muscle force is 746 N, which is greater than 696 N because the brachialis has a relatively small moment arm.


Table 2.1
Data for the three elbow muscles used for flexion
























Muscle
Moment arm $$d_{i}$$ (cm)
Physiological cross-section (PCA) (cm$$^{2}$$)

Biceps (muscle 1)
4.6
4.6

Brachialis (muscle 2)
3.4
7.0

Brachioradialis (muscle 3)
7.5
1.5


Using data from [18, 39]

Life is a bit more complex than this result suggests because we assumed that k has the same value for each muscle. Really $$M_{i}=k_{i}A_{i}$$, and all the $$ k_{i}$$’s need not be the same, as long as $$k_{i}$$ is less than the maximum that can be exerted by the muscles. Unfortunately, if all the $$k_{i}$$s are not assumed to be equal, we do not have enough information to solve this problem uniquely as posed. The body may solve the indeterminate nature of this problem (with more variables than conditions) by minimizing energy or optimizing the force distribution (shifting the load from one muscle to another) to rest specific muscles or by keeping the weight balanced better (so it will not tip in the hand).


2.3.2 Hip Problems


The hip (pelvis) is not a single bone, but several bones that are fused together (Figs. 2.11 and 2.12). The pelvis is composed of the pelvic girdle and two parts of the spinal cord, the sacrum and coccyx. The pelvic girdle itself is composed of the right and left coxal (or hip) bones. Each coxal (hip) bone is composed of three bones: the ilium (at the top $$=$$ superior), pubis (bottom front $$=$$ inferior, anterior), and ischium (iss-kee’-um) (bottom back $$=$$ inferior, posterior). The acetabulum (a-si-tab-yoo’-lum) is the socket area where the femur of the leg (Fig. 2.13) is attached (“hip joint”). Actually, the head of the femur is in this socket, and is maintained there by the muscles attached at the greater trochanter. These muscles are collectively called the hip abductor muscles . (The hip abductor muscles are not the only one attached at the greater trochanter (see below), but they are the ones that contribute to the force needed for the equilibrium condition in this problem.)

First we will determine the force on the head of the femur and in the hip abductor muscles while the subject is standing on one leg, say the right leg. This is actually a good model for determining these forces during slow walking . The upper and lower leg and the foot are treated as a rigid body. This problem is solved in two steps. First, the forces in the whole body are analyzed and then the rigid leg is treated as a free rigid body, whose only interaction with the rest of the body will be the normal force from the hip [3, 27]. The bones in the leg are shown in Figs. 2.14 and 2.15.

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Fig. 2.11
Right hip bone in adult. (From [3])


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Fig. 2.12
Front and side views of the hip . (From [3])


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Fig. 2.13
Anterior view of right femur . (From [3])


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Fig. 2.14
Bones of the leg and hip, anterior view, with names of bones in bold. (From [21]. Used with permission)


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Fig. 2.15
Bones of the leg and hip, posterior view, with names of bones in bold. (From [21]. Used with permission)

Total Body Equilibrium

There are only two forces on the body. The body weight $$W_{\mathrm {b}}$$ acts downward, and as if it all originated at the center of mass of the body, which is in the midline in the hip (Fig. 2.16). The foot feels an upward normal force from the floor of magnitude N. There are no forces in the x direction, and these two forces in the y direction must balance in equilibrium, so $$N=W$$. In equilibrium the body cannot start to rotate, so the torques are zero. It is clear from the Fig. 2.16 that this occurs when the foot is directly below the hip, in the midline. If we choose the pivot axis at the center of mass, the torque from the center of mass is zero because the distance term (from the axis to the center of mass) is zero and the torque from the normal force is zero because the normal force is antiparallel to the distance vector ($$\theta =180^{\circ }$$). (You can prove for yourself that the total toque is zero for any other axis normal to the xy plane.)

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Fig. 2.16
Anatomical diagram of the leg and hip for someone standing on one leg, or during slow walking , showing the forces on them and relevant dimensions, including the force exerted on the head of the femur by the acetabulum R and the net force exerted by the hip abductor muscles. (From [3])


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Fig. 2.17
Force diagram for a leg for someone standing on one foot, using Fig. 2.16. (From [3])

Equilibrium of the Individual Body Component

There are four external forces on the leg (Figs. 2.16 and 2.17):

(a)

N is the normal force on the leg from the floor, and we know that $$ N=W_{\mathrm {b}}$$.

 

(b)

$$W_{\mathrm {leg}}$$ is the weight of the leg. Table 1.​7 shows that this is about 0.16$$W_{\mathrm {b}}$$. It acts as if it were applied at the center of mass of the leg, which is approximately halfway down the leg.

 

(c)

R is the reaction force on the leg from the hip, and it is normal to the hip socket. We will define the x and y components $$R_{x}$$ and $$R_{y}$$ so they are positive (Fig. 2.17). Equivalently, we could define the magnitude of R and the angle it makes with the vertical, $$\varphi $$.

 

(d)

M is the force (of magnitude M) due to the hip abductor muscles . There are actually three muscles involved here: the tensor fascia (fash-ee’-uh) latae (la-tuh) (see Fig. 3.​3a), gluteus (gloo’-tee-us) minimus (see Fig. 3.​3c), and the gluteus medius (see Fig. 3.​3a, b). (The gluteus maximus muscle is what the author is sitting on as he is typing this.) These three muscles have a mass ratio of about 1:2:4 and, as we will see, this is roughly the ratio of the forces each can exert. The hip abductor muscle structure we consider is a composite of the three muscles. It has been shown that the effective action of this system is $$\simeq $$70$$^{\circ }$$ to the horizontal, acting on the greater trochanter.

 

We have three equations with three unknowns: $$R_{x}$$, $$R_{y}$$, and M. The two force equations are


$$\begin{aligned} \sum F_{x}= & {} M\cos 70^{\circ }-R_{x}=0,\end{aligned}$$

(2.25)



$$\begin{aligned} \sum F_{y}= & {} M\sin 70^{\circ }-R_{y}-0.16W_{\mathrm {b}}+W_{\mathrm {b}}=0, \end{aligned}$$

(2.26)
where $$W_{\mathrm {leg}}$$ has been replaced by 0.16$$W_{\mathrm {b}}$$.

We will choose the rotation axis to emanate from the center of the head of the femur because the reaction force from the acetabulum passes through this point. This makes the analysis easier, but, of course, the solution would be the same if we chose any other parallel axis. The relevant distances of interest are shown in the diagram (obtained from anatomical dimensions and geometry), as needed for torque analysis.



(a)

The component of the distance vector perpendicular to the normal force ($$ r^{\prime }$$) is 10.8 cm, so the normal force causes a torque of (10.8 cm)$$ W_{\mathrm {b}}$$. This is a positive torque because the normal force induces a counter clockwise rotation about the chosen z-axis (see Fig. 2.2).

 

(b)

The component of the distance vector normal to the force of the weight of the leg is 3.2 cm and this force tends to induce a clockwise rotation, so it contributes a torque of $$-(3.2\,\mathrm{cm})W_{\mathrm {leg}} = -(3.2\,\mathrm{cm}) (0.16 W_{\mathrm {b}}) = -(0.5\,\mathrm{cm})W_{\mathrm {b}}$$.

 

(c)

With the choice of the axis, the torque from the reaction force from the hip is zero, because the distance vector and normal force are antiparallel.

 

(d)

The component of the force from the hip abductor muscles normal to the horizontal distance vector (of magnitude 7.0 cm) is $$M\sin 70^{\circ }$$. Since this causes a clockwise rotation, the torque is $$-(7.0\,\mathrm{cm})M\sin 70^{\circ }$$. So we see that


$$\begin{aligned} \sum \tau _{z}&=(10.8\,\mathrm {cm})W_{\mathrm {b}}-(3.2\,\mathrm {cm}) (0.16 W_{\mathrm {b}})+0-(7.0\,\mathrm {cm})M\sin 70^{\circ }=0, \end{aligned}$$

(2.27)



$$\begin{aligned} M&=\frac{10.8-0.5}{7.0 \sin 70^{\circ }}W_{\mathrm {b}}=1.57\mathrm { }W_{ \mathrm {b}}. \end{aligned}$$

(2.28)

 

We see that torque provided by the hip abductor muscles is needed to counter the torques from the normal force from the floor and the weight of the leg. This normal force torque is much more important than that due to the leg, because of the greater magnitude of the force and the larger moment arm.

Using this value for the muscle force, the force balance in the x direction gives $$R_{x}=M\cos 70^{\circ }=0.54W_{\mathrm {b}}$$. From the balance in the y direction, $$R_{y}=M\sin 70^{\circ } +0.84W_{\mathrm {b}}=2.31W_{\mathrm {b}}$$. The magnitude of $$R = (R_{x}^{2}+R_{y}^{2})^{1/2}=2.37W_{\mathrm {b}}$$, and $$\tan \phi =R_{x}/R_{y}=0.54/2.31=0.23$$, so $$\phi =13^{\circ }$$.

Because $$M\simeq 1.6W_{\mathrm {b}}$$ and $$R\simeq 2.4W_{\mathrm {b}}$$, for $$m_{ \mathrm {b}}=90$$ kg we have $$W_{\mathrm {b}}=880$$ N (200 lb), and so $$M\simeq $$ 1,400 N (320 lb) and $$R\simeq $$ 2,100 N (470 lb). The origin of hip problems is clear: The force from the hip is much greater than the body weight because of the large moment arms.

We next examine a variation of this problem. The person now uses a cane to provide support on the left side while standing on his or her right leg (Fig. 2.18). As shown in Fig. 2.19, the cane is 30.5 cm (1 ft) from the body midline. It is supported by and pushed down by the left arm or shoulder. Consequently, there is a normal force $$N_{\mathrm {c}}$$ from the floor. We assume $$N_{ \mathrm {c}}=W_{\mathrm {b}}/6$$. We will see that the use of a cane has two immediate consequences. The right foot is no longer directly in the body midline but is displaced a distance L to the right (in the reference of the body) and the normal force felt by the right foot $$N_{\mathrm {f}}$$ is no longer the body weight.

A114622_2_En_2_Fig18_HTML.gif


Fig. 2.18
Forces on entire person for someone walking with a cane. (From [3])


A114622_2_En_2_Fig19_HTML.gif


Fig. 2.19
Force diagram for a person using a cane for some support. (From [3])

Whole body equilibrium gives


$$\begin{aligned} N_{\mathrm {f}}+N_{\mathrm {c}}-W_{\mathrm {b}}=N_{\mathrm {f}}+W_{\mathrm {b}}/6-W_{\mathrm {b}}=0 \end{aligned}$$

(2.29)
or $$N_{\mathrm {f}}=5W_{\mathrm {b}}/6.$$ Using the same axis as before, the torque balance is


$$\begin{aligned} \sum \tau _{z}=(30.5\,\mathrm {cm})N_{\mathrm {c}}-L(N_{\mathrm {f}})=(30.5\,\mathrm {cm}) (W_{\mathrm {b}}/6)-L(5W_{\mathrm {b}}/6)=0 \end{aligned}$$

(2.30)
or $$L=(N_{c}/N_{\mathrm {f}})30.5\,\mathrm{cm} =(1{/}5)30.5\,\mathrm{cm} = 6.1\,\mathrm{cm}$$.

For the same leg as in Fig. 2.17, the corresponding distances are different because the angle of the leg is now different (as obtained from anatomical dimensions and the new geometry, see Problem 2.6) (Fig. 2.20). We will examine the whole leg equilibrium again. The cane is not involved explicitly, but implicitly through the changes in the leg position and the load borne by the leg. Now


$$\begin{aligned} \sum F_{x}=M\cos 70^{\circ }-R_{x}=0, \end{aligned}$$

(2.31)

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Jun 11, 2017 | Posted by in GENERAL & FAMILY MEDICINE | Comments Off on Statics of the Body

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