5.1 Introduction

An object that either moves with constant velocity or remains at rest is said to be in a state of static equilibrium. A building is in static equilibrium because its weight is balanced vertically by the upward ground force exerted on it. A ballerina keeps a delicate balance by positioning her center of mass on a vertical line that passes through the tip of her feet in contact with the floor. What are the conditions of static equilibrium? How do we use the equations of static equilibrium to determine some of the unknown forces acting on an object? These questions are addressed in this chapter.

5.2 Equations of Static Equilibrium

We have previously shown that the motion of a body B is governed by the following equations:

in which 2F is the resultant external force acting on B, m is its mass, and ac is the acceleration of its center of mass. The symbol 2M° denotes the resultant moment acting on B with respect to point O, and Ho is the moment of momentum of B with respect to the same point. Equation 5.1 is called the equation of motion of the center of mass. According to this equation, the resultant force acting on an object must be equal to the mass of the object times the acceleration of its center of mass. Equation 5.2 is the mathematical expression of the principle of conservation of moment of momentum with respect to a fixed point O:

in which r and v denote, respectively, the position and the velocity of the mass element dm, and the integration is over the mass of body B.

The acceleration of the center of mass a body is equal to zero when the body is at rest or in constant motion. Hence, the resultant force acting on the body must be equal to zero:

This equation is called the condition of force balance. The balance of external forces acting on an object may in certain cases be sufficient to ensure static equilibrium. For example, the weight of an elevator hanging from a cable is supported by the pull of the cable. Thus, the tension in the cable must be equal to the weight of the elevator. Another example is the force balance in a game of tug-of-war. If two boys can pull on a rope with the same intensity, they and the rope remain in static equilibrium.

An object, however, is not necessarily in static equilibrium even if the resultant force acting on the object is equal to zero. That is because if there is a resultant moment acting on the object, it will, in accordance with Eqn. 5.2, alter the moment of momentum of the object, resulting in rotational motion. Moment of momentum of an object with respect to a fixed point O must be equal to zero as long as the object remains at rest. To assure static equilibrium we must also satisfy the following condition:

The term on the left-hand side of Eqn. 5.5 is the resultant moment with respect to any point fixed on earth. Thus, if the moment of forces acting on an object can be shown to be zero with respect to one point, then it is zero for any other point fixed in an inertial reference frame.

Equations 5.4 and 5.5 comprise the equations of static equilibrium. They are valid for living systems as well as nonliving objects. When the forces and the moments acting on an object are three-dimensional, the vector equations, Eqns. 5.4 and 5.5, correspond to six independent scalar equations. If the forces acting on an object all lie in a plane, then Eqn. 5.4 reduces to two independent scalar equations. Because the moment of forces in a plane is perpendicular to that plane, Eqn 5.5 yields a single scalar equation in this case. Thus, there are three independent equations in planar statics.

According to the equations of static equilibrium, if only two forces act on an object, not only that the forces must be equal in magnitude and opposite in direction but also they must have the same line of action (Fig. 5.1). The condition of force balance is satisfied when the two forces are equal in magnitude but in opposite direction. The condition of balance of moment of momentum requires that they must have the same line of action. Otherwise, the moment of one of these forces with respect to the

Figure 5.1. An object subjected to two point forces. For this object to be in equilibrium, forces involved must be of same magnitude, opposite sense of direction, and must also share the same line of action.

point of application of the other force will not be equal to zero. This result has implications on the forces carried by the long bones of the human body. The weight of a bone acts at its center of mass but may be negligible in comparison with the forces acting on the joints. This is illustrated in Fig. 5.2, in which a rod is shown to be positioned in space using strings tied to its ends. When the tensions in the two strings are large enough, the rod aligns along the tension line as if it were weightless and all the forces acting on it were at its endpoints. Note, however, that in the case of a long bone, muscle forces do not act right at the ends of the bone but have small lever arms. Thus, the assumption that the bone transmits force along the direction of its long axis may not be reasonable under a variety of loading conditions.

Another simple result directly derived from the equations of static equilibrium concerns objects under the application of three forces (Fig. 5.3). If the lines of action of two of these forces intersect each other at some point in space, then the line of action of the third force must also pass through that point. This follows from the condition of balance of external moments. The only way the resulting moment on the object will be equal to zero is if the line of action of the third force also passes through this point.

Investigation of static equilibrium requires computations of moments created by external forces. As we have already demonstrated, the moment of a force with respect to point A is defined as

in which rP/A is the vector connecting the point of application of force F to the point A (Fig. 5.4). We could determine moment MA by going through the formal procedures of vector multiplication. However, sometimes it is easier to adopt an approach that employs only scalar algebra. From the definition of the vector product we know that the direction of

Figure 5.2. A rod connected by strings to two hooks that are fixed on earth. The direction of the strings specifies the directions of the forces acting at the ends of the rod. Note that when the tensions in the strings are comparable to the weight of the rod, then the rod and the string forces are not aligned. However, as the string connecting the rod to the hook D is pulled with increasing force, the rod acts as if it is under the influence of just two forces. The weight of the rod can then be neglected in the condition of force bal-

the moment must be perpendicular to the plane created by rP/A and F. Also, the magnitude of MA (MA) is given by the following equation:

in which 6 is the angle between rP/A and F. The sense of direction of the moment is such that counterclockwise rotation is considered positive whereas clockwise rotation is considered negative (Fig. 5.4).

Figure 5.3. An object under the influence of three force vectors. If two of these vectors intersect at a point in space, then the third force vector must also pass through the same point to ensure static equilibrium.

Figure 5.4. The moment produced by force F about point A. The magnitude of the moment is equal to the products of magnitude of the force, the length between A to the point of application of the force, and the sine of the angle 6 the force makes with the position vector connecting A to the point of application of the force.

5.3 Contact Forces in Static Equilibrium

Humans and animals spend a good proportion of their times at static equilibrium, sitting, sleeping, resting. Even when they are in motion, sometimes the inertial terms (mass times acceleration) are small compared to the magnitude of some of the forces acting on them. In such situations, the equilibrium analysis serves as a valid first approximation of the contact forces involved in the movement.

Example 5.1. Stretching of the Achilles Tendon. Consider an athlete who is pushing against a wall to stretch his Achilles tendons during a warm-up before an athletic event (Fig. 5.5). The athlete has a mass m = 80 kg and a height L of 183 cm. The distance DB between the top of his head and his

D |
I |

B |
^ HW |

C. | |

l-80kg | |

e2 | |

O^ , mL; | |

Hc-^G | |

VG |

shoulders is 30 cm. The distance AC between the bottom of his foot and his center of gravity is 101 cm, and the distance AG between the heel and the toes is 25 cm. Assume further that the friction between the hands of the athlete and the wall is negligible. Determine the forces exerted on the athlete by the wall and the ground as a function of the angle of inclination 6.

Solution: This is a two-dimensional problem in which all the forces acting on the individual lie in a plane. The free-body diagram shown in Fig. 5.5 indicates that the forces acting on the individual are the force of gravity (the weight of the individual), the horizontal force exerted by the fric-tionless wall, and the vertical and horizontal forces exerted by the ground. Notice that because the person is stretching the Achilles tendon, the heels barely touch the ground, if they touch at all. Thus the point of application of the ground force is positioned somewhere on the toes.

The condition of static force balance (Eqn. 5.4) yields two scalar equations: the resultant force in both the horizontal (e1) and vertical (e2) directions must be equal to zero:

Vg - mg = 0 => Vg = 80 kg-9.81 m/s2 = 784.8 N (5.7b)

Thus the condition of force balance enables us to compute the vertical force exerted by the ground on the athlete. To determine the horizontal contact forces acting on the athlete, we need to set the resultant external moment equal to zero. For simplicity in the computations, we choose to evaluate the moments about point A. Thus, the moment created by the horizontal ground force HG will be equal to zero. The condition of balance of external moments about A then leads to the following expression:

-m g ■ AC • cos 6 + VG • AG + HW • AB • sin 6 = 0 (5.8)

in which 6 denotes the angle between the long axis of the athlete's body and the horizontal e1 direction. Note that all moments acting on the athlete are in either positive or negative e3 direction.

Equation 5.8 can now be used to compute the unknown force HG = HW as a function of 6. Using a hand calculator, one can show that HG = HW = 641 N, 336 N, and 151 N, respectively, when 6 = 30°, 45°, and 60°. Notice that the horizontal force exerted by the ground and by the wall on the athlete increases sharply with decreasing 6. If at some low value of 6 the ground and the athlete's shoes cannot generate enough friction as required by equilibrium, then the athlete will slip, rotate clockwise, and fall, reducing the vertical distance from the center of gravity to the ground.

The magnitude of the resultant force exerted by the ground on the individual can be found by using the equation that relates the magnitude of a vector to its projections along the coordinate axes:

Fg = 1013 N, 854 N, and 799 N, respectively, for 6 = 30°, 45°, and 60°.

The direction at which the ground force acts can be found by using the following equation:

We find tan $ = 0.816, 0.428, and 0.194, respectively, for d = 30°, 45°, and 60°. To determine the force exerted on each foot (or arm), we simply have to divide the contact forces thus computed by the factor 2.

Example 5.2. Toppling a Chair. A child weighing 40 kg is seated on a chair (Fig. 5.6). The legs of the chair are 0.4 m apart, and the back of the chair is 1.2 m high. Assuming that the frictional forces at the front legs are large enough to prevent slipping, what is the maximum horizontal force one could exert on the top of the back of the chair without lifting the back legs?

Solution: External forces acting on the child and the chair are either in the e1 or e2 direction. We assume that all these forces lie in the (e1, e2) plane that passes through the center of mass of the child. We then compute the moment of external forces with respect to point B, which is marked in the figure:

Note that at the beginning of impending rotation of the chair, the legs in the back will lose contact with the floor and, therefore, the ground forces at the back legs of the chair must be equal to zero at that instant. Thus, they contribute nothing to the external moment.

This example shows that a weight of 6.7 kg acting at an elevation of 1.2 m can lift 40 kg of weight. This is the essence of the principle of the lever. Archimedes articulated this principle in Greece in 250 b.c.: "Give me a point of support and a lever long enough and I will lift the earth with my right hand."

Example 5.3. A Woman on Crutches. A woman with a knee injury is using crutches for walking and standing. Determine the contact forces tan $ = (Hg/Vg)

40 • 9.81 • 0.20 + F • 1.2 = 0 => F = 65.4 N

Figure 5.6. A child sitting on a chair. The 0 figure depicts the free-body diagram of the child and the chair.

acting on the crutches and on her foot at the standing configuration shown in Fig. 5.7.

Solution: The forces acting on the woman standing with the help of crutches are shown in the figure. There are four contact forces of unknown magnitude acting on the woman and crutches. The equations of statics relating these forces are

Using the force balance in the vertical direction, we find the ground force acting on the foot as

Thus, the crutches used in the form shown in the figure reduce the vertical load on the feet by about 27%. If the woman had kept the crutches closer to her body, the force carried by the crutches would increase and the vertical ground force acting on the foot would decrease.

How do we determine the horizontal forces H1 and H2? The condition of force balance in the horizontal direction states that they should be opposite in direction but equal in magnitude. We need additional information to determine the horizontal contact forces uniquely. If one assumes that the contact forces at B create no moment with respect to the shoulder joint of the woman, we obtain the following relationship between H2 and V2:

Admittedly, this assumption is ad hoc. However, one could defend it by arguing that a finite moment at the shoulder for long durations would result in the excessive use of shoulder muscles, and thus the woman would position the crutch to prevent aching of the shoulder muscles.

Example 5.4. Human Cable. A youth weighing W kg lies on the floor and two other students, each weighing Ws kg, pick him up by the hands and the feet. The arms of the supporting students in combination with the body of the hanging student form a parabola-like curve, which is in tension. Let D denote the span (the horizontal distance) between the shoulders of the supporting students and sag d be the distance from a line between their shoulders to the bottom of the hanging youth (Fig. 5.8).

The supporting students lean backward to be able to keep the youth off the ground. Determine the angle of inclination of the supporting students and the force exerted on them by the ground.

Solution: First, we determine the force exerted on the arms of the supporting students. Then we evaluate the ground force. The curve created by the arms of the supporting students and the body of the hanging student is assumed to be given by the parabolic expression:

where x and y denote horizontal and vertical distances along the e1 and e2 directions. The angle 6 that the human cable makes with the e1 axis at the shoulders of the supporting students is found by taking the derivative of y with respect to x at x = D/2:

Thus, for d = 0.5 m and D = 2 m, we have 6 = 45°, and for d = 0.25 m and D = 2.4 m, we obtain 6 = 23°.

The condition of force balance for the human cable can be used to compute the tension carried by the arms of the supporting students:

According to this equation, the tension T increases with decreasing 6 and therefore with decreasing sag. When 6 = 0, the tension T becomes infinity, indicating that it would be impossible to hold the hanging student in a fully horizontal position. Notice also that the tension T or thrust must be the same on both supporters because there is equilibrium in the horizontal direction and there are only two horizontal forces on the cable: the pull from the right supporter and the equal pull from the left supporter.

Let us next compute horizontal (HG) and vertical (VG) components of the ground force acting on each foot of the supporting students. The conditions of force balance in e1 and e2 directions yield the following equations:

-2T cos 6 + 2Hg = 0 => Hg = T cos 6 = (W/4) cot 6 (5.11a)

To compute the angle $ between the body axis of a supporting student and the e1 direction, we consider the moment of forces with respect to the point of application of the ground forces:

where H is the length between the shoulder and the sole of the foot and h is the length between the center of gravity of the supporting student and the sole of his feet.

An interesting note about cables, chains, and strings: like living creatures and unlike other solids, they drastically change shape in response

Figure 5.9. The shape of a cable changes in response to weights attached to it.

Figure 5.9. The shape of a cable changes in response to weights attached to it.

to an external force. Consider, for example, a weightless string that is stretched between two poles. With each addition of weight, the string changes shape in such a way that the portions of the string remains straight between the weights, and between the weights and the supporters (Fig. 5.9). When human-made solid structures undergo significant changes in shape to carry a varying load, they are considered unstable.

A body is said to be in stable equilibrium if a small displacement from the equilibrium position does not result in larger displacements. In other words, when a body is slightly displaced from a stable equilibrium position, the forces brought into play move the body back toward the equilibrium position. On the other hand, if these perturbation forces move the body further from the equilibrium position, the equilibrium is said to be unstable. If the perturbation from the equilibrium position brings about no new forces, the equilibrium is neutral. These three cases of equilibrium are illustrated in Fig. 5.10a. As shown in the figure, a rod standing on its end with its center of gravity vertically above the point of support is in unstable equilibrium. A small disturbance force will result in an unbalanced moment, taking the rod further away from the equilibrium configuration. The circular cylinder shown in the figure is in neutral equilibrium because disturbance from equilibrium position in the horizontal plane does not bring out new forces. If on the other hand the elliptical cylinder is tilted a little, the new contact force that arises will rotate it back to its equilibrium configuration. The elliptical cylinder is therefore in stable equilibrium.

Figure 5.10a-c. Classification of static equilibrium of objects. The top row (a) presents examples of unstable (U), neutral (N), and stable (S) equilibrium of rigid objects. The middle row (b) classifies the various configurations of a rod under static equilibrium. The bottom row (c) illustrates that for nonrigid structures static equilibrium does not neccessarily correspond to the configuration where the center of gravity is at a minimum height.

Figure 5.10a-c. Classification of static equilibrium of objects. The top row (a) presents examples of unstable (U), neutral (N), and stable (S) equilibrium of rigid objects. The middle row (b) classifies the various configurations of a rod under static equilibrium. The bottom row (c) illustrates that for nonrigid structures static equilibrium does not neccessarily correspond to the configuration where the center of gravity is at a minimum height.

A rigid body is in stable equilibrium when its center of gravity is at a minimum height (Fig. 5.10b). All three configurations of the rod shown in the figure are in equilibrium. The height d of the center of mass of the rod for the configurations from left to right are given by the equations:

The configuration on the far right corresponds to the smallest value of d and is, therefore, in stable equilibrium. For structures other than a rigid body, stable equilibrium does not necessarily correspond to the configuration that brings the center of gravity to minimum height. Consider, for example, the structure composed of two flat plates and a spring (Fig. 5.10c). The plates are hinged at one edge and are also connected by a spring at midlength. The natural length of the spring is (L/2) where L is the width of the plates. The free ends of the plates are free to slide on the flat surface shown in the figure. If the gravitational force and the reacting contact force were the only external loads acting on the structure, stable equilibrium would have corresponded to the configuration where the two plates lay flat on the horizontal plane. In that configuration, however, the spring would have been stretched drastically and therefore could snap with the smallest of the disturbance. How do we determine which configuration corresponds to static equilibrium in this case? The answer to the question is that the potential energy of the structure is minimum at stable equilibrium. The potential energy V for this structure is given by the relation:

V = mg (L/2) sin 0 + mg (L/2) sin 0 + (k/2) S2 (5.13)

The first two terms correspond to the gravitational potential energy of the two plates. The last term is the energy stored into the spring as a result of its stretch, with S denoting the extension of the length of the spring. It can be shown that

To compute the minimum potential energy, we take the derivative of V with respect to 0 and set it equal to zero:

The term on the left is a dimensionless parameter. It is a measure of the strength of the spring force relative to the force of gravity. Solution of this algebraic equation corresponds to minimum potential energy when the second derivative of V with respect to 0 (d 2 V/d20) is positive. When kL/(mg) = 9.5, 0 = 55° corresponds to the minimum potential energy and therefore to stable equilibrium. Note that for very stiff springs the angle 0 that corresponds to stable equilibrium will be slightly less than 60°. As the spring stiffness is decreased toward zero, the structure will flatten at static equilibrium, with 0 reducing toward zero. Although the structure discussed here does not look anything like the human body or any part of the body, there are resemblances. Muscle-tendon complexes of the human body store energy like the spring of the two-rod structure. When a calf muscle goes into contraction, the stable equilibrium of the leg will be much different than when the muscle is relaxed and therefore has much less stiffness. The reader might have experienced a muscle spasm and how it can distort the resting configuration of a leg.

Example 5.5. Stability of the Human Shoulder. In the human shoulder, the glenoid fossa region of the scapula supports the humerus of the upper arm much like the nose of a seal balancing a ball (Fig. 5.11). The equilibrium of a uniform rod in vertical configuration is an unstable one. Because the humerus is not uniform, it is much more difficult to keep it balanced. How does a seal balance a ball on its nose? What are the implications for the shoulder joint?

Solution: Consider a uniform rod of length L and mass m that is in unstable equilibrium (Fig. 5.11b). Let us apply a small perturbation to the bar in the form of a horizontal force Sf. Because the rod will tend to move in the direction of the unbalanced force, the rough substrate on which the rod is resting will exert a frictional force in the direction opposite to Sf. Both the perturbation force Sf and the frictional force f will produce counterclockwise moment with respect to the center of the rod. The rod will gain angular acceleration of the magnitude given by the equation:

(Sf + f) (L/2) = (mL2/12) a => a = 6 (Sf + f)/(mL)

Thus, the rod would begin to rotate in the counterclockwise direction. If, however, the surface on which the rod rests was given a horizontal acceleration a in the direction of Sf, the rotation of the rod can be prevented. First, the rough plane moving in the direction of Sf will pull the rod with a frictional force in the same direction. Therefore, an imposed acceleration on the surface could alter the direction of the frictional force. Second, if the acceleration is chosen such that a = 2Sf/m the resultant couple with respect to the mass center will be equal to zero, and the rod will translate in the direction of the force of perturbation Sf. The nose of the seal is certainly capable of imposing lateral movement on a ball it is balancing. Similarly, the scapula is a highly mobile shoulder bone and therefore the glenoid fossa can be laterally displaced through coordinated muscle action. This example illustrates how skeletal muscles can transform an unstable equilibrium into a stable equilibrium.

The role of supporting structures in joint stability can be studied further by considering the two-link system shown in Fig. 5.12a. The two links articulate at a ball-and-socket joint. An ecccentric load will cause rotation

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The use of dumbbells gives you a much more comprehensive strengthening effect because the workout engages your stabilizer muscles, in addition to the muscle you may be pin-pointing. Without all of the belts and artificial stabilizers of a machine, you also engage your core muscles, which are your body's natural stabilizers.

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