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Forces examples
1) The coefficient of static friction between a tennis player's hand and her
racket is 0.45. How hard must she squeeze the racket if she wants to exert a
force of 200N along its longitudinal axis?
F = μN
F = 200N
μ = 0.45
200N = 0.45(Normal Force)
Normal force = 200N/0.45 = 444N
2) The coefficient of static friction between the sole of an athletic shoe and
the basketball court floor is 0.67. Tyler wears these shoes when he plays
basketball. If Tyler exerts a normal contact force of 1400 N when he pushes
off the floor to run down the court, how large is the friction force exerted by
Tyler’s shoes on the floor?
Frictional force = F
Coefficient of static friction = 0.67
Normal force = 1400N
F = μN
F = 0.67 (1400N) = 938N
3) A sprinter is just coming out of the starting block, and only one foot is
touching the block. The sprinter pushes back (horizontally) against the
block with a force of 500 N and down (vertically) against the ground with a
force of 800 N. What is the resultant of these forces?
Pythagorean Theorum
a2 + b2 = c2
c = sqrt(a2 + b2)
Resultant = sqrt((500N)2 + (800N)2) = 943N
tan θ = opp/adj
θ = tan-1 (opp/adj)
θ = tan-1 (-800N/-500N) = 238°
A 60-kg skier is in a tuck and moving straight down a 30o slope. Air
resistance pushes backward on the skier with a force of 10 N (this force
acts in a direction upward and parallel to the 30o slope). The coefficient of
dynamic friction between the skis and the snow is 0.08. What is the
resultant force that acts on the skier?
W = mg
F = μN
W = 60kg(-9.81m/s2) = 588.6 N
μ = 0.08
Normal Force
Normal force = 588.6N(cos 30°) = 509.7N
F = 0.08(509.7N) = 40.8N
Force along slope due to gravity = 588.6N(sin 30°) = 294.3N
Force along slope = 294.3N -40.8N - 10N = 243.5N
5) The ground reaction force acting on a long jumper is 4500 N acting
forward and upward at an angle of 78° from horizontal.
a. What is the vertical component of this ground reaction force?
b. What is the horizontal component of this ground reaction force?
Part A
Sin (θ) = opposite/hypotenuse
Sin (78°) = (vertical force)/4500N
Vertical force = 4402N
Part B
Cos (θ) = adjacent/hypotenuse
Cos (78°) = (horizontal force)/4500N
Horizontal force = 936N
6- The Achilles tendon inserts on the calcaneus at a distance of 8cm from
the axis of the ankle joint. If the force generated by the muscles attached to
the Achilles tendon is 3000 N and the moment arm of this force about the
ankle joint axis is 5 cm, what torque is created by these muscles about the
ankle joint?
The 5 cm is the moment arm (effort arm).
T = F X d = 3000N(0.05m) = 150 Nm
7- An athlete is doing a knee extension exercise using a 100N dumbbell
strapped to his ankle 40 cm from his knee joint. He holds his leg so that the
horizontal distance from his knee joint to the dumbbell is 30 cm.
a. For this position, what torque is created by the dumbbell about the axis
through his knee joint.
Since gravity works vertically, the 30 cm distance should be used since it
will be perpendicular to the force of gravity working through the weight.
T = F X d =100N(0.30m) = 30 Nm
b. If the moment arm of the knee extensor muscles is 4 cm about the knee
joint axis, what amount of force must these muscles produce to hold the leg
in the position described? Ignore the weight of the leg.
The torque due to the weight must be equal in magnitude, but opposite in
30Nm = Fmuscle X (0.04m)
Fmuscle = 30Nm / 0.04m= 750 N
8- What can a hurdler do to minimize the amount of vertical effort in
clearing a hurdle?
By positioning the body in a way to keep the center of mass clearance as
low as possible, the hurdler will not need to jump as high. This can be done
by bringing the lead leg up parallel to the ground, the trunk and arms
relatively low, and getting the trail leg to have a minimal clearance of the
9- A hurdler’s knee is 0.97 m above ground level. The ankle is 0.91 m
above ground level. What is the height of the center of mass of the shank
(lower leg)? The length percent for the shank 47.5%.
10- When a discus thrower begins his throwing motion, he spins with an
angular velocity of 5 rad/s. Just before he releases the discus, Anthony’s
angular velocity is 25 rad/s. If the time from the beginning of the throw to just
before release is 1 s, what is Anthony’s average angular acceleration?
11- A hammer thrower spins with an angular velocity of 540o/s. The distance
from his axis of rotation to the center of gravity of the 7.26 kg hammer is
1.9 m.
a) What is the linear velocity of the hammer head?
b) What is the centripetal acceleration of the hammer head?
c) What is the centripetal force created by the hammer head?
12- What advantages does a longer-limbed individual have in throwing and
striking activities?
A longer limb means the radius will be larger. Considering that linear
velocity equals the product of angular velocity and radius, an athlete should
be able to obtain higher linear velocities. However, this will only be true if
the athlete can generate similar angular velocities. As you consider Major
League Baseball, it seems that the most effective pitchers are relatively tall.
13- A volleyball player jumps into the air to spike the ball. He leaves the
ground without any angular momentum. As he swings his arm forward to
spike the ball, what happens to his legs if his trunk and non-hitting arm do
not rotate?
His legs, trunk, and non-hitting arm will rotate in the opposite direction.
So, if we think of the hitting arm rotating clockwise, the rest will rotate
counter-clockwise since angular momentum will be conserved at least until
the ball is contacted.
14- The mass of a discus is 2 kg, and its radius of gyration is 12 cm. The
average net torque exerted on a discus about its axis of spin is 100 Nm
during a throw. If the discus is not spinning at the start of the throwing
action, and the throwing action lasts for 0.20 s, how fast is the discus
spinning when it is released?
T= Ixα , but I= mk2 and α=Δɷ/Δt
100=2x(0.12)2 x (ɷ/0.2)
When a force is applied to a body, the resulting motion of the body is
dependent not only on the magnitude of the applied force but also on the
duration of force application.
The product of force (F) and time of its application (t) is
known as impulse (J): J = Fxt
A large change in an object’s state of motion may result from a small force
acting for a relatively long time or from a large force acting for a relatively
short time. A golf ball rolling across a green gradually loses speed because of
the small force of rolling friction. The speed of a baseball struck vigorously
by a bat changes because of the large force exerted by the bat during the
fraction of a second it is in contact with the ball. When a vertical jump is
executed, the larger the impulse generated against the floor, the greater the
jumper’s takeoff velocity and the higher the resulting jump.
Muscle forces, gravitational force, and bone-breaking force such as that
encountered in a skiing accident all affect the human body differently.
The effect of a given force depends on its direction and duration as well as
its magnitude, as described before.
Another factor that affects the outcome of interactions between two bodies is
momentum, a mechanical quantity that is particularly important in situations
involving collisions. Momentum may be defined generally as the quantity of
motion that an object possesses. More specifically, linear momentum is the
product of an object’s mass and its velocity:
M = mv
A static object (with zero velocity) has no momentum; that is, its momentum
equals zero. A change in a body’s momentum may be caused by either a
change in the body’s mass or a change in its velocity. In most human
movement situations, changes in momentum result from changes in velocity.
Units of momentum are units of mass multiplied by units of velocity,
expressed in terms of kg.m/s. Because velocity is a vector quantity,
momentum is also a vector quantity and is subject to the rules of vector
composition and resolution.
When a head-on collision between two objects occurs, there is a tendency for
both objects to continue moving in the direction of motion originally
possessed by the object with the greatest momentum.
For example, if a 90 kg hockey player traveling at 6m/s to the right collides
head-on with an 80kg player traveling at 7m/s to the left, the momentum of
the first player is the following: M = mv = (90 kg) (6 m/s)= 540 kg.m/s
The momentum of the second player is expressed as follows:
M = mv= (80 kg) (7 m/s)= 560 kg.m/s
Since the second player’s momentum is greater, both players would tend to
continue moving in the direction of the second player’s original velocity
after the collision. Actual collisions are also affected by the extent to which
the players become entangled, by whether one or both players remain on
their feet, and by the elasticity of the collision.
Neglecting these other factors that may influence the outcome of the
collision, it is possible to calculate the magnitude of the combined velocity
of the two hockey players after the collision using a modified statement
of Newton’s first law of motion which may be restated as the principle of
conservation of momentum:
In the absence of external forces, the total momentum of a given
system remains constant.
The principle is expressed in equation format as the following:
M1 = M2 or (mv)1 = (mv)2
Subscript 1 designates an initial point in time and subscript 2 represents a later time.
Applying this principle to the hypothetical example of the colliding hockey players,
the vector sum of the two players’ momenta before the collision is equal to their
single, combined momentum following the collision as in example 16 . In reality,
friction and air resistance are external forces that typically act to reduce the total
amount of momentum present.
The type of collision that occurs between a struck baseball and a bat is
known as an impact. An impact involves the collision of two bodies over an
extremely small-time interval during which the two bodies exert relatively
large forces on each other. The behavior of two objects following an impact
depends not only on their collective momentum but also on the nature of the
For the hypothetical case of a perfectly elastic impact, the relative velocities
of the two bodies after impact are the same as their relative velocities before
impact. The impact of a superball with a hard surface approaches perfect
elasticity, because the ball’s speed diminishes little during its collision with
the surface. At the other end of the range is the perfectly plastic impact,
during which at least one of the bodies in contact deforms and does not
regain its original shape, and the bodies do not separate. This occurs when
modeling clay is dropped on a surface.
Most impacts are neither perfectly elastic nor perfectly plastic, but
somewhere between the two. The coefficient of restitution describes the
relative elasticity of an impact. It is a unitless number between 0 and 1.
The closer the coefficient of restitution is to 1, the more elastic is the impact;
and the closer the coefficient is to 0, the more plastic is the impact.
The coefficient of restitution governs the relationship between the relative
velocities of two bodies before and after an impact. This relationship, which
was originally formulated by Newton, may be stated as follows:
When two bodies undergo a direct collision, the difference in their
velocities immediately after impact is proportional to the difference in
their velocities immediately before impact.
This relationship can also be expressed algebraically as the following:
The differences in two balls’
velocities before impact
is proportional to the
difference in their velocities
after impact. The factor
of proportionality is the
coefficient of restitution.
In this formula, e is the coefficient of restitution, u1 and u2 are the velocities of
the bodies just before impact, and v1 and v2 are the velocities of the bodies
immediately after impact.
In tennis, the nature of the game depends on the type of impacts between ball and
racket and between ball and court. All other conditions being equal, a tighter grip
on the racket increases the apparent coefficient of restitution between ball and
When a pressurized tennis ball is punctured, there is a reduction in the coefficient
of restitution between ball and surface of 20%. Other factors of influence are
racket size, shape, balance, flexibility, string type and tension, and swing
The nature of impact between the bat and the ball is also an important factor in
the sports of baseball and softball. The hitting surface of the bat is convex, in
contrast to the surface of the tennis racquet, which deforms to a concave shape
during ball contact. Consequently, hitting a baseball or softball in a direct, rather
than a glancing, fashion is of paramount concern.
Research has shown that aluminum baseball bats produce significantly higher
batted ball speeds than do wood bats, which suggests that the coefficient of
restitution between an aluminum bat and baseball is higher than that between a
wood bat and baseball.
In the case of an impact between a moving body and a stationary one,
Newton’s law of impact can be simplified because the velocity of the
stationary body remains zero. The coefficient of restitution between a ball
and a flat, stationary surface onto which the ball is dropped may be
approximated using the following formula:
In this equation, e is the coefficient of restitution, hd is the height from
which the ball is dropped, and hb is the height to which the ball bounces
(Example 17).
The coefficient of restitution describes the interaction between two bodies
during an impact; it is not descriptive of any single object or surface. Dropping a
basketball, a golf ball, a racquetball, and a baseball onto several different surfaces
demonstrates that some balls bounce higher on certain types of surfaces.
The coefficient of restitution is increased by increases in both impact velocity and
temperature. In sports such as baseball and tennis, increases in both incoming ball
velocity and bat or racket velocity increase the coefficient of restitution between bat
or racket and ball and contribute to a livelier ball rebound from the striking
instrument. In racquetball and squash, where the ball is constantly being deformed
against the wall, the ball’s thermal energy (temperature) is increased over the course
of play. As ball temperature increases, its rebound from both racquet and wall
becomes livelier.
Bounce heights of a basketball,
golf ball, racquetball, and baseball
all dropped onto the same surface
from a height of 1 m.
Human Body Moment of Inertia
Moment of inertia can only be defined with respect to a specific axis of
rotation. The axis of rotation for a body segment in sagittal and frontal
plane motions is typically an axis passing through the center of a body
segment’s proximal joint. When a segment rotates around its own
longitudinal axis, its moment of inertia is quite different from its moment
of inertia during flexion and extension or abduction and adduction, because
its mass distribution, and therefore its moment of inertia, is markedly
different with respect to this axis of rotation. The figures below illustrates
the difference in the lengths of the radii of gyration for the forearm with
respect to the transverse and longitudinal axes of rotation.
The moment of inertia of the human body as a whole is also different with
respect to different axes. When the entire human body rotates free of
support, it moves around one of three principal axes: the transverse (or
frontal), the anteroposterior (or sagittal), or the longitudinal (or vertical)
axis, each of which passes through the total body CG. Moment of inertia
with respect to one of these axes is known as a principal moment of
The radius of gyration (k) of the forearm for
flexion/extension movements is much larger
than for pronation/supination.
Principal moments of inertia
of the human body in different
positions with respect to
different principal axes:
(1) principal axis; (2) moment
of inertia (kg.m2).
Angular momentum
Since moment of inertia is the inertial property for rotational movement, it
is an important component of other angular kinetic quantities. As discussed
before, the quantity of motion that an object possesses is referred to as its
momentum. Linear momentum is the product of the inertial property (mass)
and linear velocity. The quantity of angular motion that a body possesses is
likewise known as angular momentum.
Angular momentum, represented as H, is the product of the angular inertial
property (moment of inertia) and angular velocity:
For linear motion: M = mv
For angular motion: H = Iɷ or H= mk2 ɷ
Three factors affect the magnitude of a body’s angular momentum:
(a) Its mass (m), (b) the distribution of that mass with respect to the axis of
rotation (k), and (c) the angular velocity of the body (ɷ). If a body has no
angular velocity, it has no angular momentum. As mass or angular velocity
increases, angular momentum increases proportionally. The factor that most
dramatically influences angular momentum is the distribution of mass with
respect to the axis of rotation, because angular momentum is proportional to
the square of the radius of gyration.
Units of angular momentum result from multiplying units of mass, units
of length squared, and units of angular velocity, which yields kg.m2/s.
For a multisegmented object such as the human body, angular momentum
about a given axis of rotation is the sum of the angular momenta of the
individual body segments. During an airborne somersault, the angular
momentum of a single segment, such as the lower leg, with respect to the
principal axis of rotation passing through the total body CG consists of
two components: the local term and the remote term. The local term is
based on the segment’s angular momentum about its own segmental
CG, and the remote term represents the segment’s angular
momentum about the total body CG. Angular momentum for this
segment about a principal axis is the sum of the local term and the remote
H= Is ɷs + mr2 ɷg
In the local term, Is is the segment’s moment of inertia and ɷs is the
segment’s angular velocity, both with respect to a transverse axis through
the segment’s own CG. In the remote term, m is the segment’s mass, r is
the distance between the total body and segmental CGs, and ɷg is the
angular velocity of the segmental CG about the principal transverse axis
as shown below.
The sum of the angular momenta of all the body segments about a principal
axis yields the total-body angular momentum about that axis.
19- At the instant of takeoff, a 60-kg diver’s angular momentum about his
transverse axis is 20kg.m2/s. His radius of gyration about the transverse axis
is 1.0 m at this instant. During the dive, the diver tucks and reduces his
radius of gyration about the transverse axis to 0.5 m.
a) At takeoff, what is the diver’s angular velocity about the transverse axis?
b) After the diver tucks, what is his angular velocity about the transverse
Notice how decreasing the radius of gyration by half, the angular velocity increased by four
Conservation of angular momentum
Whenever gravity is the only acting external force, angular momentum is
conserved. For angular motion, the principle of conservation of momentum
may be stated as follows:
The total angular momentum of a given system remains constant in the
absence of external torques.
Gravitational force acting at a body’s CG produces no torque because the
perpendicular distance d equals zero and so it creates no change in angular
The principle of conservation of angular momentum is particularly useful in
the mechanical analysis of diving, trampolining, and gymnastics events in
which the human body undergoes controlled rotations while airborne. In a
one-and-a-half front somersault dive, the diver leaves the springboard with a
fixed amount of angular momentum. According to the principle of
conservation of angular momentum, the amount of angular momentum
present at the instant of takeoff remains constant throughout the dive. As the
diver goes from an extended layout position into a tuck, the radius of
gyration is decreased, thus reducing the body’s principal moment of inertia
about the transverse axis.
Because angular momentum remains constant, a compensatory increase in
angular velocity must accompany the decrease in moment of inertia (Figure
below). The tighter the diver’s tuck, the greater the angular velocity. Once
the somersault is completed, the diver extends to a full layout position,
thereby increasing total-body moment of inertia with respect to the axis of
Change in angular momentum
When an external torque does act, it changes the amount of angular
momentum present in a system predictably. Just as with changes in linear
momentum, changes in angular momentum depend not only on the
magnitude and direction of acting external torques but also on the length of
the time interval over which each torque acts:
linear impulse = F X t
angular impulse =T X t
When an angular impulse acts on a system, the result is a change in the
total angular momentum of the system. The impulse–momentum
relationship for angular quantities may be expressed as the following:
T X t = ΔH= (Iɷ)2 - (Iɷ)1
As before, the symbols T, t, H, I, and ɷ represent torque, time, angular
momentum, moment of inertia, and angular velocity, respectively, and
subscripts 1 and 2 denote initial and second or final points in time.
Because angular impulse is the product of torque and time, significant
changes in an object’s angular momentum may result from the action of a
large torque over a small time interval or from the action of a small torque
over a large time interval.
Since torque is the product of a force’s magnitude and the perpendicular
distance to the axis of rotation, both of these factors affect angular impulse.
The effect of angular impulse on angular momentum is shown in example 21.
In the throwing events in track and field, the object is to maximize the angular
impulse exerted on an implement before release, to maximize its momentum
and the ultimate horizontal displacement following release.
As discussed before, linear velocity is directly related to angular velocity, with
the radius of rotation serving as the factor of proportionality.
As long as the moment of inertia (mk2) of a rotating body remains constant,
increased angular momentum translates directly to increased linear momentum
when the body is projected. This concept is particularly evident in the hammer
throw, in which the athlete first swings the hammer two or three times around
the body with the feet planted, and then executes the next three or four wholebody turns while facing the hammer before release. Some hammer throwers
perform the first one or two of the whole body turns with the trunk in slight
flexion (called countering with the hips), thereby enabling a farther reach with
the hands (Figure below).
This tactic increases the radius of rotation, and thus the moment of inertia of
the hammer with respect to the axis of rotation, so that if angular velocity is
not reduced, the angular momentum of the thrower/hammer system is
increased. For this strategy, the final turns are completed with the entire body
leaning away from the hammer, or countering with the shoulders.
Researchers have suggested that, although the ability to lean forward
throughout the turns should increase the angular momentum imparted to the
hammer, a natural tendency to protect against excessive spinal stresses or
shoulder strength limitations may prevent the thrower from accomplishing
this technique modification.
A hammer thrower must counter the
centrifugal force of the hammer to
avoid being pulled out of the throwing
ring. Countering with the shoulders A
results in a smaller radius of rotation
for the hammer than countering with
the hips B.
When muscles develop tension, pulling on bones to support or move the
resistance created by the weight of the body segment(s) and possibly the
weight of an added load, the muscle and bone are functioning mechanically as
a lever. A lever is a rigid bar that rotates about an axis, or fulcrum.
Force applied to the lever moves a resistance. In the human body, the bone
acts as the rigid bar; the joint is the axis, or fulcrum; and the muscles apply
force. The three relative arrangements of the applied force, resistance, and
axis of rotation for a lever are shown in Figure below
A. First-class levers.
B. Second-class levers.
C. Third-class levers.
that the paddle and shovel function as third-class levers only when the top hand does
not apply force but serves as a fixed axis of rotation.
Most levers within the human body are third class.
A. The biceps at the elbow.
B. The patellar tendon at the knee.
C. The medial deltoid at the shoulder.
The elbow flexors contract eccentrically to apply a braking resistance and to control
movement speed during the down phase of a curl exercise. In this case, the muscle–
bone lever system is second class.
A lever system can serve one of two purposes. Whenever the moment arm of the
applied force is greater than the moment arm of the resistance, the magnitude of the
applied force needed to move a given resistance is less than the magnitude of the
resistance. Whenever the resistance arm is longer than the force arm, the resistance
may be moved through a relatively large distance. The mechanical effectiveness of a
lever for moving a resistance may be expressed quantitatively as its mechanical
advantage, which is the ratio of the moment arm of the force to the moment arm of
the resistance:
Mechanical advantage
ratio of force arm to resistance arm for a given lever.
mechanical advantage= moment arm of force/moment arm of resistance
Anatomical lever
Muscle force can be resolved into rotary and dislocating components.
Solved Examples