Transcript further force and motion considerations

Advanced Programming for 3D
Applications
CE00383-3
Forces in Human Motion
Lecture 4
Bob Hobbs
Staffordshire university
Modelling forces
Types of force
• Applied Load
• Tension
Tension
• Contact:
• Friction:
• Bouyancy:
• Drag:
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F
FT
Fc = kx
Ff = mN
Fb = rVg
Fd = ACdv
Applied loads
• Normally act on one point
• Typically
– Objects hanging from ropes
– Objects being pushed
– Objects being pulled
F
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Example 1
A mass of 25 kg is hung from a bar by a hanger.
Determine the applied load?
+y
F = (m*g)
= (25 * -9.81)
= -245.25 N
25 kg
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Tension
• The internal force acting within a
– rope
– cable
– wire
– component in a framework
T
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Always PULLING
Always +ve
Example 2
A mass of 25 kg is hung from a bar by a hanger.
Determine the tension in the hanger?
FT
+y
25 kg
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FT = (m*g)
= (25 * -9.81)
= 245.25 N
25 kg
Contact
• Whenever two bodies touch each other
there is a contact force
– Always normal to the surface
– A REACTION force
– Sometimes modelled using stiffness
R
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Example 3
A block of mass 75 kg is placed on a floor. What is the
reaction force?
Applied force, F = mg
= (75 * -9.81) = -735.75 N
Reaction force = -F = 735.75 N
+y
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Block “pushing”
on the floor
Floor pushing
on the block
Friction
• Another force associated with contact
– Parallel with the surface
– Opposes motion
– Ff = mN
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Example 4
What is the maximum friction force exertable when
a car, of mass 2000 kg, has been fitted with tyres
of m =0.3?
Reaction, N = -F = 20.36 kN
+y
Using
Ff = mN
= 0.3 * 20.36x103
= 6.1 kN
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F = mg
= 2000 * -9.81 = -20.36 kN
m?
• Coefficient of friction
Ff
Block starts to slide when
gravity overcomes friction
q
m = tan(q)
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Example 5
A block is placed on a sliding table. The block begins
to slide when the table reaches an angle of 25o.
Determine the coefficient of friction.
Using
m = tan(q)
= tan(25)
= 0.466
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Stiffness
• All objects act as springs
• The restoring force is proportional to
deflection
x
Force pulling the spring
Spring pulling on the object
Fs = k*x
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Stiffness coefficient
+
Applied
force, F
(N)
+
+
+
+
+
+
Fax
F = k*x
k = DF/Dx (N/m)
+
+
+
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deflection, x
(m)
Example 6
A spring is subjected to an applied load of 750 N.
The spring extends by 25 mm. Determine the
stiffness of the spring.
Unloaded spring
25 mm
750 N
Using k = DF/Dx
= 750 / 0.025
= 30 kN/m
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Stiffness
• A spring under Tension PULLS
• A spring under Compression PUSHES
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String, Cables and Ropes
• They can only PULL
• You may assume constant tension
throughout.
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Dampers (Viscous)
Opposing force generated
proportional to v
I.e.
F = cv
where c is the damping coefficient
Body moves with
velocity v
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Impact forces
F = constant; F =Dp/t
Dp = change in momentum
General Area = Dp
Triangular form
F = 2Dp/t
Sine wave
F = p/2 Dp/t
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Compression- Tension Cycle
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Ground Reaction
Force
• Newton's Law of Gravitation:
– any two objects with masses attract each other
and the magnitude of this attracting force is
proportional to the product of the masses and
inversely proportional to the square of the
distance. This also holds for the gravitation
between the earth and an object on the earth. The
gravitational force acted upon an object by the
earth is called gravity or weight of the object.
• Newton's Law of Reaction:
– there is an equal and opposite reaction to every
action. In other words, the action to the ground is
always accompanied by a reaction from it. This
reaction force from the ground is called the
ground reaction force (R). The ground reaction
force is an important external force acting upon the
human body in motion. We use this force as
propulsion to initiate and to control the movement.
• Newton's Law of Acceleration:
– the external forces acting on the body causes an
acceleration:
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Ground Reaction Force
F = m·a [1]
where F = sum of external forces (vector), m = mass of the subject
(scalar), and a = acceleration (vector) of the subject's centre of
mass (CM).
F=R+W
[2a]
Here, note that weight W always acts downward. Considering only
the vertical forces and their directions:
Fz = Rz - W
[2b]
where Fz = net vertical force acting on the body, Rz = the vertical
ground reaction force, and W = weight of the body. From [1] and
[2b]:
Rz - W = maz
[2c]
where az = magnitude of the vertical acceleration. Rearranging [2c],
we obtain
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az = ( Rz - W ) / m
[3]
Impulse and Momentum
• During step sequence
•
•
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– Average net Force = mass * average acceleration
– Acceleration = change of velocity
Force over a period of time causes a change of
momentum over that time.
– Average net force over a period of time = m(Vf – Vi)
Therefore changing momentum over a short period of
time generates a greater force than over a longer period
of time
– When landing from a height the start and finish
velocities are the same - as is the mass – whether
you land stiff-legged or ‘bounce’
Impact
• The first peak is called the
impact peak (P1) while the
second is called the propulsion
peak (P2). The impact peak is
associated with the impact of the
foot to the ground during early
foot contact phase. The propulsion
peak is associated with the
propulsion of the body forward. It
has always been the main focus of
the shoe engineers that how to
design the shoe-sole to reduce the
impact peak while maintaining the
propulsive characteristics.
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Reduce impact by
1. Cushioning
2. Alter gait
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