1. Forces in and on the Body
Chapter 2
Forces in and on the Body

2. Forces in and on the body
1.Muscular forces that cause the blood to circulate and
the lungs to take in air .
2. Molecular forces
(in bone , calcium atom ) .
3. Electric forces .
4. Gravitational forces .

3. Gravitational forces there is a force of attraction
Newton laws state that :
there is a force of attraction
between any two objects
( our weigh is due to the
attraction between the earth
and our bodies )

4. Gravitation force F = mg Where F is the force of gravity m is the mass
g is the acceleration due
to the gravity

5. Medical effects of gravitation forces
Medical effects of gravitation forces is the formation of varicose veins in the legs as the venous blood travels against the force of gravity on its way to the heart , and the second effects is on the bone .

6. If a person becomes weightless such as in orbiting satellite , he may lose bone mineral .
Long term bed rest removes much of the force of the body weight from bones .

7. Forces on the Body 1. Static force : where the body is in equilibrium
2. Dynamic force : where the body
is accelerated .

8. Static Force When objects stationary ( static )
they are in state of equilibrium .
The sum of the forces in any
direction is equal to zero .
The force is a vector quantity

9. Torque The sum of the torque's about any axis is equals to zero .
Torque = force x length
The sum of the torque's about
any axis is equals to zero .

10. For example :
In the body , many of the
muscles and bones systems acts as levers .

11. first , second, and third - class systems as shown in figure 1.
Levers are classified as
first , second, and
third - class systems
as shown in figure 1.

12. Figure 1

13. Example of a levers system in the body
A simple class of a levers in the body is the case of the biceps muscle and the radius bone acting to support a weight W in the hand , as shown in figure 2.

14. Figure 2

15. For example , we can find the force supplied by the biceps, if we sum the torque's about the pivot point at the joint (figure 2b ).

16. 1.due the weight W acting clock wise , for example: torque =30 W
There are only two torque's ;
1.due the weight W acting clock wise ,
for example: torque =30 W
2. produced by the muscle force M,
which is counterclockwise,
for example = 4M

17. when the arm is in equilibrium , we find that
where M is the muscular force ,
W is weight of the body
M = 7.5 W
If W = 20 Newton
4 M – 30 x 20 = 0
M = 30 x 20 /4
M = 15 Newton

18. gravity for the weight of the forearm and hand H , we
In case of finding the center of
gravity for the weight of the
forearm and hand H , we
consider all the weight at point
as shown in figure 2 c.

19. For example a typical value of hand is H = 15 N
By assuming torque's about the joint we obtain
4 M = 30 W + 14 H
M =7.5 W H
W = weight = 20 N
H = hand weight = 15 N
M =7.5 x x 15
M =
M = 67.5Newton

20. Frictional Force
Friction and energy loss due to friction appear every day in our life .
The maximum force of friction F is
F = µ N
Where N is a Normal force .
µ Is the coefficient between
the two surfaces. .

21. The value of µ depends upon the two materials in contact , and it is essentially independent of the surface area , as shown in
Table 1.

22. Table 1

23. When a person is walking , as the
heel of the foot touches the ground a force is transmitted from the foot to the ground (figure 3).

24. Figure 3

25. we can resolve this force into horizontal and vertical components
we can resolve this force into horizontal and vertical components . the vertical reaction force is applied by the surface and is labeled
N (normal force ) .
the horizontal reaction component
must be applied by frictional forces ,
as shown in figure 3.

26. Measurements have been made of the horizontal force component of the heel as it strikes the ground when a person is
walking , and it has been
to be = 0.15 W ,
Where W is the person ´s weight .

27. The frictional force is large enough both when the heel touches down and when the toe leaves the surface to prevent a person from slipping .

28. this how large the frictional force
must be in order to prevent
the heel from slipping .

29. The coefficient of friction in bone joints is very small (Table 1) .
If a disease of the joint exists, the friction may become large .
The synovial fluid in the joint is involved in the lubrication .

30. The saliva we add when we chew food acts as a lubricant (to reduce the friction force ).
For example , if you swallow a piece of dry toast you become painfully aware of this lack of lubricant .

31. Dynamics force According to second law of Newton , the force is equal
F = ma
momentum = mv
The change in momentum Δ(mv) over a short interval of time is
F = Δ(mv ) Δt

32. Example 1 for dynamic force
A 60 Kg person walking at 1 m/sec bumps into a wall and stops in a distance of 2.5 cm in about 0.05 sec . what is the force developed on impact ?

33. the force developed on impact is F = Δ(mv ) Δt F = 60Kg m/sec 0.05
Δ(mv) = (60 Kg ) (1m/sec) – (60 Kg ) ( 0 m/sec)
= 60 Kg m/sec
the force developed on impact is
F = Δ(mv ) Δt
F = 60Kg m/sec
0.05
F = 1200 Kg m/sec²
F = 1200 Newton

34. Example 2 A. A person walking at 1 m/sec
hits his head on a steel beam Assume his head stops in
0.5 cm in about 0.01 sec .
If the mass of his head is
4Kg , What is the force
developed ?

35. Q.2 a. Δ(mv)=(4Kg)(1m/sec)-(4Kg)(0m/sec) = 4 Kg m/sec F = Δ(mv ) Δt
F = 4 kg m/sec
0.01
F = 400 Newton

36. b. if the steel beam has 2 cm of
padding and Δt is increased to
0.04 sec , what is the force
developed ?

37. Q.2 b.
F = Δ(mv) Δt
F = ( 4 Kg m/sec)
0.o4 sec
F = 100 Newton