Chapter 2
Forces in and on the BodyForces 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 .Gravitational forces
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 )Gravitation force
F = mg Where F is the force of gravity m is the mass g is the acceleration due to the gravityMedical 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 .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 .
Forces on the Body
1. Static force : where the body is in equilibrium 2. Dynamic force : where the body is accelerated .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 quantityTorque
Torque = force x length The sum of the torque's about any axis is equals to zero .For example : In the body , many of the muscles and bones systems acts as levers .
Levers are classified as first , second, and third - class systems as shown in figure 1.
Figure 1
Example of a levers system in the bodyA 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.
Figure 2
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 ).
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
when the arm is in equilibrium , we find that 4M = 30W 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
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.
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 + 3.5 H W = weight = 20 N H = hand weight = 15 N M =7.5 W + 3.5 H M =7.5 x 30 + 3.5 x 15 M = 15 + 52.5 M = 67.5NewtonFrictional Force
Table 1When 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).
Figure 3
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.
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 .
this how large the frictional force must be in order to prevent the heel from slipping .
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 .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 .
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 ) ΔtExample 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 ?Δ(mv) = (60 Kg ) (1m/sec) – (60 Kg ) ( 0 m/sec) = 60 Kg m/secthe force developed on impact isF = Δ(mv ) ΔtF = 60Kg m/sec 0.05F = 1200 Kg m/secІ F = 1200 Newton
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 ?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 Newtonb. if the steel beam has 2 cm of padding and Δt is increased to 0.04 sec , what is the force developed ?