4.1 Summary of Newton's three laws of motion:

• Newton's First Law is the Law of Inertia
  Inertia, resistance to a change in motion, is a property of mass.
  The law of inertia states that the velocity of a body is constant without a force acting on the body.
  
• Newton's Second Law is the Law of Acceleration: F = m a  
  F is the force exterting externally on the body of mass m;
  a  is the acceleration of the body resulting from the external force.
  The law of inertia is a special case of the law of motion because if F = 0,
  then a = 0 which means the velocity of a body is unchanged.
  
• Newton's Third Law is the Law of Interaction FAB = - FBA

4.2  Force

In the MKS (m-kg-s) system, force in unit of Newton (N) is defined as by

     1 (N) = 1 (kg) x 1 (m/s**2)

A force of 1 N pushing on a mass of 1 kg results in an acceleration of 1 m/s**2

Dimensionally, the unit for force in [N] is [kg] x [m] / [s]**2

The force unit (N) is a derived unit from a combination of the fundamental units of length (m), mass (kg), and time (s).

Force is a Vector quantity specified by Magnitude and Direction.

            Figure 1a                                               Figure 1b

4.3 The Second Law: F = m a

The second law enable us to "calculate the acceleration from the force acting on a body of mass m." From the acceleration we can calculate the velocity. From the velocity we can calculate the position of the body as a function of time.

Conversely, if the acceleration of a body can be calculated, one can calculate the force exerted on the body. Study example 4.3.

Acceleration was defined in Chapter 3 as the time rate of change of the velocity,   
                  a = dV/dt

Momentum P of a body with mass m and velocity V is defined by P = mV.

The second law can be rewritten as    F = d P/dt.

4.4 The Third Law: two-body interaction -- FAB = ­ FBA

Body A exerts a force on body B, called the action force FAB.
Body B reacts to the action force by exerting a force acting on body A, called the reaction force FBA.

Note: the action force FAB is on body B and reaction force FBA is on body A.

4.5 The Effect of Force: Newton's Laws

Example 1: P31
A golf club strikes a 0.046 kg ball. The ball attains a speed of 70 m/s during the 0.5 ms collision.    Find the average force exerted by the club on the ball.

Given: m = 0.046 kg, DV = 70 m/s, DT = 0.5 x 10**-3 s

a = 70 (m/s) / [0.5x10**-3 (s)] = 140 x10**3 m/s**2 so that

f = ma = 0.046 kg x 140 x10**3 (m/s**2) = 6.4 kN

The Free-Body Diagram: graphically showing all the forces acting on a body

****** Study Example 4.4: Fig. 4.14 ********

where g = 9.8 m/s**2 (-r) is the gravitational acceleration pointing downward toward the center of the earth as indicated by the radial vector r pointing away from the center of the earth. The minus sign revereses the direction of the vector.

Examine carefully Figure 4.19 showing the forces acting on a jumping person. Fw is the gravitational force (body weight) downward. The floor exerts an upward force -FF on the feet to support the body. When the person jumps, her feet exert a downward force Fj on he floor; the floor exert a reaction force upward on the feet -Fj.  The net force acting on the body is F = Fw + (-FF) + (-Fj) = -Fj  because Fw + (-FF) = 0

Free-body diagram for a block of mass m on an inclined plane as shown below: 

The pulley in Fig. 4.21a is weightless and frictionless. M1 weighs 300N and M2 weighs 100N. A person hold on to M1 so that the system is motionless. What is the tention in the rope and the acceleration of M1? How much force must the person exert and in what direction?

If F is the force required to hold M1 motionless, i.e.,

M1 g - FT - F = 0       (1)

FT = M2 g = 100 N     (2)

solve (1) and (2) for F = 200N upward