Freely falling objects are objects not supported by anything and not acted on by any forces except the gravitational force.  Near the surface of the earth such objects are accelerating.  This acceleration is due to the gravitational force acting between the objects and the earth.  The direction of the gravitational force exerted on by the earth on any object is towards the center of the earth.  Its magnitude decreases as one over the square of the distance from the center of the earth.

Link:  Gravity Force Lab  (PhET)

The radius of the earth is 6368 km.  If you climb a 1000 m high mountain, your distance from the center of the earth changes by (1/6368)*100% = 0.016% and the magnitude of the gravitational force acting on you changes by (1/6368)²*100% = 2.4*10^-6%.  For all objects near the surface of the earth the distance from the center is nearly constant, and the magnitude of the gravitational force vector is therefore approximately constant.  Over small distances, when the curvature of the earth's surface can be neglected, the direction of the gravitational force vector is also nearly constant.  It points straight downward towards the center of the earth. The force of gravity acting on an object is called its weight. 

Gravitational acceleration

Assume you are standing on a 20 m high platform with a ball in your outstretched hand.  At t = 0 you let go of the ball and it starts falling towards the ground below.  At t = 0 the ball has zero velocity.  At some later time, but before it hits the ground, its velocity is in the downward direction.  Its speed is increasing as it falls.  The ball is accelerating.  Why is a falling ball accelerating?  Which force is acting on it?

The force of gravity is acting on the falling ball.  On the surface of the earth, the direction of this force is always downward, towards the ground.  It pulls on all objects with mass.

As the object gains speed, other forces also act on it.  The drag force on a moving object is always directed opposite to the object's velocity relative to the air.  It always tries to slow down the object.  The magnitude of this force depends on the shape of the object, its speed, and the medium in which it is moving.  For many smooth, dense objects the magnitude of the drag force at low speeds in air is very small compared to the gravitational force and we can safely neglect it.

Assume we are dropping two smooth, spherical objects of different masses, such as a bowling ball and a marble, at the same time.  If the force of gravity acting on the two objects had the same magnitude, then the bowling ball would accelerate less and gain less speed in the same amount of time.  The marble would hit the floor first.  In an experiment, however, the two objects hit the floor at nearly the same time.  They gain the same speed in the same time.  This mean that the force of gravity acting on the bowling ball must have a greater magnitude than the force of gravity acting on the marble.  From Newton's second law we have a = F_g/m, where F_g is the gravitational force.  If a does not depend on the mass of the object, than the mass m must drop out the equation.  The force of gravity acting on an object must be proportional to the mass of the object.  We write

F_g = mg.

Then a = g, where g is called the acceleration due to gravity.  Near the surface of the earth its magnitude is g = 9.8 m/s² and its direction is downward.

Near the surface of the earth all objects experience the same acceleration due to gravity in the downward direction, regardless of their mass.  The acceleration due to frictional forces is always in the direction opposite the object's relative velocity, and differs from object to object.  However, when we are justified to neglect friction, then we can say that all dropped objects accelerate at the same rate.  Freely falling objects are therefore objects, which are moving with constant acceleration g.

Links:

• The Elephant and The Feather - Free Fall
• The Elephant and The Feather - with Air Resistance

Mass and weight

Mass and weight are different quantities.  Mass is a scalar.  It is an inherent property of an object, independent of where and how it is measured.  It tells us how hard it is to accelerate the object.  Weight is a vector.  It is the gravitational force acting on the object.  It depends on the location of the object.  On the surface of the moon the weight of an object points towards the center of the moon and its magnitude is approximately 1/6 the magnitude of its weight on the surface of the earth.  The mass of the object, i.e. its resistance to acceleration, is the same everywhere.  The magnitude of the gravitational acceleration is therefore smaller on the surface of the moon than on the surface of the earth.  If you drop an object near the surface of the moon, its velocity changes less rapidly then the velocity of a similar object dropped near the surface of the earth.

The force of gravity on an object is always proportional to the mass of the object.  But its magnitude and direction depends on where the object is located in the universe.  As you rise above the surface of the earth, the magnitude of the gravitational acceleration diminishes.

Problem:

How much does a 60 kg person weigh on earth?

Solution:

• Reasoning:
  Weight (magnitude) = F_g = mg.
• Details of the calculation:
  Weight = F_g = mg = (60 kg)*(9.8 m/s²) = 588 N.
  (588 N)*(1 lbf/4.448 N) = 132 lbf

Problem:

A pitcher throws a baseball of weight 1.4 N with velocity v = 32 i m/s by uniformly accelerating her arm for 0.09 s.  If the ball starts from rest,
(a)  through what distance does the ball accelerate before its release?
(b)  What vector force does she exert on it?

Solution:

• Reasoning:
  ∆x = v_xavgt is the distance through which the ball accelerates.   v_xavg = (v_f + v_i)/2,
  The ball is thrown from rest in the x-direction.
  The net force on the ball is F_x = ma_x,  a_x = (v_f - v_i)/t.
  F_net = F_pitcher + F_g = F_pitcher  - mg j, with the y-axis pointing up.
  F_pitcher = F_net + mg j,
• Details of the calculation:
  (a)  ∆x = v_xavgt = (16 m/s)0.09 s = 1.44 m is the distance through which the ball accelerates.
  (b)  F_net = ma is the total force on the ball.  a = (v_f - v_i)/t = (32/0.09) i m/s² = 356 i m/s²,
  therefore F_net  = i [1.4 N/(9.8 m/s²)]*356 m/s² = i 50.8 N is the total force on the ball. 
  The force of gravity is 1.4 N (-j), which must be balanced by the pitcher.
  She therefore must exert a force of (50.8 i + 1.4 j) N on the ball.

Problem:

A man weighing 800 N stands in an elevator that starts from rest and accelerates at a rate of 2 m/s² upward to a final speed of 10 m/s. 
(a)  What is the force exerted on the man by the floor of the elevator while it is accelerating?
(b)  What is the force exerted on the man by the floor of the elevator while it is moving upward with constant speed of 10 m/s?

Solution:

• Reasoning:
  This is a one-dimensional problem.  Let the upward direction be positive.
  The net force on the man is  F_net = -mg + F_floor.  The acceleration of the man is a = F_net/m.
• Details of the calculation:
  (a)  The man is accelerating upward, his acceleration is 2 m/s².
  The net force on the man is  F_net = -mg + F_floor = m*2m/s². 
  F_floor = mg + m*2m/s² = m(10 + 2) m/s².  (Take g = 10 m/s² to simplify the calculation.)
  The man's mass is m = 800 N/g = 80 kg.
  F_floor = 80*12 N = 960 N pointing up.
  (b)  While the elevator and the man are moving with constant velocity, F_net = 0, F_floor = mg = 800 N, pointing up.

Support forces

Solid objects are incompressible.  In general, we cannot push a solid object into another solid object.

Examples:

A table rests on the kitchen floor.  Gravity is acting on the table, pulling it downward.  But it does not accelerate into the ground.  No acceleration implies no net force.  The net force on the table is zero.  The ground exerts a support force on the table which is equal in magnitude and opposite in direction to the table's weight.  The support force direction is perpendicular or normal to the ground surface.

A book rests on the table.  Gravity is pulling it downward, but the table exerts a support force on the book which is equal in magnitude and opposite in direction to the book's weight.  The support force direction is perpendicular or normal to the table surface.