When a mass [ measured in kilograms ( kg ) ] is under the influence of balanced forces, it will either sit still with respect to an observer that is also ” at rest ” within the same frame of reference ( Ex. A television set across a room from a viewer. ), or it will move with a constant velocity with respect to a stationary observer until it’s acted upon by an unbalanced force ( Newton’s First Law of Motion ). A constant velocity is one in which the same distance in meters ( m ) is travelled during each second ( s ) of time transpires ( v = m/s ).
Think for a second about a car travelling at 37 m/s. The car’s engine generates a force that propels the car forward, but that force is balanced out by an equal and opposite force created by wind resistance ( Newton’s Third Law of Motion ). Forces always occur in pairs: The aforementioned television set exerts a downward force upon the frame in which it sits, and the frame in turn exerts an equal force upward upon the television. In regard to the car, the tires exert forces upon the roadway beneath them, and the roadway beneath exerts frictional forces back onto the tires ( preventing slippage ). Likewise, the car exerts a force upon the atmosphere it encounters, and the atmosphere exerts and equal and opposite force back onto the car. As we shall see, even when forces are not balanced, these force pairs are still at play ( just not equal in magnitude to one another ).
When the forces acting upon an object are not balanced, the object will experience a net force ( F ) that is not equal to zero. Under these conditions, the object will experience an acceleration ( a ) as opposed to being at rest or moving with a constant velocity ( Newton’s Second Law ). If the aforementioned television were pushed off of its resting frame, the downward force due to gravity would no longer be balanced by the frame upon which it rests, and the magnitude of the downward motion that ensued would increase over each second that transpires until it reaches the earth’s surface. In this circumstance, the television exerts a downward force upon the atmosphere, and the atmosphere in turn exerts an upward force upon the television set, BUT THE FORCES THAT ARE PAIRED TOGETHER ARE NO LONGER EQUAL IN MAGNITUDE. Since the velocity is no longer constant, there is a net change in velocity ( Δv ) over each second that transpires. The acceleration due to velocity changing over time is quantified as follows:
a = Δv/s
Recall that velocity ( v ) = m/s, so…
a = ( m/s )/s = ( m/s )( 1/s ) = m/s^2
When dealing with acceleration due to gravity near the earth’s surface, the acceleration constant is represented with a value of ” g “:
a = g = 9.86 m/s^2
So how was this value of ( g ) determined??? Let’s first observe that more generic expression of acceleration, where the acceleration caused by a force applied to an object is inversely proportional to how massive the object is:
F/m ∝ a
F = ma
The larger the mass ( m ) upon which a given force is applied, the less acceleration that will be experienced relative to the same force acting upon a less massive object.
Note: The confusion with using the letter ” m ” to represent mass as opposed to meters will become more familiar as usage of equations become second nature. For example, momentum ( p ) is expressed a p = mv, where ” m ” is equal to meters, and ” v ” is equal to velocity. Recall that velocity = m/s. See how confusing it would be if momentum was expressed as p = ( m )( m/s )??? Things will become less confusing when metric units are substituted into such equations:
mass = ( kg )
meters = ( m )
seconds = ( s )
velocity = ( m/s )
p = mv = ( kg )( m/s )
Getting back to the topic of force, Isaac Newton quantified the gravitational force of attraction ( Fg ) between two masses as follows:
Fg = Gm1m2/r^2
Recall that the generic expression for force:
F = ma
Setting these expressions equal to one another, we have the following:
Fg = Gm1m2/r^2 = ma
Since one of the two masses will always be in reference to the mass of the earth ( M ), we have…
Fg = GMm/r^2
GMm/r^2 = ma
The ” G “, ” M “, and earth’s radius ” r ” are constants that can be grouped together and combined as a constant ( g ):
Fg = ( GM/r^2 )( m )
G = 6.6743 x 10^-11 m^3*kg^-1*s^-2
M = 5.98 x 10^24 kg
r = 6.371 x 10^6 m
r^2 = 4.05 x 10^13 m^2
g = [ ( 6.6743 x 10^-11 m^3*kg^-1*s^-2 )( 5.98 x 10^24 kg ) ]/( 4.05 x 10^13 m^2 ) ]
g = 9.86 m/s^2
And now…
Fg = ( GM/r^2 ) = mg
Setting the gravitational force of attraction equal to the generic express of force, we have…
Fg = F
mg = ma
g = a
g = 9.86 m/s^2
RENAISSANCE PHYSICS 