Physics
PHS 5043 Forces & Energy
Newton’s Laws
Force:
Any agent capable of
changing the shape of an
object or changing its state of
rest or motion
Symbol: F
Units: N (newton)
PHS 5043 Forces & Energy
Newton’s Laws
Force:
Vector quantity
(magnitude and direction)
Forces can be added as any
other vector (polygon,
parallelogram and component
method)
PHS 5043 Forces & Energy
Newton’s Laws
Force:
At any given moment in time,
all objects are subjected to
the simultaneous action of
many forces
PHS 5043 Forces & Energy
Newton’s Laws
Equilibrium:
Objects are said to be in equilibrium when they
are “at rest” or when they have “uniform
rectilinear motion”
PHS 5043 Forces & Energy
Newton’s Laws
Equilibrium:
 When an object is in equilibrium all external
forces add up to zero.
 If we add all forces vectors (regardless of the
method), we should obtain a zero value resultant
PHS 5043 Forces & Energy
Newton’s Laws
Practice:
Given the following forces values, determine if the object is in
equilibrium.
Fa = 30N, 270°
Fb = 10N, 45°
Fc = 20N, 135°
_Coordinates: Fa (0, -30); Fb (10,10); Fc (-20,20)
_Resultant Force: Fr (-10,0)
No, the system is not in equilibrium for the resultant vector
(component method) does not have (0.0) coordinates
Try with the polygon method!
PHS 5043 Forces & Energy
Newton’s Laws
Equilibrant force:
 Is the force that balance the resultant force
 It brings the system (object) to a state of
equilibrium
PHS 5043 Forces & Energy
Newton’s Laws
Practice:
Given the following forces values, determine the
magnitude and direction of the equilibrant force
Fa = 3N, 90°
Fb = 4N, 0°
_Coordinates: Fa (0, 3); Fb (4, 0), therefore Fr (4, 3)
_Magnitude Fr = 5N (Pythagoras), direction = 37°
_Feq = Fr but in opposite direction
Therefore, the equilibrant force would be Feq = 5N, 217°
PHS 5043 Forces & Energy
Newton’s Laws
Inertia:
Property that causes an object to remain in its
state of rest or uniform rectilinear motion, that is,
to remain in equilibrium.
PHS 5043 Forces & Energy
Newton’s Laws
Newton’s First Law:
“If no external or unbalanced force acts on an
object, it maintains its state of rest or its constant
velocity in a straight line”
PHS 5043 Forces & Energy
Newton’s Laws
Newton’s Second Law:
“The change of momentum of an object is
proportional to the net applied force”
PHS 5043 Forces & Energy
Newton’s Laws
Newton’s Second Law:
F=ma
F = Δp / Δt
F = mΔv / Δt
F=ma
F: Net force (N)
m: Total mass (kg)
a: acceleration (m/s2)
PHS 5043 Forces & Energy
Newton’s Laws
Practice:
A 15 kg sled is pulled horizontally with a constant
force of 60N. What acceleration would this force
impart on the sled if the frictional force (sled-snow)
is 10N?
F=ma
Fa – Ff = m a
a = (Fa – Ff ) / m
a = (60N – 10N) / 15kg
a = 3.3 m/s2
PHS 5043 Forces & Energy
Newton’s Laws
Practice:
A 15 kg sled is pulled horizontally with a constant
force of 60N at 30° with respect to ground. What
acceleration would this force impart on the sled if
the frictional force (sled-snow) is 10N?
F=ma
Fa – Ff = m a
a = (Fax cos 30°– Ff ) / m
a = (60N)(cos 30°) – 10N / 15kg
a = 2.8 m/s2
PHS 5043 Forces & Energy
Newton’s Laws
Impulse:
 It can be consider the cause
of the motion of an object
 Its effect is the change of
Momentum of a moving object
 It is defined as the product of
Force and time, or more conveniently
as the variation of momentum
PHS 5043 Forces & Energy
Newton’s Laws
Practice:
A 168 g hockey puck hits the boards of the ice rink
perpendicularly at a velocity of 28 m/s and rebounds
along same trajectory at 20 m/s. What impulses did
the boards impart to the puck?
Δp = m Δv
Δp = 0.168 kg (-20 m/s – 28 m/s)
Δp = 0.168 kg (-48 m/s)
Δp = – 8.1 kg m/s
Δp = – 8.1 N s
PHS 5043 Forces & Energy
Newton’s Laws
Newton's second law & momentum:
You (50 kg) ride a bike (10 kg) at 10 m/s and speed
up to 20 m/s, after 100 m. If frictional force (air &
ground) is 30 N, what force did you apply during
the acceleration?
F=ma
V22 – V12 = 2ad
Fa – Ff = m a
a = (V22 – V12 ) / 2d
Fa = m a + Ff
a = 1.5 m/s2
Fa = (60 kg) (1.5 m/s2) + 30 N
Fa = 120 N
PHS 5043 Forces & Energy
Newton’s Laws
Newton’s Third Law:
“If object A exerts a force on object B (action),
then object B exerts a force equal in magnitude,
but opposite in direction, on object A (reaction)”
PHS 5043 Forces & Energy
Newton’s Laws
Newton’s Third Law:
 Rockets and airplanes are examples of the
application of Newton’s third law
 Both forces act simultaneously (always)