Resolving Forces
Into Vector
Components
Physics
Montwood High School
R. Casao
Resolving Weight
The weight vector Fw for a mass
can be resolved into an x- and ycomponent.
The x-component is called the
parallel force and is represented
by Fx or Fp.
The y-component is called the
perpendicular or normal force and
is represented by Fy or FN.
Resolving Weight
 = 0°
 The Fw vector is
perpendicular (normal)
to the surface: Fw = Fy.
 Don’t worry about the
sign of Fw or Fy.
 Fy represents the
amount of the object’s
weight supported by the
surface.
 The normal force:
FN – Fw = 0; FN = Fw
Normal Force
The normal force is a force that keeps
one object from penetrating into
another object.
The normal force is always
perpendicular a surface.
The normal exactly cancels out the
components of all applied forces that
are perpendicular to a surface.
Normal Force on Flat Surface
The normal force is equal to the
weight of an object for objects resting
on horizontal surfaces.
FN = FW = m·g
FN
m·g
Resolving Weight
 For  > 0°
 The weight vector Fw always points
straight down. To make things easier,
rotate the x and y axes so that the x-axis
is parallel to the surface. The y-axis will
be perpendicular to the surface.
Resolving Weight
 Fx is the part of the
weight that causes
the mass to slide
down the inclined
plane.
 Fx will be positive
because we take the
direction of the
motion to be positive.
 Fx is the accelerating
force; Fx = m·a.
Resolving Weight
Fy is the part of the
weight that presses
the mass to the
surface.
The surface exerts an
equal and opposite
upward force to
balance Fy – the
normal force FN.
The normal force:
FN – Fy = 0; FN = Fy
Normal Force on Ramp
The normal force
is
perpendicular
FN = m·g·cos 
to inclined ramps
as well. It’s
FN
always equal to
the component of
Fx = m·g·sin  weight
perpendicular to
the surface.
Fy = m·g·cos 

Fw = m·g

Resolving Weight
In friction problems, FN is the
force that presses the mass to
the surface.
The angle  between Fw (the
hypotenuse) and Fy is always
equal to the angle of the
inclined surface.
Resolving Weight
Complete a
right triangle:
cos θ 
Fy
Fw
Fx
sin θ 
Fw
Fy  Fw  cos θ
Fx  Fw  sin θ
Resolving Weight
General equations (if N are given):
Fx  Fw  sin θ
Fy  Fw  cos θ
General equations (if kg are given):
Fx  m  g  sin θ
Fy  m  g  cos θ
Accelerating force: Fx=m·a
How Does the Incline Affect the Components?
mg sin 
mg sin 
mg cos 
mg
mg
The steeper the incline, the greater  is, and the greater
sin  is. Thus, a steep incline means a large parallel
component and a small horizontal one. Conversely, a
gradual incline means a large horizontal component and a
small vertical one.
Extreme cases: When  = 0, the ramp is flat; red = mg; blue = 0.
When  = 90, the ramp is vertical; red = 0; blue = mg.
Inclined Plane: Normal Force
FN = mg cos
mg sin
Recall normal force is perpen-dicular to
the contact surface. As long as the ramp
itself isn’t accelerating and no other forces
are lifting the box off the ramp or pushing
it into the ramp, N matches the
perpendicular component of the weight.
This must be the case, otherwise the box
would be accelerating in the direction of
red (mg cos  downward) or green (mg cos
 upward).

mg cos
mg
FN > mg cos  would mean the box is jumping
off the ramp.
FN < mg cos  would mean that the ramp is
being crushed.
Acceleration on a Ramp
FN = m·g·cos 
FN
Fx = m·g·sin 
Fw = m·g

Fy = m·g·cos 
What will the
acceleration be in
this situation?
SF = m·a
Fx = m·a
m·g·sin  = m·a
g·sin  = a

Acceleration on a Ramp
FN = m·g·cos 
FN
F
Fx = m·g·sin 

Fw = m·g
How could you keep the
block from accelerating?
Supply a pulling force F
that is equal in magnitude
to Fx and is opposite in
direction to Fx.
Fy = m·g·cos 

Pulling an Object at an
Angle wrt the Horizontal
 The pulling force F has
an x and y component.
 Construct a right
triangle to determine
Fx and Fy.
 Fx is the force that is
moving the object
forward along the
surface.
 Fy is the upward pull
of the force F.
Pulling an Object at an
Angle wrt the Horizontal
Fx
cos  
F
Fy
sin  
F
Fx  F  cos 
Fy  F  sin 
Pulling an Object at an Angle
wrt the Horizontal
 Fx is the accelerating force: Fx = m·a.
 To determine FN, use forces up = forces down.
 FN + Fy = Fw, therefore: FN = Fw - Fy
Pushing an Object at an
Angle wrt the Horizontal
 The pushing force F
has an x and y
component.
 Construct a right
triangle to determine
Fx and Fy.
 Fx is the force that is
moving the object
forward along the
surface.
 Fy is the force that is
pushing the object to
the surface F.
Pushing an Object at an
Angle wrt the Horizontal
Fx
cos  
F
Fy
sin  
F
Fx  F  cos 
Fy  F  sin 
Pushing an Object at an Angle
wrt the Horizontal
Fx is the
accelerating
force; Fx = m·a.
To determine
FN, use forces
up = forces
down.
FN= Fy + Fw
Pulley Problems
Tension is determined by examining one block
or the other.
SF  (m1+m2)·a
m2·g – FT + FT – m1·g·sin  = (m1 + m2)·a
FT
FT
FN
m1·g

m2
m2·g
Pulley Problems
Tension is determined by examining one block
or the other.
m2·g - FT = m2·a
FT - m1·g·sin  = m1·a
FT
FT
FN
m1·g

m2
m2·g