1. Making Pretty Pictures
Free Body Diagrams and Net Force

2. Free Body Diagrams VECTOR diagrams!
Shows ALL FORCES acting on an object
Must be properly labeled
Motionless
Equilibrium FN Fg

3. unbalanced force, it is also Fnet Not moving up or down, so…
Net Force
X Y
Net Force = *Equilibrium*
Motionless or Constant Velocity
Net Force ≠ 0
Accelerating or Decelerating FA Fg FN
Total = FA
Total = 0
Total = FA
Since this is the only
unbalanced force, it is also Fnet
Not moving up or down, so…
Fg + FN = 0
Fg = -FN FN m Fg
frictionless

4. Example #1
A man pushes a 50 kilogram crate across a frictionless surface with a constant force of 100 Newtons.
What is the normal force that pushes on the crate?
Draw a free-body diagram of the crate.
What is the net force on the crate?
What is the crate’s acceleration?
What is the weight of the crate?
Fnet will only be
the 100N horizontal
force FA FN Fg
Fg = mg
Fg = (50 kg)(9.81 m/s2)
Fg = N
FN = Fg
FN = N
a = Fnet / m
a = (100 N) / (50 kg)
a = 2 m/s2

5. Example #2
A horse pulls a 500 kilogram sled with a constant force of 3,000 Newtons. The force of friction between the sled and the ground is 500 Newtons.
What is the normal force that pushes on the sled?
Draw a free-body diagram of the sled.
What is the net force on the sled?
What is the sled’s acceleration?
What is the weight of the sled?
Fnet = ΣFx
Fnet = 3000 N – 500 N
Fnet = 2500 N
Fg = mg
Fg = (500 kg)(9.81 m/s2)
Fg = 4905 N FA FN Fg Ff
FN = Fg
FN = 4905 N
a = Fnet / m
a = (2500 N) / (500 kg)
a = 5 m/s2

6. This applied force (FA) The total vertical force must
Pulling on an Angle
A block is pushed along a frictionless, horizontal surface with a force of 100 newtons at an angle of 30° above horizontal.
This applied force (FA)
can be broken into
COMPONENTS
X Y
FAX FAY Fg FN
Total = FAX Total = 0 FN
The total vertical force must
be 0, so
FN = Fg - FAY FA
FAY
30˚
FAX
Acceleration depends only on
FAX Fg

7. Example #3
A man pulls a 40 kilogram crate across a smooth, frictionless floor with a force of 20 N that is 45˚ above horizontal.
What is the net force on the sled?
How could the acceleration be increased?
Fnet = FA cos θ
Fnet = (20 N)(cos 45°)
Fnet = N
Pushing at a smaller angle will make Fnet greater and
therefore increase acceleration.
What is the crate’s acceleration?
a = Fnet / m
a = (14.14 N) / (40 kg)
a = 0.35 m/s2

8. This applied force (FA) The total vertical force must
Pushing on an Angle
A block is pushed along a frictionless, horizontal surface with a force of 100 newtons at an angle of 30° below horizontal.
This applied force (FA)
can be broken into
COMPONENTS
X Y
FAX FAY Fg FN
Total = FAX Total = 0 FN
The total vertical force must
be 0, so
FN = Fg + FAY
FAX
FAY
-30˚ FA
Acceleration depends only on
FAX Fg

9. Example #4
A girl pushes a 30 kilogram lawnmower with a force of 15 Newtons at an angle of 60˚ below horizontal.
Assuming there is no friction, what is the
acceleration of the lawnmower?
Fnet = FA cos θ
Fnet = (15 N)(cos 60°)
Fnet = 7.5 N
a = Fnet / m
a = (7.5 N) / (30 kg)
a = 0.25 m/s2
What could she do to reduce her acceleration?
Push at an greater angle