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Forces
Forces
Free Body Diagrams
• Shows all forces as vectors acting on an object
• Push or pull on an object
• Causes acceleration
• Measured in Newtons N =
• Vectors always point away from object
Kg m
s2
• Used to help find net force
Contact Forces
Field Forces
Applied Force Gravitational Force Frictional Force Electrical Force
Tensional Force Magnetic Force
Normal Force Drag Force Spring Force FN
Fpull
Ff
Fg
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Find the unknown forces!!
Ex. 1
100 N
Ex. 2
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Newton’s First Law
Law of Inertia – Resistance to change motion
FA
• Objects in motion stay in motion 75 N
• Objects at rest stay at rest
50 N
Fnet = ? Fnet = 10 N Downward
Equilibrium – balanced forces, net force = 0
Fnet = 100N – 75 N
Fnet = FA – 50 N = ‐10 N
Net force – sum of all forces
Fnet = 25 N Upward
FA = 40 N
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Newton’s Second Law
Drag Force
A net force will cause acceleration
• “friction” force from a fluid (gases and liquids)
mass
Terminal Velocity – constant velocity of falling F = ma
force
Gravity force →
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when Fdrag = Fg
acceleration
F = mg
Mass and weight are not the same!!!
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Forces
Newton’s Third Law
Solving Tips
• Each action has an opposite and equal reaction
FA on B = ‐ FB on A 1. Draw the problem and choose coordinates
• Interaction Pair – action / reaction forces
2. Determine known and unknown forces.
3. Create a free‐body diagram showing the net force.
4. Use Newton’s laws to link acceleration and net force.
5. Solve equations for the unknowns
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Combining Forces
Normal Force
+X
FN = mg
FN = mg + Fhand
FN = mg ‐ Fstring
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Friction Factor
Friction Force
• Always against motion
• Two branches of friction (3 Types)
– Kinetic (Moving)
• Sliding
• Rolling
– Static (Stationary)
• Friction of fluids is called viscosity
Kinetic Friction
Ff = k Fn
 = friction factor
Fn = Normal Force
Static Friction
Ff,static ≤ s Fn
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2
Forces
Static Friction
Fn
Fn
• The force of static friction is not constant!
Static FFriction
• The maximum static friction is equal to 
q
sFn
Ff,static,max ≥ Fapplied → stays still
Ff,k
Ff,s
Kinetic FFriction
• Static friction is equal to pulling force until the object begins to move
Ff
Ff,s,max
Ff = sFn,
Ff = kFn,
Ff,static,max = Fapplied → constant speed, a=0
Ff,static,max ≤ Fapplied → accelerates
FApplied
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Static Friction and Motion
Thrust from Friction
What is the maximum acceleration a car can achieve if the tires/road friction coefficient is equal to 0.7? Fn
(ignore drag)
FDrag
Fnet,x = Fthrust ‐ Fdrag = ma
Fthrust
Fg
The maximum thrust cannot exceed road friction h
h
d
df
Fthrust = Ff,s,max = sFn= s mg
From Fnet,x
ma = s mg
a = s mg = (0.7)(9.8m/s2) = 6.9m/s2
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Static Equilibrium
Static Equilibrium
The most important rule:
ALWAYS FOLLOW THESE STEPS:
F = 0
This means that:
Fx = 0
and
1.
Draw a labeled free body diagram
2.
g
forces into components
p
Break angled
3.
Write net equations ( Fnet,x = … )
Fy = 0
Only use the components of angled forces!!
Fnet,x = 0 and Fnet,y = 0
4.
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Solve for unknowns one at a time
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Forces
Vector Direction ( 2 Common Ways)
Working with Forces at an Angle
• Labeled degrees north or south of x‐axis
• Degrees from east direction (0°).
80° N of East
45° S of East
O 80°
Or 80°
30° N of West
O 315°
Or 315°
O 150°
Or 150°
When a force is at an angle:
•break into x and y components
–Do not use the original force again!!
•Add x and y components separately
•Find the new resultant force and its angle
•Find the new resultant force and its angle
80°
30°
45°
O
A
 = Tan‐1
F = Fx2+Fy2
Using the angle from the x‐axis:
X‐Component
y‐Component
Fx = F cos 
Fy = F sin 
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SOH
Right Triangle Help
O
A
A = H Cos θ
 = angle
Opposiite
 = Sin‐1
O = H Sin θ
 = Tan‐1
90°
Adjacent
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Forces at an Angle ‐ Breaking into components

FPy
m
FN
Find FFP Components
Find
FPx = FP Cos 
FPy = FP Sin 
Use components for net equations
Fnetx = FPx
Fnety = FN + FPy ‐ Fg
CAH
TOA
A
O
Cos θ =
Tan θ =
H
A
A
O
= Cos‐1
= Tan‐1
H
A
O
Sin θ =
H
90° ‐ 
You will typically want to work with the angle from the x‐axis.
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O
H
Adjacent
Opposite
O = H Sin θ
A = H Cos θ
O = A Tan θ
A =
O = H2 ‐ A2
O
Tan θ
A = H2 ‐ O2
Hypotenuse
O
H =
Sin θ
A
H =
Cos θ
H = A2+O2
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Forces on a Ramp ‐ Breaking into components
Find Fg Components
Fgx = Fg Cos (90‐)
Fgy = Fg Sin (90‐)
FP

FPx

Or just use SOH CAH TOA
Fgx = Fg Sin 
Fgy = Fg Cos 
Fg = mg
FN should not equal Fg !!
FN = Fg ‐ Fpy if ay = 0
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Use components for net equations
Fnetx = ‐Fgx
Fnety = FN ‐ Fgy

Fg = mg
FN should not equal Fg !!
Many cases, FN will equal Fgy
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Forces
Free Body Diagrams
Tension Force
Spring Force
Drag Force
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