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AP Physics – Newton’s Laws
Force
Any push or pull
Newton’s 1st Law
Law of inertia (Restatement of Galileo’s principle of inertia).
Objects in motion stay in motion, objects at rest stay at rest unless
acted upon by an outside force.
Newton’s 2nd Law
 F  ma
Newton’s 3rd Law
Equal and opposite forces. For every action force there is an equal
& opposite reaction force. Forces come in action - reaction pairs.
F
Key to all problems.
 Fx is the sum of all forces acting in the x direction.
 Fy is the sum of all forces acting in the y direction.
 F  ma
This is the sum of all forces. Known as the Net Force. You may need
to solve for a using the kinematic equations, then solve for force, or given force
you solve for a and then use it in the kinematic equations to find v, x, or t.
F 0
For objects at rest or moving with constant motion. Very important
concept. Make sure that you pick up on it immediately – save you a lot of grief.
Strategy on force Problems:
1. Draw FBD (free body diagram).
2. Set direction of motion. What would the object do if it could? Considered this the
positive direction.
3. Using the forces listed below write the F equations relevant to the problem. In what
direction is the problem moving? What matters, the x or the y direction? The parallel or
the perpendicular direction? Any force vectors in the FBD pointing in the direction of
motion are positive while any vectors the other way are negative.
4. Substitute known equation, (forces like Fw becomes mg).
5. Substitute for F. Ask yourself what the sum of force should be based on the chart below.
Is the object standing still, moving at constant velocity, or accelerating. Substitute zero or
ma for F.
1
v=0
v = 0
a=0
F = 0
2
v = constant value
(could be +/-)
v = 0
a=0
F = 0
3
v increasing or
decreasing
v = constant value
a = constant value
F = m a
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6. Plug in and solve. (All values including 9.8 m/s2 are entered as positives. The negative
signs were decided when setting up the sum of force equation. Plugging in – 9.8 m/s2 will
just turn a vector assigned as – Fg into a positive. You decided its sign based on the way it
was pointing relative to the problems direction of motion. Don’t reverse it now!)
FP
Push or Pull.
Fg
Force of gravity.
T
Tension is a force acting in a rope, string, etc. Tension has no equation.
solve for it, or it cancels, or it’s given. Tension is always a pull.
FW  mg
Fw is called the “weight”.
You either
n or N Normal force. A contact force, always perpendicular to the surface.
f
Far
Fc
Friction force. f   N Always opposes motion. Static friction: not moving.
Kinetic friction: object moving.
Force of air resistance. This force has no equation. You either solve for it, or it
cancels, or it’s given.
mv 2
Force Centripetal. It is the F in circular motion problems. FC 
r
Fc can be any force that keeps an object in circular motion.
It can also be two or more of these added together. The direction of motion is toward
the center. So any force directed toward the center is positive and any force directed
outward is negative.
The key in using any of these equations is to ask yourself: 1. What is causing the
circular motion? 2. Then set up the equality. 3. Substitute known equations. 4.
Solve.
FB
Force due to a magnetic field. This force is perpendicular to the field and
perpendicular to the velocity of the particle. So any charged particle will move in a
circle. Use the right hand rule for positive charges or positive current.
Fany subscript that make ssense to solve the problem
n
mg sin 
mg cos 
FW= mg
Normal force: Gravity pulls the object down the slope and into the slope. If we only
consider the motion into the slope (perpendicular), the object has no perpendicular velocity.
So the F= 0. Then the surface must push upward, equal and opposite to the perpendicular
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gravity component. Named the normal force, it is a contact force and operates perpendicular
to any surface. It must counter only the component of gravity perpendicular to the surface.
n  FW cos
Where  is the angle of the slope. It is also the tilt angle of the
surface measured from the ground. Substituting mg for Fg.
n  mg cos
nN  FW
Flat surfaces  = 0o,
or n  mg
Problems with multiple objects:
Draw free body diagram for each object.
Develop equation showing sum of forces on each object.
Work to eliminate unknowns. Often the tension will cancel out.
T
T
m
1
m2
m2 g
m g
1
The forces on the rising mass, use up as positive:
F
y
F
 m1a  T  m1 g
y
 m2 a  m2 g  T
For the falling mass, down is positive
Add the 2 equations: m1a  m2a  T  m1g  m2 g  T
m1a  m2a  T  m1g  m2 g  T
a  m1  m2   m2 g  m1g
ag
 m2  m1 
 m1  m2 
n
fs
F
Friction:
The frictional force always opposes the motion.
fs = F
if block not moving
Fg
fs  static force of friction
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fs can have max value of F
Block moves if F is greater.
Moving:
Fk  kinetic force of friction
Kinetic friction is always less than static friction
fs
 s n
fk
 k n
s  coefficient of static friction
k  coefficient of kinetic friction
Values depend on surfaces
Friction when pulling at an angle:
F
m
O
m
O
F
Fsin O
n
Fcos O
mg
Friction on a Ramp:
n
fs
mg cos O
mg sin O
O
O
Fg = mg
Block rests on a ramp
fs must be exceed the force component down the ramp: fs > mg sin 
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If  is increased until block just slides, then fs must equal component down ramp
fs = mg sin 
If block slides down ramp at constant speed, then fk must equal component down ramp
fk = mg sin 
If block accelerates down ramp, then net force down ramp is the component minus the kinetic
friction which equals ma.
F = mg sin  - fk = ma
m2 g  T  m2a
+
T
T  m1 g sin   f k  m1a
T
m2
add the equations:
+
m2 g  m1 g sin   f k  m1a  m2 a
Solve for m2:
m2 g
fk
m 1 g sin O
m2  g  a   m1a  m1g sin  f k
m a  m1 g sin   f k
f k  k m1 g cos
m2  1
 g  a
m2 g  m2 a  m1a  m1 g sin   f k
f k  k m1 g cos
m2 
m1a  m1 g sin   k m1 g cos
 g  a
m2 
m1  a  g sin   k g cos 
 g  a
Equations on test
 F  Fnet  ma
F fric   N
a = acceleration
F = Force
= coefficient of friction
N = normal force
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