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AP Physics Review
Newton’s Laws and Forces
Basics of Forces
 Balanced forces lead to a net force = 0 and no change in an objects
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motion. (constant velocity, or stopped)
Unbalanced forces lead to a net force that causes an acceleration and
therefore a change in the objects motion. (increase or decrease in
velocity, circular motion etc.)
An object with a net force = 0 and that is at rest is said to be in static
equilibrium.
An object with a net force = 0 and that is motion at constant velocity
is said to be in dynamic equilibrium.
Units of any force is a Newton (N) which is equivalent to a kg m/s2
The force of gravity can also be called the weight of an object. W=mg
Since forces are vectors, any force directed down is said to be negative
and typically any force against motion is negative as well.
Newton’s Laws
 1st Law – inertia. An object at rest stays at rest and an object in motion
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stays in motion unless acted upon by an outside force.
2nd Law – an objects acceleration is proportional to the net force on the
object and inversely proportional to the objects mass
a = F/m or F = ma
3rd Law – For every action there is an equal but opposite reaction.
You are attracted to Earth by the same magnitude force that Earth is
attracted to you. The objects have different masses and therefore behave
differently. Remember the bug and the windshield.
Every force in the universe has an equal and opposite force, it just
depends on how far you expand the system you are observing.
Net Force
 The net force is the sum of all the forces acting on an object.
 This may equal 0 in which case the objects motion does not change.
 If the sum does not equal 0 then the object will accelerate in the
direction of the net force.
 If the force is at an angle. Components must be used and you will have a
net force in the x direction and a net force in the y direction which will
be used to find the resultant.
Static Equilibrium
 The sum of the forces in the x and y directions must be equal to
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zero.
Start with a free body diagram. On the problem below, in the x
direction you have simply the x components of the tension in
each string. -T1x + T2x = 0 or T2x = T1x since the x
component of T1 is in the negative direction, this will work out.
T1x = T1cos(70) and T2x = T2 cos(50)
In the y direction the weight of the mass is acting downward and
the y components of each tension is acting upward..
W+T1y+T2y = 0 since W is down it will be negative, so
T1y+T2y = W
Solve for the tension in wires 1 and 2.
T2 = 78N
T1 = 148N
Inclined Planes
 In the top picture, the forces acting on the object are the
normal force from the surface, gravity, and friction. The
normal force is a component of gravity and the force
opposing friction is the other component of gravity.
 If all forces balance out, the block remains at rest or
sliding down at a constant rate.
 If the they do not balance out then the block accelerates
either up the ramp or down the ramp depending on the
direction of the net force.
 It is helpful to rotate your standard axis so that it lines up
with the surface of the ramp.
Friction
 Friction opposes motion and is dependant on the surfaces in
contact and the Normal force.
 Friction can be static or dynamic depending on if the object is
moving or not.
 µ is the coefficient of friction and strictly dependant on the
surfaces in contact.
 In the case of the person pushing the lawnmower, the force
applied by the person is at an angle. The friction would act
against the x component of this force only!
Attwood Machines
 A situation involving mass around a pulley.
 Theses masses may be static or dynamic.
 They may be on a horizontal surface or on an incline.
 They may be frictionless or experience friction.
 They are the culmination of all force laws and strategies put
together in one.
 Create a FBD for each
object in the system and
then use net force
equations to solve for
unknowns.
Example
Masses m1 = 4.00 kg and m2 = 9.00 kg are connected by a light string that passes over a frictionless pulley.
As shown in the diagram, m1 is held at rest on the floor and m2 rests on a fixed incline of angle 40 degrees.
The masses are released from rest, and m2 slides 1.00 m down the incline in 4 seconds. Determine (a) The
acceleration of each mass (b) The coefficient of kinetic friction and (c) the tension in the string.
T
FN
T  m1 g  m1a  T  m1a  m1 g
Ff
m2gcos40
m2 g sin   ( F f  T )  m2 a
40
T
m2 g
m1
m2gsin40
m1 g
FNET  ma
40
FNET  ma
Example
T  m1 g  m1a  T  m1a  m1 g
m2 g sin   ( F f  T )  m2 a
x  voxt  1 at 2
2
1  0  1 a ( 4) 2
2
a  0.125 m / s 2
T  4(.125)  4(9.8)  39.7 N
m2 g sin   F f  T  m2 a
m2 g sin   F f  (m1a  m1 g )  m2 a
m2 g sin    k FN  m1a  m1 g  m2 a
m2 g sin    k m2 g cos   m1a  m1 g  m2 a
m2 g sin   m1a  m1 g  m2 a   k m2 g cos 
k 
m2 g sin   m1a  m1 g  m2 a
m2 g cos 
k 
56.7  0.5  39.2  1.125
 0.235
67.57
A mass, m1 = 3.00kg, is resting on a frictionless horizontal table is connected
to a cable that passes over a pulley and then is fastened to a hanging mass,
m2 = 11.0 kg as shown below. Find the acceleration of each mass and the
tension in the cable.
FNet  ma
FN
m2 g  T  m2 a
T  m1a
T
T
m1g
m2 g  m1a  m2 a
m2 g  m2 a  m1a
m2 g  a (m2  m1 )
m2g
a
m2 g
(11)(9.8)

 7.7 m / s 2
m1  m2
14
FNet  ma
m2 g  T  m2 a
T  m1a
FNET
FNet  ma 
m
a
Rise
Slope 
Run
T  (3)(7.7)  23.1 N