Download Newton`s 2nd Law

Chapter Four – Force, 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).
7
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
2
v = constant value
(could be +/-)
v increasing or
decreasing
3
v = 0
v = 0
a=0
v = constant
a = constant
value
a=0
value
F = 0
F = 0
F = m
a
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!)
FA
Push or Pull.
Fg /FW/W
Force of gravity.
FW  mg Fw is called the “weight”.
T
Tension is a force acting in a rope, string, etc. Tension has no equation.
You either
solve for it, or it cancels, or
it’s given. Tension is always a pull.
n/ FN Normal force. A contact force, always perpendicular to the surface.
Ff
Far
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.
Fany subscript that make ssense to solve the problem
n
mg sin 
mg cos 
FW= mg
8
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 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 g
m g
m2
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 
9
Friction:
The frictional force always opposes the motion.
fs = F
n
fs
if block not moving
F
fs  static force of friction
Fg
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
F
Fsin O
n
O
m
Fcos O
mg
10
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 
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
fk
add the equations:
m2 g  m1 g sin   f k  m1a  m2 a
+
m2 g
m 1 g sin O
Solve for m2:
11
m2  g  a   m1a  m1g sin  f k
m2 g  m2 a  m1a  m1 g sin   f k
f k  k m1 g cos
m2 
m2 
m1a  m1 g sin   k m1 g cos
 g  a
m1a  m1 g sin   f k
 g  a
m2 
f k  k m1 g cos
m1  a  g sin   k g cos 
 g  a
Equations
 F  Fnet  ma
F fric   N
a = acceleration
F = Force
µ = coefficient of
friction
FN = normal force
12