Transcript 212 Lecture 9

Lecture 9
Newton’s Laws and Forces
(Cont….)
Inclined plane with “Normal” and Frictional
Forces (Cont….)
“Normal” means perpendicular
1. Static Equilibrium Case
Normal
Force
2. Dynamic Equilibrium
3. Dynamic case with non-zero acceleration
SF=0
Fx= 0 = mg sin q – f
Friction
f Force
mg sin q
Fy= 0 = mg cos q – N
mg cos q
with mg sin q = f ≤ mS N
if mg sin q > mS N, must slide
Critical angle ms = tan qc
q
Block weight is mg
q
q
y
x
Inclined plane with “Normal” and Frictional
Forces
1. Static Equilibrium Case
“Normal” means perpendicular
Normal
Force
2. Dynamic Equilibrium
Friction opposite velocity
v
(down the incline)
SF=0
Fx= 0 = mg sin q – fk
fK
Friction
Force
mg sin q
Fy= 0 = mg cos q – N
fk = mk N = mk mg cos q
Fx= 0 = mg sin q – mk mg cos q
mk = tan q
mg cos q
q
(only one angle)
mg
q
q
y
x
Inclined plane with “Normal” and Frictional
Forces
3. Dynamic case with non-zero acceleration
Result depends on direction of velocity
Normal
Force
Friction Force
Sliding Down
v
mg sin q
q
Fx= max = mg sin q ± fk
Fy= 0 = mg cos q – N
fk = mk N = mk mg cos q
Fx= max = mg sin q ± mk mg cos q
ax = g sin q ± mk g cos q
fk Sliding
q
Weight of block is mg
Up
“Free” Fall


Terminal velocity reached when Fdrag = Fgrav (= mg)
For 75 kg person with a frontal area of 0.5 m2,
vterm  50 m/s, or 110 mph
which is reached in about 5 seconds, over 125 m of fall
Newton’s Third Law
If object 1 exerts a force on object 2 (F2,1 ) then object 2
exerts an equal and opposite force on object 1 (F1,2)
F1,2 = -F2,1
For every “action” there is an equal and opposite “reaction”
IMPORTANT:
Newton’s 3rd law concerns force pairs which
act on two different objects (not on the same object) !
Example (non-contact)
Consider the forces on an object undergoing projectile motion
FB,E = - mB g
FB,E = - mB g
FE,B = mB g
FE,B = mB g
EARTH
Question: By how much does g change at an altitude of
40 miles? (Radius of the Earth ~4000 mi)
Example
Consider the following two cases (a falling ball and ball
on table),
Compare and contrast Free Body Diagram
and
Action-Reaction Force Pair sketch
Example
The Free Body Diagram
mg
FB,T= N
mg
Ball Falls
For Static Situation
N = mg
Normal Forces
Certain forces act to keep an object in place.
These have what ever force needed to balance all others (until
a breaking point).
FB,T
FT,B
Main goal at this point : Identify force pairs and apply
Newton’s third law
Example
First: Free-body diagram
Second: Action/reaction pair forces
FB,E = -mg
FB,T= N
FT,B= -N
FE,B = mg
FB,E = -mg
FE,B = mg
Example: Friction and Motion

A box of mass m1 = 1 kg is being pulled by a horizontal string
having tension T = 40 N. It slides with friction
(mk= 0.5) on top of a second box having mass m2 = 2 kg, which in
turn slides on a smooth (frictionless) surface.
 What is the acceleration of the second box ?
Central question: What is force on mass 2?
(A) a = 0 N (B) a = 5 N (C) a = 20 N (D) can’t tell
v
T
a=?
m1
m2
slides with friction (mk=0.5 )
slides without friction
Solution

First draw FBD of the top box:
v
N1
T
fk = mKN1 = mKm1g
m1
m1g
Solution

Newtons 3rd law says the force box 2 exerts on box 1 is
equal and opposite to the force box 1 exerts on box 2.

As we just saw, this force is due to friction:
Reaction
f2,1 = -f1,2
Action
m1
f1,2 = mKm1g = 5 N
m2
(A) a = 0 N
(B) a = 5 N
(C) a = 20 N (D) can’t tell
Solution

Now consider the FBD of box 2:
N2
f2,1 = mkm1g
m2
m1g
m2g
Solution

Finally, solve Fx = ma in the horizontal direction:
mK m1g = m2a
m1m k g 5 N
a

m2
2 kg
= 2.5 m/s2
f2,1 = mKm1g
m2