Transcript Physics 131: Lecture 9 Notes

Physics 151: Lecture 8

Reaminder:
Homework #3 : (Problems from Chapter 5)
due Fri. (Sept. 22) by 5.00 PM

Today’s Topics :
Review of Newton’s Laws 1 and 2 - Ch. 5.1-4
Newton’s third law: action and reaction - Ch. 5.6
Physics 151: Lecture 8, Pg 1
See text: 5.1-4
Review
Newton’s Laws 1 and 2

Isaac Newton (1643 - 1727) published Principia Mathematica
in 1687. In this work, he proposed three “laws” of motion:
Law 1: An object subject to no external forces is at rest or moves
with a constant velocity if viewed from an inertial
reference
frame.
Law 2: For any object, FNET = F = ma
Physics 151: Lecture 8, Pg 2
See text: 5-6
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”
This is among the most abused concepts in physics.
REMEMBER:
Newton’s 3rd law concerns force pairs which
act on two different objects (not on the same object) !
Physics 151: Lecture 8, Pg 3
An Example
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
Physics 151: Lecture 8, Pg 4
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
Physics 151: Lecture 8, Pg 5
Force Pairs
Newton’s 3rd law concerns force pairs. Two members of
a force pair cannot act on the same object.
Don’t confuse gravity (the force of the earth on an object)
and normal forces. It’s an extra part of the problem.
FB,T
FB,E = -mg
FT,B
FE,B = mg
Physics 151: Lecture 8, Pg 6
An Example
Consider the following two cases
Physics 151: Lecture 8, Pg 7
An Example
The Free Body Diagrams
mg
FB,T= N
mg
Ball Falls
For Static Situation
N = mg
Physics 151: Lecture 8, Pg 8
An Example
The action/reaction pair forces
FB,E = -mg
FB,T= N
FT,B= -N
FE,B = mg
FB,E = -mg
FE,B = mg
Physics 151: Lecture 8, Pg 9
Lecture 8, Act 1
Newton’s 3rd Law

Two blocks are being pushed by a finger on a horizontal
frictionless floor. How many action-reaction pairs of forces
are present in this system?
a
(a) 2
(b) 4
b
(c) 6
Physics 151: Lecture 8, Pg 10
Lecture 8, Act 1
Solution:
Fa,f
Fb,a
Ff,a
a
FE,a
Fa,b
bF
Fg,a
Fg,b
Fa,g
Fb,g
Fa,E
E,b
Fb,E
2
4
6
Is Fa,f = Fb,a?
(A) YES
(B) NO
Fb,a = Fa,f [mb/(mb+ma)]
Physics 151: Lecture 8, Pg 11
Lecture 8, Act 2

You are going to pull two blocks (mA=4 kg and mB=6 kg) at
constant acceleration (a= 2.5 m/s2) on a horizontal frictionless
floor, as shown below. The rope connecting the two blocks can
stand tension of only 9.0 N. Would the rope break ?

(A) YES
(B) CAN’T TELL
A
rope
(C) NO
a= 2.5 m/s2
B
Physics 151: Lecture 8, Pg 12
Lecture 8, Act 2
Solution:

What are the relevant forces ?
mB
mA
mA
rope
T
Fapp = a (mA + mB)
Fapp = 2.5m/s2( 4kg+6kg)
= 25 N
a= 2.5 m/s2
Fapp
aA = a = 2.5 m/s2
T
-T
-T
T = a mA
T = 2.5m/s2 4kg = 10 N
T > 9 N, rope will brake
ANSWER (A)
a = 2.5 m/s2
a = 2.5 m/s2
mB
total mass !
Fapp
Fapp - T = a mB
T = 25N - 2.5m/s2 6kg=10N
T > 9 N, rope will brake
THE SAME ANSWER -> (A)
Physics 151: Lecture 8, Pg 13
See text: Example 5.7
Exercise: Inclined plane

A block of mass m slides down a frictionless ramp that makes angle  with
respect to horizontal. What is its acceleration a ?
m
a

Physics 151: Lecture 8, Pg 14
See text: Example 5.7
Inclined plane...

Define convenient axes parallel and perpendicular to plane:
 Acceleration a is in x direction only.
j
m
a

i
Physics 151: Lecture 8, Pg 15
See text: Example 5.7
Inclined plane...

Consider x and y components separately:
i: mg sin  = ma
a = g sin 

j: N - mg cos  =0.

m
N = mg cos 
ma
j
mg sin 
N 
mg cos 
mg
i

Physics 151: Lecture 8, Pg 16
See text: Example 5.7
Angles of an Inclined plane
m
a = g sin 
N

mg

Physics 151: Lecture 8, Pg 17
Free Body Diagram
A heavy sign is hung between two poles by a rope at each
corner extending to the poles.
Eat at Bob’s
What are the forces on the sign ?
Physics 151: Lecture 8, Pg 18
Free Body Diagram
T2
T1
2
1
Eat at Bob’s
mg
Add vectors :
Vertical (y):
mg = T1sin1 + T2sin2
Horizontal (x) :
T1cos1 = T2cos2
y
T2
T1
1
2
x
mg
Physics 151: Lecture 8, Pg 19
Example-1 with pulley


Two masses M1 and M2 are connected by
a rope over the pulley as shown.
Assume the pulley is massless and
frictionless.
Assume the rope massless.
If M1 > M2 find :
Acceleration of M1 ?
Acceleration of M2 ?
Tension on the rope ?
T1
T2
Video
M2
Animation
M1
a
Free-body diagram for each object
Physics 151: Lecture 8, Pg 20
Example-2 with pulley


A mass M is held in place by a force F.
Find the tension in each segment of the
rope and the magnitude of F.
Assume the pulleys massless and
frictionless.
Assume the rope massless.
T4
T1
We use the 5 step method.
F
Draw a picture: what are we looking for ?
What physics idea are applicable ? Draw
a diagram and list known and unknown
variables.
Newton’s 2nd law : F=ma
T3
T2
<
T5
M
Free-body diagram for each object
Physics 151: Lecture 8, Pg 21
Pulleys: continued

FBD for all objects
T4
T2
T3
T4
T1
T3
T2
5
F
<
T
T5
T3
F=T1
T5
M
T2
M
Mg
Physics 151: Lecture 8, Pg 22
Pulleys: finally

Step 3: Plan the solution (what are the relevant equations)
F=ma , static (no acceleration: mass is held in place)
T5
T5=Mg
M
T2
T1+T2+T3=T4
T4
Mg
F=T1
T3
T2+T3=T5
T3
F=T1
T
5
T2
Physics 151: Lecture 8, Pg 23
Pulleys: really finally!

Step 4: execute the plan (solve in terms of variables)
We have (from FBD):
F=T1
T5=Mg
T2+T3=T5
T1+T2+T3=T4
Pulleys are massless and frictionless
T1=T3
T4
T2=T3
T2+T3=T5 gives T5=2T2=Mg
T1
T2=Mg/2

T2
F
T1=T2=T3=Mg/2 and T4=3Mg/2
T5=Mg
T3
and F=T1=Mg/2
Step 5: evaluate the answer (here,
dimensions are OK and no numerical values)
<
T5
M
Physics 151: Lecture 8, Pg 24
Lecture 9, ACT 1
Gravity and Normal Forces
A woman in an elevator is accelerating upwards
The normal force exerted by the elevator on the woman is,
A) greater than
B) the same as
C) less than
the force due to gravity acting on the woman
Physics 151: Lecture 8, Pg 25
Lecture 9, ACT 1
Gravity and Normal Forces
The free body diagram is,
N
mg
For the woman to accelerate upwards, the normal force
on the woman must be A) greater than the force due to
gravity acting on the woman
Note, both of these forces act on the woman, they cannot
be an action/reaction pair
Physics 151: Lecture 8, Pg 26
Lecture 9, ACT 1b
Gravity and Normal Forces
A woman in an elevator is accelerating upwards
The normal force exerted by the elevator on the woman is,
A) greater than
B) the same as
C) less than
the force the woman exerts on the elevator.
Physics 151: Lecture 8, Pg 27
Lecture 9, ACT 1b
Gravity and Normal Forces
The action/reaction force diagram for the woman and
elevator is,
N = FW,E
FE,W
By Newton’s third law these must be (B) equal.
Physics 151: Lecture 8, Pg 28
Recap of today’s lecture

Newton’s 3 Laws

Free Body Diagrams

Action/Reaction Force pairs

Reading for Friday, Ch 5.7-8, pp. 123-139
Applications of Newton’s Laws and Friction
Homework #3 (due next Wed. / Sept. 21 by 11:59 pm)
Physics 151: Lecture 8, Pg 29