Physics 151: Lecture 8, Pg 1
Physics 151: Lecture 8Physics 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 2
ReviewReviewNewton’s Laws 1 and 2Newton’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, FFNET = FF = maa
See text: 5.1-4
Physics 151: Lecture 8, Pg 3
Newton’s Third Law: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
See text: 5-6
This is among the most abused concepts in physics.
REMEMBER: Newton’s 3rd law concerns force pairs whichact on two different objects (not on the same object) !
For every “action” there is an equal and opposite “reaction”
Physics 151: Lecture 8, Pg 4
An ExampleAn Example
FB,E = - mB g
EARTH
FE,B = mB g
Consider the forces on an object undergoing projectile motion
FB,E = - mB g
FE,B = mB g
Physics 151: Lecture 8, Pg 5
Normal ForcesNormal Forces
Certain forces act to keep an object in place. These have what ever force needed to balance all others
(until a breaking point).
FT,B
FB,T
Physics 151: Lecture 8, Pg 6
Force PairsForce 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.
FT,B
FB,T
FB,E = -mg
FE,B = mg
Physics 151: Lecture 8, Pg 7
An ExampleAn Example
Consider the following two cases
Physics 151: Lecture 8, Pg 8
An ExampleAn Example
The Free Body Diagrams
mg
mg
FB,T= N
Ball FallsFor Static Situation
N = mg
Physics 151: Lecture 8, Pg 9
An ExampleAn Example
The action/reaction pair forces
FB,E = -mg FB,T= N
FE,B = mg
FB,E = -mg
FE,B = mg
FT,B= -N
Physics 151: Lecture 8, Pg 10
Lecture 8, Lecture 8, Act 1Act 1Newton’s 3rd LawNewton’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) 2 (b) 4 (c) 6
a b
Physics 151: Lecture 8, Pg 11
246
Lecture 8, Lecture 8, Act 1Act 1Solution:Solution:
a bFFa,f FFf,a FFb,a FFa,b
FFg,ag,a
FFa,ga,g
FFg,bg,b
FFb,gb,g
FFE,aE,a
FFa,Ea,E
FFE,bE,b
FFb,Eb,E
Is Fa,f = Fb,a? (A) YES (B) NO Fb,a = Fa,f [mb/(mb+ma)]
Physics 151: Lecture 8, Pg 12
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 (C) NO
Lecture 8, Lecture 8, Act 2Act 2
A Ba= 2.5 m/s2rope
Physics 151: Lecture 8, Pg 13
What are the relevant forces ?
Lecture 8, Lecture 8, Act 2Act 2 Solution:Solution:
mAmB
Fapp
a= 2.5 m/s2
mA
rope
mB
T
-T T
-T
Fapp = a (mA + mB)Fapp = 2.5m/s2( 4kg+6kg) = 25 N
total mass !
aA = a = 2.5 m/s2
T = a mA T = 2.5m/s2 4kg = 10 NT > 9 N, rope will brakeANSWER (A)
Fapp - T = a mB T = 25N - 2.5m/s2 6kg=10NT > 9 N, rope will brake
a = 2.5 m/s2
Fapp
a = 2.5 m/s2
THE SAME ANSWER -> (A)
Physics 151: Lecture 8, Pg 14
Exercise: Inclined planeExercise: Inclined plane
A block of mass m slides down a frictionless ramp that makes angle with respect to horizontal. What is its acceleration a ?
ma
See text: Example 5.7
Physics 151: Lecture 8, Pg 15
Inclined plane...Inclined plane...
Define convenient axes parallel and perpendicular to plane:
Acceleration a is in x direction only.
ma
ii
jj
See text: Example 5.7
Physics 151: Lecture 8, Pg 16
Inclined plane...Inclined plane...
Consider x and y components separately: ii: mg sin = ma a = g sin
jj: N - mg cos . N = mg cos
mgg
NN
mg sin
mg cos
maa
ii
jj
See text: Example 5.7
m
Physics 151: Lecture 8, Pg 17
Angles of an Inclined planeAngles of an Inclined plane
a = g sin
mg
N
See text: Example 5.7
m
Physics 151: Lecture 8, Pg 18
Free Body DiagramFree 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 19
Free Body DiagramFree Body Diagram
Eat at Bob’s
T1
mg
T2
Add vectors :
x
yT1
T2
mg
Vertical (y):mg = T1sin1 + T2sin2
Horizontal (x) :T1cos1 = T2cos2
Physics 151: Lecture 8, Pg 20
Example-1 with pulleyExample-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 ?
Free-body diagram for each object
M1
T2T1
M2
aAnimation
Video
Physics 151: Lecture 8, Pg 21
Example-2 with pulleyExample-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.
M
T5
T4
T3T2
T1
F We use the 5 step method.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
Free-body diagram for each object
Physics 151: Lecture 8, Pg 22
Pulleys: continuedPulleys: continued FBD for all objects
M
T5
T4
T3T2
T1
F
T4
F=T1
T2
T3
T2 T3
T5
M
T5
Mg
Physics 151: Lecture 8, Pg 23
Pulleys: finallyPulleys: finally
Step 3: Plan the solution (what are the relevant equations)F=ma , static (no acceleration: mass is held in place)
M
T5
Mg
T5=Mg
T2 T3
T5
T2+T3=T5
T4
F=T1
T2
T3
F=T1
T1+T2+T3=T4
Physics 151: Lecture 8, Pg 24
Pulleys: really finally!Pulleys: really finally! Step 4: execute the plan (solve in terms of variables)
We have (from FBD):
T5=MgF=T1 T2+T3=T5 T1+T2+T3=T4
M
T5
T4
T3T2
T1
F
T2=T3T1=T3
T2=Mg/2
T2+T3=T5 gives T5=2T2=Mg
F=T1=Mg/2
T1=T2=T3=Mg/2 and T4=3Mg/2
T5=Mg and
Pulleys are massless and frictionless
Step 5: evaluate the answer (here, dimensions are OK and no numerical values)
Physics 151: Lecture 8, Pg 25
Lecture 9, Lecture 9, ACT 1ACT 1Gravity and Normal ForcesGravity and Normal Forces
A woman in an elevator is accelerating upwards
The normal force exerted by the elevator on the woman is,A) greater thanB) the same asC) less than
the force due to gravity acting on the woman
Physics 151: Lecture 8, Pg 26
Lecture 9, Lecture 9, ACT 1ACT 1Gravity and Normal ForcesGravity and Normal Forces
The free body diagram is,
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 womanNote, both of these forces act on the woman, they cannot be an action/reaction pair
N mg
Physics 151: Lecture 8, Pg 27
Lecture 9, Lecture 9, ACT 1bACT 1bGravity and Normal ForcesGravity and Normal Forces
A woman in an elevator is accelerating upwards
The normal force exerted by the elevator on the woman is,A) greater thanB) the same asC) less than
the force the woman exerts on the elevator.
Physics 151: Lecture 8, Pg 28
Lecture 9, Lecture 9, ACT 1bACT 1bGravity and Normal ForcesGravity and Normal Forces
The action/reaction force diagram for the woman and elevator is,
By Newton’s third law these must be (B) equal.
N = FW,EFE,W
Physics 151: Lecture 8, Pg 29
Recap of today’s lectureRecap 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)