7 problems of newton law

Physics 111: Lecture 5, Pg 1
Physics 111: Lecture 5Physics 111: Lecture 5
Today’s AgendaToday’s Agenda
More discussion of dynamics
Recap
The Free Body DiagramFree Body Diagram
The tools we have for making & solving problems:
» Ropes & Pulleys (tension)
» Hooke’s Law (springs)

Review: Newton's LawsReview: Newton's Laws
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 = maa
Where FFNET = Σ FF
Law 3: Forces occur in action-reactionaction-reaction pairs, FFA ,B = - FFB ,A.
Where FFA ,B is the force acting on object A due to its
interaction with object B and vice-versa.

Gravity:Gravity:
What is the force of gravity exerted by the earth on a typical
physics student?
Typical student mass m = 55kg
g = 9.81 m/s2
.
Fg = mg = (55 kg)x(9.81 m/s2
)
Fg = 540 N = WEIGHT
FFE,S = -= -mgg
FFS,E = F= Fg == mgg

Lecture 5,Lecture 5, Act 1Act 1
Mass vs. WeightMass vs. Weight
An astronaut on Earth kicks a bowling ball and hurts his foot.
A year later, the same astronaut kicks a bowling ball on the
moon with the same force.
His foot hurts...
(a) more
(b) less
(c) the same
Ouch!

SolutionSolution
Ouch!
The masses of both the bowling
ball and the astronaut remain the
same, so his foot will feel the same
resistance and hurt the same as
before.

SolutionSolution
Wow!
That’s light.
However the weights of the
bowling ball and the astronaut are
less:
Thus it would be easier for the
astronaut to pick up the bowling
ball on the Moon than on the
Earth.
W = mgMoon gMoon < gEarth

The Free Body DiagramThe Free Body Diagram
Newton’s 2nd Law says that for an object FF = maa.
Key phrase here is for an objectfor an object..
So before we can apply FF = maa to any given object we
isolate the forces acting on this object:

The Free Body Diagram...The Free Body Diagram...
Consider the following case
What are the forces acting on the plank ?
P = plank
F = floor
W = wall
E = earth
FFW,P
FFP,W
FFP,F FFP,E
FFF,P
FFE,P

Consider the following case
What are the forces acting on the plank ?
Isolate the plank from
the rest of the world.
FFW,P
FFP,W
FFP,F FFP,E
FFF,P
FFE,P

The forces acting on the plank should reveal themselves...
FFP,W
FFP,F FFP,E

Aside...Aside...
In this example the plank is not moving...
It is certainly not accelerating!
So FFNET = maa becomes FFNET = 0
This is the basic idea behind statics, which we will
discuss in a few weeks.
FFP,W + FFP,F + FFP,E = 0
FFP,W
FFP,F FFP,E

ExampleExample
Example dynamics problem:
A box of mass m = 2 kg slides on a horizontal frictionless
floor. A force Fx = 10 N pushes on it in the xx direction.
What is the acceleration of the box?
FF = Fx ii aa = ?
m
yy
xx

Example...Example...
Draw a picture showing all of the forces
FF
FFB,F
FFF,B
FFB,E
FFE,B
yy
xx

Draw a picture showing all of the forces.
Isolate the forces acting on the block.
FF
FFB,F
FFF,B
FFB,E = mgg
FFE,B
yy
xx

Draw a free body diagram.
FF
FFB,F
mgg
yy
xx

Draw a free body diagram.
Solve Newton’s equations for each component.
FX = maX
FB,F - mg = maY
FF
FFB,F
mgg
yy
xx

FX = maX
So aX = FX / m = (10 N)/(2 kg) = 5 m/s2
.
FB,F - mg = maY
But aY = 0
So FB,F = mg.
The vertical component of the force
of the floor on the object (FB,F ) is
often called the Normal ForceNormal Force (N).
Since aY = 0 , N = mg in this case.
FX
N
mg
yy
xx

Example RecapExample Recap
FX
N = mg
mg
aX = FX / m
yy
xx

Normal ForceNormal Force
A block of mass m rests on the floor of an elevator that is
accelerating upward. What is the relationship between
the force due to gravity and the normal force on the block?
m
(a)(a) N > mgN > mg
(b)(b) N = mgN = mg
(c)(c) N < mgN < mg
a

SolutionSolution
m
N
mg
All forces are acting in the y direction,
so use:
Ftotal = ma
N - mg = ma
N = ma + mg
therefore N > mg
a

Tools: Ropes & StringsTools: Ropes & Strings
Can be used to pull from a distance.
TensionTension (T) at a certain position in a rope is the magnitude
of the force acting across a cross-section of the rope at that
position.
The force you would feel if you cut the rope and
grabbed the ends.
An action-reaction pair.
cut
T
T
T

Tools: Ropes & Strings...Tools: Ropes & Strings...
Consider a horizontal segment of rope having mass m:
Draw a free-body diagram (ignore gravity).
Using Newton’s 2nd law (in xx direction):
FNET = T2 - T1 = ma
So if m = 0 (i.e. the rope is light) then T1 = T2
T1 T2
m
a xx

An ideal (massless) rope has constant tension along the
rope.
If a rope has mass, the tension can vary along the rope
For example, a heavy rope
hanging from the ceiling...
We will deal mostly with ideal massless ropes.
T = Tg
T = 0
T T
2 skateboards

The direction of the force provided by a rope is along the
direction of the rope:
mg
T
m
Since ay = 0 (box not moving),
T = mg

Force and accelerationForce and acceleration
A fish is being yanked upward out of the water using a fishing
line that breaks when the tension reaches 180 N. The string
snaps when the acceleration of the fish is observed to be is
12.2 m/s2
. What is the mass of the fish?
m = ?
a = 12.2 m/s2
snap ! (a) 14.8 kg
(b) 18.4 kg
(c) 8.2 kg

Solution:Solution:
Draw a Free Body Diagram!!
T
mg
m = ?
a = 12.2 m/s2
Use Newton’s 2nd law
in the upward direction:
FTOT = ma
T - mg = ma
T = ma + mg = m(g+a)
m
T
g a
=
+ ( )
kg28
sm21289
N180
m 2
.
..
=
+
=

Tools: Pegs & PulleysTools: Pegs & Pulleys
Used to change the direction of forces
An ideal massless pulley or ideal smooth peg will
change the direction of an applied force without altering
the magnitude:
FF1
ideal peg
or pulley
FF2
| FF1 | = | FF2 |

Tools: Pegs & PulleysTools: Pegs & Pulleys
Used to change the direction of forces
An ideal massless pulley or ideal smooth peg will
change the direction of an applied force without altering
the magnitude:
mg
T
m T = mg
FW,S = mg

SpringsSprings
Hooke’s Law:Hooke’s Law: The force exerted by a spring is
proportional to the distance the spring is stretched or
compressed from its relaxed position.
FX = -k x Where x is the displacement from the
relaxed position and k is the constant
of proportionality.
relaxed position
FX = 0
x

Springs...Springs...
relaxed position and k is the
constant of proportionality.
relaxed position
FX = -kx > 0
x
x < 0

Springs...Springs...
relaxed position and k is the
constant of proportionality.
FX = - kx < 0
x
x > 0
relaxed position
Horizontal
springs

Scales:Scales:
Springs can be calibrated to tell us the applied force.
We can calibrate scales to read Newtons, or...
Fishing scales usually read
weight in kg or lbs.
0
2
4
6
8
1 lb = 4.45 N
Spring/string

m m m
(a)(a) 0 lbs. (b)(b) 4 lbs. (c)(c) 8 lbs.
(1) (2)
?
Force and accelerationForce and acceleration
A block weighing 4 lbs is hung from a rope attached to a
scale. The scale is then attached to a wall and reads 4 lbs.
What will the scale read when it is instead attached to
another block weighing 4 lbs?
Scale
on a
skate

Solution:Solution:
Draw a Free Body Diagram of one
of the blocks!!
Use Newton’s 2nd Law
in the y direction:
FTOT = 0
T - mg = 0
T = mg = 4 lbs.
mg
T
m T = mg
a = 0 since the blocks are
stationary

Solution:Solution:
The scale reads the tension in the rope, which is T = 4 lbs in
both cases!
m m m
T T T T
TTT

Problem: AccelerometerProblem: Accelerometer
A weight of mass m is hung from the ceiling of a car with a
massless string. The car travels on a horizontal road, and
has an acceleration a in the x direction. The string makes
an angle θ with respect to the vertical (y) axis. Solve for θ
in terms of a and g.
a
θ
ii

Accelerometer...Accelerometer...
Draw a free body diagram for the mass:
What are all of the forces acting?
m
TT (string tension)
mgg (gravitational force)
θ
ii

Using componentsUsing components (recommended):
ii: FX = TX = T sin θ = ma
jj: FY = TY − mg
= T cos θ − mg = 0
TT
mgg
θ
m
maa
jj
ii
TX
TY
θ

Using componentsUsing components :
ii: T sin θ = ma
jj: T cos θ - mg = 0
Eliminate T :
mgg
θ
m
maaT sin θ = ma
T cos θ = mg
tan θ =
a
g
TX
TY
jj
ii
TT

Alternative solution using vectorsAlternative solution using vectors (elegant but not as(elegant but not as
systematic):systematic):
Find the total vector force FFNET:
TT
mgg
FFTOT
θ
m
TT (string tension)
θ

Alternative solution using vectorsAlternative solution using vectors (elegant but not as(elegant but not as
systematic):systematic):
Find the total vector force FFNET:
Recall that FFNET = ma:
So
maa
tan θ = =
ma
mg
a
g
TT
mgg θ
tan θ =
a
g
m
TT (string tension)
θ

Let’s put in some numbers:
Say the car goes from 0 to 60 mph in 10 seconds:
60 mph = 60 x 0.45 m/s = 27 m/s.
Acceleration a = Δv/Δt = 2.7 m/s2
.
So a/g = 2.7 / 9.8 = 0.28 .
θ = arctan(a/g) = 15.6 deg
tan θ =
a
g
a
θ
Cart w/
accelerometer

Problem: Inclined planeProblem: Inclined plane
A block of mass m slides down a frictionless ramp that
makes angle θ with respect to the horizontal. What is its
acceleration a ?
θ
m
a

Inclined plane...Inclined plane...
Define convenient axes parallel and perpendicular to plane:
Acceleration a is in x direction only.
θ
m
a
ii
jj

Consider x and y components separately:
ii: mg sin θ = ma. a = g sin θ
jj: N - mg cos θ = 0. N = mg cos θ
mgg
NN
mg sin θ
mg cos θ
θ
maa
ii
jj
Incline

Alternative solution using vectors:
m
mg
N
a = g sin θ ii
N = mg cos θ jj
θ
ii
jj

Angles of an Inclined planeAngles of an Inclined plane
θ
θ
ma = mg sin θ
mg
N
The triangles are similar, so the angles are the same!

Forces and MotionForces and Motion
A block of mass M = 5.1 kg is supported on a frictionless
ramp by a spring having constant k = 125 N/m. When the
ramp is horizontal the equilibrium position of the mass is at
x = 0. When the angle of the ramp is changed to 30o
what
is the new equilibrium position of the block x1?
(a) x1 = 20cm (b) x1 = 25cm (c) x1 = 30cm
x = 0
M
k
x1
= ?
M
k
θ = 30o

SolutionSolution
x1
M
k
θ
x
y
Choose the x-axis to be along downward direction of ramp.
Mg
FBD: The total force on the block is zero since it’s at rest.
N
θ
Fx,g = Mg sinθ
Force of gravity on block is Fx,g = Mg sinθConsider x-direction:
Force of spring on block is Fx,s = -kx1
Fx,s
= -kx1

SolutionSolution
x1
M
k
θ
x
y
Since the total force in the x-direction must be 0:
Fx,g
= Mg sinθ
Fx,s
= -kx1
Mg sinθ − kx1 = 0
κ
sinΜg
x1
θ
=
m20
mN125
50sm189kg15
x
2
1
.
...
=
⋅⋅
=

Problem: Two BlocksProblem: Two Blocks
Two blocks of masses m1 and m2 are placed in contact on a
horizontal frictionless surface. If a force of magnitude F is
applied to the box of mass m1, what is the force on the
block of mass m2?
mm11 mm22
F

Problem: Two BlocksProblem: Two Blocks
Realize that F = (mm11+ mm22)a :
Draw FBD of block mm22 and apply FNET = ma:
F2,1F2,1 = mm22 a
F / (mm11+ mm22) = a
m22,1
( )





+
=
m2m1
F
F
Substitute for a :
mm22
(m1 + m2)
m2
F2,1 F=

Problem: Tension and AnglesProblem: Tension and Angles
A box is suspended from the ceiling by two ropes making
an angle θ with the horizontal. What is the tension in each
rope?
mm
θθ

Problem: Tension and AnglesProblem: Tension and Angles
Draw a FBD:
θθ
T1 T2
mg
Since the box isn’t going anywhere, Fx,NET = 0 and Fy,NET = 0
T1sin θ T2sin θ
T2cos θT1cos θ
jj
ii
Fx,NET = T1cos θ - T2cos θ = 0 T1 = T2
2 sin θ
mg
T1 = T2 =Fy,NET = T1sin θ + T2sin θ - mg = 0

Problem: Motion in a CircleProblem: Motion in a Circle
A boy ties a rock of mass m to the end of a string and twirls
it in the vertical plane. The distance from his hand to the
rock is R. The speed of the rock at the top of its trajectory is
v.
What is the tension T in the string at the top of the rock’s
trajectory?
R
v
TT
Tetherball

Motion in a Circle...Motion in a Circle...
Draw a Free Body Diagram (pick y-direction to be down):
We will use FFNET = maa (surprise)
First find FFNET in y direction:
FFNET = mg +T TT
mgg
yy

FFNET = mg +T
Acceleration in y direction:
maa = mv2
/ R
mg + T = mv2
/ R
T = mv2
/ R - mg
R
TT
v
mgg
yy
F = ma

What is the minimum speed of the mass at the top of the
trajectory such that the string does not go limp?
i.e. find v such that T = 0.
mv2
/ R = mg + T
v2
/ R = g
Notice that this does
not depend on m.
R
mgg
v
T= 0
v Rg=
Bucket

Lecture 6Lecture 6,, Act 3Act 3
Motion in a CircleMotion in a Circle
A skier of mass m goes over a mogul having a radius of
curvature R. How fast can she go without leaving the
ground?
R
mgg NN
vv
(a) (b) (c) Rgv =mRgv =
m
Rg
v =
Track
w/ bump

Lecture 6Lecture 6,, Act 3Act 3
SolutionSolution
mv2
/ R = mg - N
For N = 0:
R
v
mgg NN
v Rg=

Recap of Today’s lecture:Recap of Today’s lecture:
The Free Body DiagramFree Body Diagram
The tools we have for making & solving problems:
» Ropes & Pulleys (tension)
» Hooke’s Law (springs)
Accelerometer
Inclined plane
Motion in a circle

7 problems of newton law

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7 problems of newton law