Dynamics I
Motion Along a
Line
Static Equilibrium
• Example: consider the situation below where two
ropes hold up a weight:
qleft =30o
Tleft
qright = 55o
Tright
W = 100 Nt
Dynamic Equilibrium
An elevator that is moving upward at 2 m/s. Find
the tension T in the cable, if elevator has mass 400
kg
Ramps
Why is it easier to push something up a ramp
than it is to lift it?
P
Ramps
F// = P - mg sin(q) = mg//
F = FN - mg cos(q) = mg = 0
FN
P
F//
Pbalance = mg sin(q)
q
q
F
W=mg
Ramps
F// = P – F//- Ff = ma//
FN
P
Ff =Fc
F//
W=mg
Ramps
FN = mg cos(q)
P = mg sin(q) +  mg cos(q)
= mg[sin(q) +  cos(q)]
FN
Ff =FN
P
W=mg
sin(q) +  cos(q) < 1 → P < mg
Pulleys
Pulleys
P
W = mg
P = Tension = W
Pulleys
P
T
T
W
P=T, 2T = W; so P=W/2
Pulleys
P
Pulleys
P
W
Newton’s law of gravitation
1
Fg  2
r
Fg  m1m 2
Gm1m 2
Fg 
2
r
G = 6.67 x 10-11 Nm2/kg2
Problem 1
• Three masses are each at a vertex of an
isosceles right triangle as shown. Write
an expression for the force on mass three
due to the other two.
m1
Gm3  m1  m2 
Fn 
r
m2
r
r
m3
Gravity at earth’s surface
Gm E m 2
FG 
2
r
Gm E m
mg 
2
r
Gm E
g 2
r
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
Fg = mg
Test your Understanding
• 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...
Ouch!
(a)
more
(b)
less
(c)
the same
Test your Understanding
• However the weights of
the bowling ball and the
astronaut are less:
W = mgMoon gMoon < gEarth
• Thus it would be easier for
the astronaut to pick up
the bowling ball on the
Moon than on the Earth.
Friction
(force on
box due to
table)
f
Acme
Hand
Grenades
f
v
(force on table due to box)
The forces shown are an action-reaction pair.
Friction...
• Friction is caused by the “microscopic”
interactions between the two surfaces:
Two Kinds of Friction
• Static friction
– Must be overcome in order to
budge an object
– Present only when there is no
relative motion between the
bodies, e.g., the box & table
top
• Kinetic friction
– Weaker than static friction
– Present only when objects
are moving with respect to
each other (skidding)
fs
FA
Objects are still or
moving together.
Fnet = 0.
fk
FA
Fnet is to the right.
a is to the right.
v is left or right.
Friction Facts
• Friction is due to electrostatic attraction between the
atoms of the objects in contact.
• It can speed you up, slow you down, or make you turn.
• It allows you to walk, turn a corner on your bike, warm
your hands in the winter, and see a meteor shower.
• Friction often creates waste heat.
• It makes you push harder / longer to attain a given
acceleration.
• Like any force, it always has an action-reaction
pair.
Friction Strength
The magnitude of the friction force is proportional to:
• how hard the two bodies are pressed together (the
normal force, N ).
• the materials from which the bodies are made (the
coefficient of friction,  ).
Attributes that have little or no effect:
• sliding speed
• contact area
Coefficients of Friction
• Static coefficient … s.
• Kinetic coefficient … k.
• Both depend on the materials in contact.
– Small for steel on ice or scrambled egg
on Teflon frying pan
– Large for rubber on concrete or
cardboard box on carpeting
• The bigger the coefficient of friction, the
bigger the frictional force.
Friction...
Force of friction acts to oppose motion:
Parallel to surface.
Perpendicular to Normal force.
j
N
F
i
ma
fF
mg
Model for Sliding Friction
Ff  N
Ff  N
The “heavier” something is, the greater
the friction will be...makes sense!
Dynamics
i:
j:
F  KN = ma
N = mg
so
F  Kmg = ma
j
N
ma
F
K mg
mg
i
Test your Understanding
• A box of mass m1 = 1.5 kg is being pulled by a
horizontal string having tension T = 90 N. It
slides with friction (k = 0.51) on top of a second
box having mass m2 = 3 kg, which in turn slides
on a frictionless floor.
– What is the acceleration of the second box ?
(a) a = 0 m/s2 (b) a = 2.5 m/s2 (c) a = 3.0 m/s2
T
m1
slides with friction (k=0.51 )
a=?
m2
slides without friction
Solution
• First draw FBD of the top box:
N1
T
m1
m1g
f = KN1 = Km1g
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.
m1
f2,1
m2
f1,2= Km1g
Solution
• Now consider the FBD of box 2:
N2
f2,1 = km1g
m2
m1g
m2g
Solution
Finally, solve F = ma in the horizontal direction:
Km1g = m2a
1.5kg
m1
a
k g 
 0.51 9.81m
m2
3kg
a = 2.5 m/s2
f2,1 = Km1g
m2
s
2
Inclined Plane with Friction:
• Draw free-body diagram:
ma
KN
j
N q
mg
q
i
Inclined plane...
Consider i and j components of
FNET = ma :
i
mg sin qKN = ma
j
N = mg cos q
 KN
ma
j
mg sin qKmg cos q = ma
N q
mg
i
mg sin q
q
mg cos q
a / g = sin qKcos q
Static Friction...
• So far we have considered friction acting when
something moves.
– We also know that it acts in un-moving “static”
systems:
• In these cases, the force provided by friction will
depend on the forces applied on the system.
N
F
j
i
fF
mg
Static Friction...
• Just like in the sliding case except a = 0.
F fF = 0
N = mg
i:
j:

While the block is static: fF F
N
F
j
i
fF
mg
Static Friction...
N
F
j
i
fF
mg
Static Friction...
S is discovered by increasing F until the
block starts to slide:
i : FMAX SN = 0
j : N = mg
S  FMAX / mg
N
FMAX
Smg
j
i
mg
Test Your Understanding
• A box of mass m =10.21 kg is at rest on a floor.
The coefficient of static friction between the
floor and the box is s = 0.4.
• A rope is attached to the box and pulled at an
angle of q = 30o above horizontal with tension T
= 40 N.
– Does the box move?
(a) yes (b) no (c) too close to call
T
static friction (s = 0.4 )
m
q
Solution
Pick axes & draw FBD of box:
Apply FNET = ma
y
y: N + T sin q - mg = maY = 0
N = mg - T sin q = 80 N
x: T cos q - fFR = maX
The box will move
if T cos q - fFR > 0
N
x
T
fFR
q
m
mg
Solution
y
y: N = 80 N
x: T cos q - fFR = maX
The box will move
if T cos q - fFR > 0
T cos q = 34.6 N
fMAX = sN = (.4)(80N) = 32 N
N
x
T
fMAX = sN
q
m
mg
So T cos q > fMAX and the box does move
Static Friction:
• We can also consider S on an inclined plane.
q
• In this case, the force provided by friction will
depend on the angle q of the plane.
Static Friction...
• The force provided by friction, fF , depends on q.
ma = 0
(block is not moving)
fF
mg sin qff 
(Newton’s 2nd Law along x-axis)
j
N
q
mg
q
i
Static Friction...
• We can find s by increasing the ramp
angle until the block slides:
mg sin qff
In this case:
ffSN  Smg cos qM
 SN
mg sin qMSmg cos qM
j
N
qM
mg
q
i
Stan qM
Friction Facts
• Since fF = N , the force of friction does
not depend on the area of the surfaces
in contact.
• By definition, it must be true that
S > K
for any system (think about it...).
Friction vs Applied force
fF = SN
fF = KN
fF
fF = FA
FA
Problem: Box on Truck
A box with mass m sits in the back of a
truck. The coefficient of static friction
between the box and the truck is S.
What is the maximum acceleration a that
the truck can have without the box
slipping?
m
a
S
Problem: Box on Truck
• Draw Free Body Diagram for box:
– Consider case where fF is max...
(i.e. if the acceleration were any
larger, the box would slip).
N
j
i
fF =  SN
mg
Problem: Box on Truck
Use FNET = ma for both i and j components
I : SN = maMAX
J: N = mg
aMAX = S g
N
j
aMAX
fF = SN
i
mg
Forces and Motion
• An inclined plane is accelerating with constant
acceleration a. A box resting on the plane is held in
place by static friction. What is the direction of the
static frictional force?
S
a
Ff
Ff
(a)
(b)
Ff
(c)
Solution
• First consider the case where the
inclined plane is not accelerating.
N
Ff
mg
N

All the forces add up to zero!
mg
Ff
Solution
If the inclined plane is accelerating, the normal force
decreases and the frictional force increases, but the
frictional force still points along the plane:
N
Ff
a
mg
N
All the forces add up to ma!
F = ma
The answer is (a)
mg
Ff
ma
Putting on the brakes
• Anti-lock brakes work by making sure the wheels roll
without slipping. This maximizes the frictional force
slowing the car since S > K .
• The driver of a car moving with speed vo slams on the
brakes. The coefficient of static friction between the
wheels and the road is S . What is the stopping
distance D?
ab
vo
v=0
D
Putting on the brakes
Use FNET = ma for both i and j components
i : SN = ma
j: N = mg
a = S g
N
j
a
fF = SN
mg
i
Putting on the brakes
As in the last example, find ab = Sg.
 Using the kinematic equation:
v2 - v02 = 2a( x -x0 )
 In our problem:
0 - v02 =  2ab( D )

ab
vo
v=0
D
Putting on the brakes
0 - v02 =  2ab( D )
In our problem:
 Solving for D:

v 02
D
2ab
Putting in ab = Sg
ab
vo
D
v 02
2sg
v=0
D
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 x direction. What is the acceleration of
the box?
y
F = Fx i
a =?
m
x
Solution
• Draw a picture showing all of the forces
y
FB,F
F
x
FF,B
FB,E
FE,B
Solution
• Draw a picture showing all of the forces.
• Isolate the forces acting on the block.
FB,F
F
y
x
FF,B
FB,E = mg
FE,B
Solution
• Draw a picture showing all of the
forces.
• Isolate the forces acting on the block.
• Draw a free body diagram.
FB,F
y
F
x
mg
Solution
•
•
•
•
Draw a picture showing all of the forces.
Isolate the forces acting on the block.
Draw a free body diagram.
Solve Newton’s equations for each component.
– FX = maX
– FB,F - mg = maY
y
FB,F
F
x
mg
Solution
•
FX = maX
– So aX = FX / m = (10 N)/(2 kg) = 5 m/s2.
• FB,F - mg = maY
– But aY = 0
– So FB,F = mg.
FX
N
y
x
m
g
• The vertical component of the force of the floor on
the object (FB,F ) is called the Normal Force (N).
• Since aY = 0 , N = mg in this case.
Recap
N = mg
y
FX
aX = FX / m
x
mg
Normal 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?
(a) N > mg
(b) N = mg
a
(c) N < mg
m
Solution
All forces are acting in the y direction, so use:
N
Ftotal = ma
a
m
N - mg = ma
N = ma + mg
therefore N > mg
mg
Tools: Ropes & Strings
• Can be used to pull from a distance.
• Tension (T) at a certain position in a rope is the
magnitude of the force acting across a crosssection of the rope at that position.
– The force you would feel if you cut the rope and
grabbed the ends.
– An action-reaction pair.
T
cut
T
T
Tools: Ropes & Strings...
• Consider a horizontal segment of rope having
mass m:
– Draw a free-body diagram (ignore gravity).
m
T1
a
T2
x
• Using Newton’s 2nd law (in x direction):
FNET = T2 - T1 = ma
• So if m = 0 (i.e. the rope is light) then T1 = T2
Tools: Ropes & Strings...
• An ideal (massless) rope has constant tension along
the rope.
T
T
• If a rope has mass, the tension can vary along the
rope
– For example, a heavy rope
T = Tg
hanging from the ceiling...
T=0
• We will deal mostly with ideal massless ropes.
Tools: Ropes & Strings...
• The direction of the force provided by a
rope is along the direction of the rope:
T
Since ay = 0 (box not moving),
m
mg
T = mg
Force 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?
snap !
(a) 14.8 kg
(b) 18.4 kg
a = 12.2 m/s2
(c)
m=?
8.2 kg
Solution:
• Draw a Free Body Diagram!!

Use Newton’s 2nd law
in the upward direction:
T
a = 12.2 m/s2
m=?
FTOT = ma
T - mg = ma
T = ma + mg = m(g+a)
mg
T
m=
g+a
180N
m=
= 8.2kg
2
 9.8 +12.2  m s
Tools: 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:
F1
| F1 | = | F2 |
ideal peg
or pulley
F2
Tools: 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:
FW,S = mg
T
m
mg
T = mg
Springs
• 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...
• 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 = -kx > 0
x0
x
Springs...
• 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 = - kx < 0
x>0
x
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.
1 lb = 4.45 N
0
2
4
6
8
Force 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?
?
m
m
(2)
(1)
(a)
0 lbs.
m
(b) 4 lbs.
(c)
8 lbs.
Solution:
• Draw a Free Body Diagram of one of the blocks!!

Use Newton’s 2nd Law in the y direction:
a = 0 since the blocks are stationary
T
FTOT = 0
m
T - mg = 0
T = mg = 4 lbs.
mg
T = mg
Solution:
• The scale reads the tension in the rope, which
is T = 4 lbs in both cases!
T
T
T
T
m
T
T
T
m
m
Car going around a turn
What are the forces that cause this circular
acceleration?
Gravity (weight) acts down
Contact Force acts up (perpendicular to the surface)
Friction - which way does it act (left or right)?
Fc
a=v2/r
W=mg
Car is heading into the screen.
Car going around a turn
Without friction the car will NOT make the turn, it
will continue straight - into the left ditch!
Therefore, friction by the road must push on the
car to the right. (Friction by the car on the road
will be opposite - to the left.)
The fastest the car can go around
the turn without sliding is when
Fc
Ff  Fc
the friction is maximum:
Ff = Fc .
a=v2/r
W=mg
Car going around a turn
We now apply Newton’s Second Law - in
rectangular components:
Fx = Ff = ma = mv2/r
To make the car go around the turn fastest, we need
the maximum force of friction: Ff = Fc
Fy = Fc - W = 0
which says Fc = mg, so
mg = mv2/r, or vmax = [gr]
Fc
Ff Fc
a=v2/r
W=mg
Car going around a turn
vmax = [gr]
Note that the maximum speed (without slipping)
around a turn depends on the coefficient of
friction, the amount of gravity (not usually under
our control), and the sharpness of the turn
(radius).
If we go at a slower speed around the turn, friction
will be less than the maximum: Ff < Fc.
There is one other thing we can do to go faster
around the turn - bank the road! How does this
work?
Banked turn
By banking the road, we have not added any forces,
but we have changed the directions of both the
contact force and the friction force!
Have we changed the direction of the acceleration? No
- the car is still travelling in a horizontal circle.
Fc
a=v2/r
q
W=mg
Ff
Banked turn
Since the acceleration is still in the x direction, we
will again use x and y components (rather than //
and 
Fx = Fc sin(q) + Ff cos(q) = mv2/r
Fy = Fc cos(q) - Ff sin(q) - mg = 0
If we are looking for the max
speed, we will need the max
Fc
friction: Ff = Fc .
This gives 3 equations for
a=v2/r
3 unknowns: Ff, Fc, and v.
q
W=mg
Ff
Banked turn
Fx = Fc sin(q) + Ff cos(q) = mv2/r
Fy = Fc cos(q) - Ff sin(q) - mg = 0
Ff = Fc . Using the third equation, we can eliminate Ff
in the first two:
Fc sin(q) + Fc cos(q) = mv2/r
Fc cos(q) - Fc sin(q) - mg = 0
We can now use the second equation to find Fc:
Fc = mg / [cos(q) -  sin(q)], and use this in the first
equation to get:
v = [gr {sin(q)+ cos(q)} / {cos(q) -  sin(q)} ]1/2
Banked Turn
v = [g r {sin(q)+ cos(q)} / {cos(q) -  sin(q)} ]1/2
Notice that the mass cancels out. This means that
the mass of the car does not matter! (Big heavy
trucks slip on slippery streets just like small cars.
When going fast, big heavy trucks flip over rather
than slide off the road; little cars don’t flip over
like big trucks. But flipping over is not the same
as slipping! We’ll look at flipping in Part 4 of the
course.)
Note also that when q0, the above expression
reduces to the one we had for a flat road:
vmax = [gr]1/2 .
Banked Turn
v = [g r {sin(q)+ cos(q)} / {cos(q) -  sin(q)} ]1/2
Note that as  sin(q) approaches cos(q), the
denominator approaches zero, so the maximum
speed approaches infinity!
What force really supports such large speeds (and
so large accelerations)? As the angle increases,
the contact force begins to act more and more to
cause the acceleration. And as the contact force
increases, so does friction.
Actually, there is a limit on the maximum speed
because there is a limit to the contact force.
Banked turn - Minimum Speed
Is there a minimum speed for going around a
banked turn? Consider the case where the
coefficient of friction is small and the angle of
bank is large. In that case the car, if going too
slow, will tend to slide down (to the right) so
friction should act to the left.
Can you get an equation for the minimum speed
Fc
necessary?
Ff
What changes in what we did for
max speed?
a=v2/r
q
W=mg