Ch 5 – Applications of Newton’s Laws
In this chapter, we only study sections 5.1 and 5.2. We will have a combined
shorter test on Ch 5 with some Ch 4 material, as well.
5.1 Friction
Do you remember the following notes from 10th grade?
4.2 Friction
DEF: Friction = a counter-active force that acts:
• In response to force
• In the opposite direction of the force
• Can’t be morr than the force applied
• Does not depend on speed
• Does not depend on area of contact
• Can happen in solids, liquids, and gases
2 Types of Friction
A. Static friction is
• the friction that prevents something from moving (Fs)
• equal in size and opposite in direction to the force applied (Fs = - F)
• increases / decreases in direct proportion to F
• Has a maximum value, Fs,max above which the object will begin to
accelerate
B. Sliding (Kinetic) Friction is
• Fk
• Less than static friction
• Still a slowing force
• Is proportional to the normal force on the object
• Once an object is in motion, Fnet = Fapplied - Fk
DEF: Cold welding = molecular forces between a stationary object and the
surface it’s resting on.
Because of cold welding, Fk is always less than Fs
DEF: Coefficient of Friction is
• Denoted by µ
• the ratio of friction to the normal force acting between two objects
Coefficient of Static Friction = µs = Fs, max
Fn
Coefficient of Kinetic Friction = µk = _Fk_
Fn
Your text lists the formulas this way:
What this table tells you is the
A. coefficients of static friction (the ratio of the maximum force that can be
applied for the object to remain still to the normal force)
B. coefficients of kinetic friction (the ratio of the minimum force needed to
cause acceleration of the object to the normal force)
Friction doesn’t just happen between solids, it can occur in liquids and gases,
as well. Because liquids and gases both flow, they are called fluids.
Without friction, we couldn’t walk, drive a car, play sports, et cetera.
There is one more type of coefficient of friction to consider:
Typical values of µr range from 0.01 – 0.02 for rubber tires on concrete and
from 0.001 – 0.002 for steel wheels on steel rails.
Q: How do these values for rubber on concrete and steel on steel compare to
the others in the table? Does that make sense?
A: They are substantially less. It will in a moment if it doesn’t already! ☺
Q: Why do these coefficient values have no units?
A: They are derived from a ratio of two forces so the units cancel out.
Let’s talk about the molecular source of friction for a moment.
This is a magnified section of highly
polished steel. You can see that even for
something like polished steel, there are
still a considerable number of crevices and
resultant spaces. If you have steel-onsteel, the surfaces will still only make
contact at those high points, or asperities.
So, the cold fusion we discussed earlier happens at those asperities. Cold
fusion is simply the forces of attraction that exist between the atoms that come
in contact with each other at those asperities. Cold fusion bonding is only valid
within a range of a few atoms’ diameters. The less surface area you have (thus,
the fewer asperities that are in contact with one another), the less friction you
have. That’s why it’s easier to move things that are on wheels. There is much
less contact between surface areas, and procession of motion is facilitated by
the fact that the object will roll automatically.
Q: What do you think would happen if you continue to scrape two surfaces
like these together?
A: As the normal forces increase respectively (the forces they exert on each
other), the steel slabs flatten out morr.
Q: What would this resultant flattening do to the overall friction?
A: Increase it, of course!
Let’s take a look at the graph below.
This graph shows the plot of frictional
force on a box by a floor as a function of
the applied force. The frictional force
balances the applied force until the point
at which the box is jarred out of
equilibrium and just starts to move. Then,
the frictional force is constant.
Q: Why is there a decrease in the graph at the point at which it starts to
move?
A: The bonds of cold fusion are broken. Once something is moving, it’s easier
to keep it moving.
See Ex 5-1, p 120
A cruise ship passenger uses a shuffleboard cue to push a shuffleboard disk of
mass 0,40 kg horizontally along the deck so that the disk leaves the cue with a
speed of 5.5 m/s. The disk then slides a distance of 8m before coming to rest.
Find the coefficient of kinetic friction between the disk and the deck.
∑Fx: - Fk = ma
Fk = µkFn
= µkmg
∴ - Fk = - µkmg
Also, a = F/m ∴
a = - µkmg
m
a = - µkg
Note: The m’s end up canceling here.
Q: What does that really mean?
A: A greater mass is harder to stop, but greater mass also means greater
friction. The net effect is that the mass has no effect! ☺
Now, use Ch 2 kinematics to solve for a in terms of vf and vi.
vf2 = vi2 + 2a∆
∆x
0 = (5.5 m/s)2 + 2(- µk)(-9.81 m/s2)(8 m)
- (5.5 m/s)2
= µk
2
2(-9.81 m/s )(8 m)
µk = 0.19
See Ex 5-2, p 121
A hardcover book is resting on a table top with its front cover facing upward.
You place a coin on this cover and very slowly open the book until the coin
starts to slide. The angle θmax is the angle the front cover makes with the
horizontal just as the coin starts to slide. Find the coefficient of static friction
between the book cover and the coin in terms of θmax.
∑Fy: Fn – mg cosθ
θ = m⋅⋅0
Fn = mg cosθ
θ
Fs = µsFn
∴
Fs = µsmg cosθ
θ
∑Fx: Fs – mg sinθ
θ = m⋅⋅0
Fs = mg sinθ
θ
Since both of these are equal to Fs, set them equal
to each other.
µsmg cosθ
θ = mg sinθ
θ
Because the question is asking you about θmax,
you’ve got to express this in terms of just one θ.
Remember, tan = sin/cos, ∴ divide both sides by
mg cosθ
θ.
µsmg cosθ
θ = mg sinθ
θ
mg cosθ
θ
mg cosθ
θ
µs = tan θ
KNOW this formula!!!! ☺
So, θmax = the angle of repose = the angle at which the object just begins to slip!
See Ex 5-3, p 122
Two children sitting at rest in the snow ask you to pull them. You pull the
sled’s rope by making an angle of 400 with the horizontal. The children have a
combined mass of 45 kg and the sled has a mass of 5 kg. The coefficients of
static and kinetic friction are: µs = 0.2 and , µk = 0.15. Find the frictional force
exerted by the ground on the sled, and the acceleration of the children and the
sled, starting from rest, if the tension in the rope is a) 100 N b) 140 N
What they are really asking is if these tensions are enough to accelerate these
children. ☺
Q: Why is Fn shorter than mg?
A: It will be compensated for by the y-component of T.
ΣFy:
ΣFx:
Fn – mg + T sinθ
θ = m••0
Fn = mg - T sinθ
θ
Fs = µsFn = µs(mg - T sinθ
θ)
- Fs + T cosθ
θ = m••0
Fs = T cosθ
θ
Since these are both equal to Fs, set them equal to each
other.
µs(mg - T sinθ
θ) = T cosθ
θ
µsmg - µsT sinθ
θ = T cosθ
θ
µsmg = µsT sinθ
θ + T cosθ
θ
µsmg = T(µ
µs sinθ
θ + cosθ
θ)
µsmg
= T(µ
µs sinθ
θ + cosθ
θ)
(µ
µs sinθ
θ + cosθ
θ)
(µ
µs sinθ
θ + cosθ
θ)
µsmg
=T
(µ
µs sinθ
θ + cosθ
θ)
T = 110 N
Since part “a” only gave you 100 N, this is not
Enough to accelerate the kid-sled system
∴a=0
Fs at this point = T cosθ
θ
= 100 N (cos 400) = 76.6 N
To find Fk we use Fk = µkFn = µk(mg - T sinθ
θ) = 0.15[(50 kg)(9.81 N/kg) – (140 N)(sin 400)]
= 60.1 N
ΣFx: - Fk + T cosθ
θ = m••ax ∴
ax = - Fk + T cosθ
θ = (- 60.1N + (140 N)( cos 400))
m
50 kg
ax = 0.943 m/s2
See Ex 5-5, p 125
A runaway baby buggy is sliding without friction across a frozen pond toward
a hole in the ice. (Are you kidding me?! There better not be a baby in there!
☺). You race after the buggy on skates. As you grab it, you and the buggy are
moving toward the hole with velocity v0. The coefficient of friction between
your skates and the ice is given by µk. D is the distance to the hole when you
reach the buggy, M is the total mass of the buggy, and m is your mass.
a) What is the lowest value of D such that you stop the buggy before it reaches
the hole?
b) What force do you exert on the buggy?
a)
ΣFy: Fn1 – mg = m⋅⋅0
Fn1 = mg
Fk = µkFn1
Fk = µkmg
ΣFx: -F = Max ∴
- Max - µkmg = max
∴ ax = - µkg___
1 + M/m
(a is negative, as you’d expect!)
ΣFx: F – Fk = max
F - µkmg = max
Now that you have “a” you can use Ch 2 Kinematics to solve for D
vf2 = v02 + 2ax∆x
0 = v02 + 2axD
2
D=
− v0
2a x
2
=
 M  v0
1 + 
m  2µ k g

b) F = -Max =
µ k Mg
1+
M
m
Let’s shift over to talking about the forces on a car (n’yuk, n’yuk…).
Q: What balances / cancels the weight vector for a car?
A: The normal force on each tire.
Q: If a road were frictionless, how would that affect the
motion of a car?
A: Your wheels would spin around and around, but you
wouldn’t be able to move!
For forward motion of a vehicle: When friction is present, the frictional force
of the road on the tires is in the forward direction, opposing the tendency of the
tire to slip backwards. It is that frictional force that provides the acceleration
necessary to propel the car forward. If the friction between the tire and the
road is small, the tire does not “slip” on the road. The wheels roll without
slipping and the tread touching the road at any given instant is at rest relative
to the road.
Q: What type of friction is this, then?
A: Static.
The largest frictional force that the tire can exert on the road (and the road
can exert on the tire) is: Fs, max = µsFn
For a car moving in a straight line with speed v relative to the road, the center
of each wheel also moves with speed v as shown. In the picture, the dashed
lines represent velocities relative to the body of the car while the solid lines
represent velocities relative to the ground.
If a wheel rolls without slipping, its top is moving faster than v
while its bottom is moving slower than v. That’s because
there is NO friction on the top of the wheel, but there IS
friction on the bottom. However, relative to the car, each
point on the perimeter of the wheel moves with the same
speed v.
At that point, the speed of the point on the tire momentarily
in contact with the ground is zero.
zero If it weren’t zero, the tire
would be skidding along the ground!
Given more power from the engine, the tire will slip and the wheels will spin.
At that point, the force that accelerates the car is the force of kinetic friction.
Q: What did this graph tell you about Fk
versus Fs?
A: Fk < Fs, max
So, if you’re stuck on ice or snow, you’d be better off using a light touch of the
accelerator as you’d potentially have FS,max versus Fk working for you.
Q: Is that true of braking a car, as well? In other words, would you be better
off using a light touch of the brakes for stopping, or slamming on the brakes?
A: Within reason, the lightest force you can apply to the brakes to get them to
stop on time is best. That’s what anti-lock braking systems (ABS) do!
If you slam on the brakes so that the tires skid on the road, the force working
to stop the car is Fk, the force of kinetic friction. Before ABS systems, one
would “pump” the brakes in effort to stop. The problem is that the pumping
of the brakes is applied to all of them. In ABS systems, the pumping action is
only applied to the locking wheel. This is called threshold braking. This type
of braking maximizes quick stopping ability. Pretty cool, huh?!
See Ex 5-7, p 128.
A car is traveling at 30 m/s along a horizontal road where µs = 0.5 and µk = 0.3.
How far does the car travel before stopping if
a) the car brakes using an ABS system so that threshold braking is sustained?
b) the car is braked hard with no ABS so the wheels lock?
a) Use Ch 2 kinematics to solve for ∆x
vf2 = v02 + 2ax∆x
0 = v02 + 2ax∆x
∆x = - v02
2ax
Now solve for ax
ΣFy: Fn – mg = m⋅⋅0
Fn = mg
Fs, max = µsFn = µs mg
ΣFx: - Fs, max = max
- µs mg = max
ax = - µsg
∆x = - v02 = - (30 m/s)2 = 91.8 m
2ax 2(- 0.5)(9.81)
b) When the wheels lock, ax is given by: ax = - µsg as indicated above.
∆x = - v02 = - (30 m/s)2 = 153 m
2ax
2(- 0.3)(9.81)
Note: The stopping distance is more than 50% greater when the wheels are
locked. Also note that the stopping distance is independent of mass, providing
the coefficients of friction are the same.
Do you remember covering this in 10th grade? We said that when the speed
was tripled, the stopping distance required was nine times morr. This was
based on the equation for KE. We know KE = ½ mv2. What we determined
was that in the ratio below, the m’s ended up canceling out.
KE1 = ½ m1v12
KE2 ½ m2v22
Same is true here; m’s end up canceling out! ☺
5.2 Motion Along a Curved Path
Q: What is the formula for centripetal acceleration that we learned in Ch 2?
A: ac = v2 / r where the centripetal acceleration was always toward the center
of the circle.
The tangential component of ac given by at is:
at = dv/dt in the tangential direction. We learned in 10th grade that the
centripetal force is a center-seeking force and that the velocity is always
tangent to the circle at any given instant. The evidence is that if you cut the
string (or source of the force) the object flies off tangent to the circle at that
point. There is no real centrifugal force (outward-seeking force), or the object
would fly outward, not tangent.
See Ex 5-8, p 129.
You swing a pail of water in a vertical circle of radius r. If its speed is vt at the
top of the circle, find
a) the force exerted on the water by the pail at the top of the circle
b) the minimum value of vt for the water to remain in the pail
c) the force exerted by the pail on the water at the bottom of the circle,
where the speed of the pail is vb
Water at top
Water at bottom
a) ΣFy, top: - Fp – mg = m (-vt2 / r)
 vt 2

Fp = m 
− g
 r



The pail can push on the water, but not pull on it.
The water is there via its own initial inertial state.
The minimum force it can exert is zero.
b)
 v t , min 2

0=m 
− g
 r



vt, min = rg
c) ΣFy, bottom: Fp – mg = m (vb2 / r)
 vb 2

Fp = m 
+ g
 r



Solve for vt, min
Q: Why is there no arrow for the centripetal force in the FBD’s?
A: It is a resultant force in the centripetal direction. Those never go in the
FBD, regularly. Think of the ball rolling down the ramp!
Q: Is the centripetal force still real, if it’s not in the FBD?
A: Yep. See the paragraph on the previous page for a reminder.
A couple of extra notes:
When the bucket is at the top of its path, both the bucket and g contribute to
the centripetal force on the water. At that peak, the water is moving with
minimum speed, and is in free-fall. At the bottom of the path, Fp has to be
greater than mg to still provide a net force in the centripetal direction, or up.
Q: At the bottom, when v = 0 again, what is Fp?
A: Fp, of course!
When a car rounds a curve on a horizontal road, both the centripetal and
forward forces are provided by the force of static friction exerted by the road
on the tires.
If the car travels at constant speed, then the forward component of the
frictional force is balanced by the rearward forces of air drag and rolling
friction.
Q: If air drag is negligible, what does that mean about the forward component
of frictional force in this case?
A: It is zero.
See Ex 5-10, p 131
You have a summer job as part of an automobile tire design team. You are
testing new prototype tires to see whether or not the tires perform as well as
predicted. In a skid test, a new BMW 530i was able to travel at constant speed
in a circle of radius 45.7 m in 15.2 s without skidding.
a) What was the speed, v?
b) What is the acceleration?
c) Assuming negligible air drag and rolling friction, what is the minimum
value for the coefficient of static friction between the tires and the road?
a) v = 2π
πr = 2π
π(45.7 m) = 18.9 m/s
t
15.2 s
This is just distance over time.
b) ac = v2 = (18.9 m/s)2 = 7.81 m/s2
r
(45.7 m)
at = dv = 0
dt
(Constant speed)
∴ the only acceleration is 7.81 m/s2 in the centripetal direction.
c) ΣFy: Fn – mg = m⋅⋅0
Fn = mg
Fs, max = µsFn = µsmg
ΣFr: - Fs, max = m
 − v2

 r



∴ µsmg = mv2
r
µs = v2 =
(18.9 m/s)2____ = 0.796
rg (45.7 m)(9.81 m/s2)