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Physical Sciences 2
September 17 – 22
Physical Sciences 2: Assignment for September 17 – 22
Homework #3: Forces and Motion
Due Tuesday, September 29, at 9:30AM
After completing this homework, you should…
• Be able to identify any constraints placed on acceleration in a given scenario
• Be able to identify the forces acting on an object (gravity, normal, tension, etc.)
• Be able to draw and label a free-body diagram
• Be able to apply Newton’s second law in a variety of situations to calculate unknown quantities
• Understand the difference between static friction and kinetic friction, and when to use each
• Understand the difference between viscous drag and pressure drag, and when to use each
• Be able to calculate the terminal velocity of an object
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Physical Sciences 2
September 17 – 22
Here are summaries of this lecture’s important concepts to help you complete this homework:
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1. Iris and Javier (1 pt)
Iris and Javier are playing on a playground rather than studying for their physics exam…
Iris:
It’s your turn to sit on the merry-go-round, Javier!
Javier:
Iris, I see you haven’t been holding on as you sit on the merry-go-round. Well, I can do
that too – and I won’t slide off, as you just did. You see, since I’m heavier, the force of
static friction will be greater for me than it was for you. So, pfffft.
Iris:
Actually, Javy, you might want to check your physics notes again…
What’s wrong with Javier’s argument? Support your answer with a clear diagram and a derivation.
Lets think about the situation. You might know from your own experience with merry-go-rounds that
when you sit on the merry-go-round going at a slow speed, you don’t slide off, but if the speed is fast
enough you will slide off. So there is a maximum speed above which the static friction between you and
the floor of the merry-go-round is no longer enough to keep you moving in a circle, and you have to
hold on to something to stay on. Javier claims that this speed depends on mass; that he, with a higher
mass, will not slide off riding at the same speed as Iris. Is this true? We can use FANCLAN to find out.
The free-body diagram (F) represents the forces acting on Javier, as seen from the side when he is at the
right, like in the picture above. The forces acting on Javier are: the force of the Earth’s gravity, pointing
down towards the center of the Earth (Fgrav), the force of the merry-go-round, pushing him up (FN), and
the force of static friction, keeping him from sliding off, pointing towards the center of the circular
trajectory (Ffs).
We will use the usual axes (A), with the positive x-direction pointing to the right, and the positive ydirection pointing up. Newton’s second law (N), written in components, is then:
⌃Fx = max
Ffs = max
⌃Fy = may
FN Fgrav = may
Now for the constraints (C) on the motion: We can assume that Javier is moving in a circle at constant
speed, so there will be a center-pointing acceleration, in the negative x-direction equal to v2/R, where v is
the speed of Javier and R is the radius of the circular trajectory. We can also assume that Javier’s
acceleration in the y-direction is zero (he doesn’t jump up and down while riding the merry-go-round).
The force laws (L) relevant in this situation are: Fgrav = mg and Ffs ≤ µs FN. Putting it all together, and
doing a little algebra (A), we can find the maximum speed before Javier slides off the merry-go-round:
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Physical Sciences 2
FN
FN
Fgrav = may
mg = 0
FN = mg
September 17 – 22
Ffs = max
µs.max FN =
2
mvmax
/R
2
µs.max mg = mvmax
/R
2
vmax
= Rµs.max g
The maximum speed at which Javier can go without sliding off the merry-go-round is independent of
mass! This means that, like Iris, he will have to hold on to something when going as fast as she did.
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2. Rats! (2 pts)
Coal miners often find mice in deep mines but rarely find rats; let’s see if we can figure out why. A
mouse is roughly 5 cm long by 2 cm wide and has a mass of 30 g; a rat is roughly 20 cm long by 5 cm
wide and has a mass of 500 g. Assume that both have a drag coefficient CD ≈ 0.3.
a) Estimate the terminal falling speeds reached by a mouse and a rat, respectively.
The drag force is dominated by pressure drag. As we saw in lecture, in this case the terminal speed is
given by
2mg
.
vterminal =
CD ρair A
Plugging in the numbers given (and the density of air equal to 1 kg/m3), we get vterminal = 44.3 m/s for
the mouse and 57.2 m/s for the rat. These aren’t so different—the mouse has a smaller mass, but also a
much smaller area.
b) Assume that mine shafts are deep enough that both the mouse and the rat reach terminal velocity
before hitting the bottom. Estimate the magnitude of the maximum force required to stop both the rat
and the mouse when they hit the bottom. (Hint: Over what distance will the center of mass travel
between the beginning of the collision and when the animal is at rest? What is the acceleration
required to bring the animal to rest over this distance?)
In order to estimate the force needed to stop the falling rodent, we need to guess the stopping distance
for the mouse and the rat. A reasonable rough estimate is the length of their legs, which might be about 2
cm for the mouse and 4 cm for the rat. Then assuming a constant force acts to bring the rodent to a stop
over that distance (which we’ll call d), the net force must have a magnitude of
2
vy2 − v02y
0 − vterminal
m2 g
∑ Fy = may = m 2Δy = m 2(−d) = C ρ Ad .
D air
For the mouse, we get a force of magnitude 1.47 × 103 N; for the rat, it’s a much larger 2.04 × 104 N.
c) Bones will break if they are subjected to a compressional force per unit area of more than about
1.5 × 108 N/m2. A mouse may have leg bones about 1.5 mm in diameter; for a rat, they might be
about twice as thick. Using your estimates from part b), determine if either the mouse or the rat (or
both) will suffer broken legs. Is your answer consistent with the observations of the coal miners?
Explain.
The mouse’s leg bone has a cross-sectional area of π(0.75 mm)2 = 1.8 × 10–6 m2. For the rat, the area is
four times this, or about 7.1 × 10–6 m2. Now each rodent has four legs, so we multiply this area by four
to give the total area of leg bone that must bear the force we calculated in the previous part. Then the
force per unit area of leg bone is:
mouse: 2.04 × 108 N/m2
rat: 7.18 × 108 N/m2
So using our estimates, the mouse is a bit over the limit of bone-breaking, but might still survive. The rat
is way over the limit, so it will likely break its legs. Once it has broken legs, it won’t live long because it
can’t forage for food.
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3. Post-lab Assignment for Lab 2 (2 pts)
Last week in lab, you investigated conservation of momentum in the Gauss Gun system. Some groups
indicated that momentum was conserved, or close to it, while some groups indicated that momentum
was not conserved. The vast majority of these found that there was a change in momentum in the same
direction. (To remind yourself what your own result was, go to the “Laboratory” section of the course
website.)
Let’s see if we can understand why this happened. First, we’ll review our definitions:
In the initial state, ball 1 is rolling towards the magnet with some initial x-velocity v1i,x just before the
collision. We took the final state to be the time of the first video frame after the collision (not the time of
the collision itself). In the final state, the recoiling 1-M-2 system has a (negative) x-velocity v1f,x , and the
outgoing ball has a (large positive) x-velocity v3f,x .
a) The majority of groups observed that the final state had a larger x-component of total momentum
than the initial state, i.e. pf,x − pi,x > 0 . What can you conclude about the net external force on the
system during the time between the initial (just before collision) and final (first video frame after
collision) states? Draw the free-body diagram for the system during this time. What single force do
you think is mostly responsible?
If the system’s total momentum changes, there must have been a net
external force on it during the time in question. In particular, if the
change is to the right (towards +x), there must also have been a net
external force on the system to the right.
The free-body diagram is shown at right. There may also be other,
smaller friction forces (rolling friction on ball 1 or ball 3, which
would point to the left), but the big one has got to be sliding (kinetic)
friction on the recoiling 1-M-2 system. Since it slides to the left, the
friction force points to the right.
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b) Below is a graph taken from a Logger Pro video analysis, showing the incoming motion of ball 1
(blue) and the recoil of 1-M-2 (green), with the appropriate linear and quadratic fits displayed.
Using this graph, calculate µk , the coefficient of kinetic friction between 1-M-2 and the track.
(Don’t worry about uncertainty.)
The position-vs-time graph for the recoil of 1-M-2 fits a quadratic, which corresponds to constant
acceleration. This agrees with our model that kinetic friction is responsible. We’ve already drawn the
free-body diagram (F), and we’ll use the same coordinate system (A) that we used in lab (x right, y up).
Writing down Newton’s 2nd Law in component form (N), we get:
max = ∑ Fx = Fkf
may = ∑ Fy = FN − Fg
The motion is constrained (C) to be horizontal, which means that there is no y-acceleration. We know
the force laws (L) for both gravity and kinetic friction:
Fg = mg
Fkf = µk FN
After some algebra (A), we can find the x-acceleration:
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ax =
Fkf µk FN µk ( mg )
=
=
= µk g .
m
m
m
The quadratic fit for constant acceleration should obey the equation
1
x ( t ) = x0 + v0 x t + ax t 2
2
or in the slightly different form used in the fit,
x (t ) =
1
2
a x ( t − t f ) + xf
2
In either form, the coefficient of the squared term is
1
ax . In the fit, this is labeled as the parameter A.
2
Putting in the numbers (N), we get:
A = 93.43 cm/s 2 =
1
1
ax = µ k g
2
2
So we can solve for the kinetic friction coefficient:
(
)
2
2A 2 93.43 cm/s
µk =
=
= 0.19 .
g
980 cm/s 2
There are other ways to calculate this (e.g. determine Δx and Δt from the graph and use them to
calculate ax), but this one is the most straightforward.
c) Recall that for a non-isolated system, the change in the system’s total momentum during some time
is equal to the total external impulse on the system during that time. We can’t calculate the total
impulse, because we don’t know how much time elapsed between the collision and the final state—
but we can put an upper bound on it, because we know that the time was at most one video frame
(33 milliseconds). For the experiment graphed above, calculate the maximum possible x-impulse
Δpx due to kinetic friction. You may need to know the masses: the mass of one ball was 8.36 g, and
the mass of 1-M-2 together was 22.76 g. The results of the experiment at the time were
pi,x = ( 268 ± 8 ) g ⋅ cm/s , and pf,x = ( 380 ± 40 ) g ⋅ cm/s . Could kinetic friction alone be enough to
account for the observed change in momentum?
The friction force is constant while it acts, so the impulse due to friction is just equal to the force times
the time during which it acts:


Δpdue to kf = Fkf Δt
( )
In the equation above, Δt is the time between the collision (when the friction force begins to act) and
the “final” state that we considered for momentum conservation. Certainly, kinetic friction continued to
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Physical Sciences 2
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act after this—after all, it eventually brought 1-M-2 to a stop entirely—but we are attempting to account
for a momentum change observed between the initial (pre-collision) state and the final (first video frame
after collision) state, so anything that friction did after that doesn’t matter any more. We can write the xcomponent of the equation as
Δpdue to kf,x = ( Fkf ) Δt = µk mgΔt
where m is the mass of the entire 1-M-2 object. As stated in the problem, we don’t know Δt , and can’t
really measure it with any accuracy, but it has an upper bound of 33 ms. So the upper bound on
Δpdue to kf,x itself is
(
)
Δpdue to kf,x < µk mg ( Δt max ) = ( 0.19 ) ( 22.76 g ) 980 cm/s 2 ( 0.033 s ) = 140 g ⋅ cm/s 3
The observed change in x-momentum was
Δpx = (110 ± 50 ) g ⋅ cm/s
This is less than the maximum possible impulse due to kinetic friction, so yes, kinetic friction alone
could be enough to account for the observed change.
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4. Putting Everything Together (Exam-Type Question): Newton’s Second Law Practice (2 pts)
Question from Tutorials in Introductory Physics
The table below provides information about the motion of a box in four different situations. In each
case, the information given about the motion is in one of the following forms: (1) the algebraic form of
Newton’s second law, (2) the free-body diagram for the box, or (3) a written description and picture of
the physical situation. In each case, complete the table by filling in the information that has been
omitted. Case 1 has been done as an example.
(All symbols in the equations represent positive quantities. In each case, use a coordinate system for
which +x is to the right an +y is toward the top of the page.)
KEY: B – box; C – small container; H – hand; S – surface; E – Earth; R, R1, R2 – massless ropes
a.
b
.
Net force is to the left
NBS
FBH
FBR
A rope is pulling the box at an angle to the
right while a hand is pulling the box to the
left at the same time. The hand is applying
the greater force, so the box is accelerating
to the left on a frictionless surface.
hand is pulling box
WBE
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Physical Sciences 2
c.
September 17 – 22
Net force is to the right
𝛴𝐹! : 𝑓!" = 𝑚𝑎!
𝛴𝐹! : 𝑁!" − 𝑊!" = 0
NBS
fBS
WBE
d
.
𝛴𝐹! : 𝑇!!! cos 𝜃 − 𝑇!!! cos 𝜃 = 0
The box is hung from two ropes that
cannot support its weight. It is
accelerating down towards the Earth.
𝛴𝐹! : 𝑇!!! sin 𝜃 + 𝑇!!! sin 𝜃 − 𝑊!"
= −𝑚𝑎!
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5. Putting Everything Together (Exam-Type Question): Don’t
look down! (2 pts)
Mountain goats (Oreamnos americanus) can climb steep inclines;
their hooves are designed to provide good traction against rock.
The image at right shows a mountain goat adult and kid on very
steep cliff in the Rocky Mountains in North America.
a) If a mountain goat with m = 90 kg is standing still on a
mountain with a θ = 60° incline (from the horizontal), what is
the minimum coefficient of static friction required between its
hooves and the rock?
F: First we draw the free-body
diagram for the goat. (If you squint a
little, the gray box looks like a goat…
okay, no, it doesn’t. But it represents
the goat.) There are three forces on the
goat: the gravitational force, the

normal force, and the static frictional force Fsf . A priori, we don’t know

which way Fsf points. We’ve draw it pointing uphill (in the +x direction), but
we won’t make any assumptions about the direction when we solve
algebraically.
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A: Because we know the motion will be along the plane of the mountain face, we’ll orient the axes so
€ direction (and the y-axis is perpendicular to it).
that the x-axis points along that
N: Then we can write down Newton’s Second Law in component form:
max = Fsf, x − Fgrav sin θ
may = FN − Fgrav cosθ
C: Now we can apply the constraints from what we know about the goat’s motion. In this part of the
problem, the goat is standing still, so the acceleration must be zero in both directions.
L: We use the gravitational force law Fgrav = mg .
A: Plugging in from C and L and solving, we get:
Fsf, x = mg sin θ
FN = mg cosθ
The first thing to note is that Fsf,x is positive, so the direction of static friction on our FBD was correct.
Second, we know that the static friction force has a limit: Fsf (which is equal to Fsf,x ≤ µsFN).
Substituting the expressions for the magnitudes of the two forces, we get the requirement µs ≥ tan θ.
N: So for a 50° incline, the static friction coefficient must be at least tan 60° = 1.73 in order for the goat
to stand still. That’s rather large.
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b) To escape a predator, the mountain goat accelerates (with constant acceleration) from rest to a
velocity of 3 m/s directly up the incline over a duration of 0.1 second. Which force is responsible for
accelerating the goat up the incline: static friction or kinetic friction? Explain your answer.
The free-body diagram has not changed, nor has the expression of Newton’s Second Law. You might
think that because the goat is moving, we’re now talking about kinetic friction, but as long as the goat’s
hooves do not slip against the rock face, it’s still static friction. As for the direction of the static friction
force, it still points up the incline.
On a side note, we’re given some data and can calculate the coefficient of static friction necessary in
order for the goat to accelerate. The problem is very similar to part a), except, instead of being at rest,
the goat now accelerates up the hill with a known acceleration ax = +(3 m/s) / (0.1 s) = 30 m/s2. (This is
a very large acceleration—about three times as large as g.)
L: No change here.
A: Solving for the forces, we get:
Fsf, x = max + mg sin θ
FN = mg cosθ
So yes, the static friction force still points uphill. In fact, it’s bigger than it was in part a). Applying the
static friction condition Fsf ≤ µsFN, we get
max + mgsin θ ≤ µsmgcos θ
ax
µs ≥
+ tan θ
gcosθ
N: Putting in the numbers, we get µs ≥ 7.85, a much larger coefficient than in part a). In fact, this is an
absolutely enormous coefficient of static friction; in contrast, rubber tires on pavement (which are
expressly designed for good traction) offer static friction coefficients of only about 1.7.
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