Solutions for Physics 1201 Course Review (Problems 1 through 9)
1) In an estimation problem, we must often make reasonable guesses about certain
numbers we need in order to carry out calculations. To make an estimate of how much
water a person drinks in their lifetime, we might start out with a guess as to how much
they drink in a single day. A typical contemporary recommendation is that one should
drink eight (8) cups of water each day; this is 8 cups/day · 8 ounces/cup
= 64 ounces/day · 1 pound/16 ounches = 4 pounds/day . This is a weight
corresponding to a mass of 4 pounds · ( 1 kg. / 2.2 pounds mass equivalent* ) ~ 2 kg.
The density of water is 1000 kg./m.3 = 1000 kg./1000 l. = 1 kg./l. , so the eight cups
of water have a volume of about 2 liters. The actual daily intake of water by drinking
varies considerably among people, so we will only retain one significant figure in our
calculations (this is typical for this sort of physical estimation).
* Since the pound is a unit of weight, rather than mass, it is not really correct to say that
“one pound equals 2.2 kilos”.
The lifespan of people throughout the world and history also varies a good deal,
but a fairly good estimate for the average might be about 50 years. This would give us
an estimate for the amount of water a typical person drinks in their lifetime as
2 liters/day · 365 days/year · 50 years ~ 4 · 104 liters .
⎛ 100 cm. ⎞3
1 l.
= 1000 liters in a cubic meter, this amount of
⎟ ⋅
Since there are ⎜
3
⎝ 1 m. ⎠ 1000 cm
water has a volume of 40 m3 . This is the volume of a cube 3 40 ≈ 3.4 meters
( × 3.3 ft./m. ≈ 11 ft. ) on a side. A body of water of comparable size would be a
small pond, a water tower for an office building, or a modest public swimming pool. (By
€ an Olympic-size swimming pool hold 2.5 · 106 liters!)
contrast,
€
If we were to consider water also taken in from food consumed, our estimate
might be roughly doubled; our current number, then, will be good enough to get an idea
of the scale of water consumption. By way of comparison, the daily amount of water
that a person in an industrialized nation might use or require for bathing, disposal of
wastes, growing of food, production of items for daily use, etc., would be larger by easily
a factor of 30 to 100 .
If we may take the figure that the total number of members of historical
Humanity is on the order of 300 billion or 3 · 1011 , then the total amount of water that
humans have drunk is 3 · 1011 persons · 4 · 104 liters/person ~ 1 · 1016 liters
= 1 · 1013 m3 . All of this water eventually finds its way to the oceans, where it is
mixed with water already there, some of which will be drunk by someone else later.
About 70% of the Earth’s surface is covered by bodies of water; if the water down to a
depth of 100 meters has been kept thoroughly mixed by weather and ocean currents,
then the amount available to human use is around
0.7 · ( 4π R2Earth ) · 100 m. ≈ 0.7 · 4π · ( 6400 km. · 1000 m./km. )2 · 100 m
~ 4 · 1016 m3 .
The fraction of this water that humans have ever drunk is then on the order of
1 ⋅1013 m3
~ 3 ⋅10−4 ; this is rather approximate, in that we are ignoring the real
4 ⋅1016 m3
complexities of the re-use of water and the circulation of water through the earthly
environment.
€
If all of this water is thoroughly mixed, then this fraction is also the fraction of
any water you drink now that has been previously drunk by someone else. The
molecular weight of water is 18 amu, so one mole of water has a mass of 18 grams. A
liter bottle of drinking water contains 1000 grams of water (as we have mentioned
above), or ( 1000 gm. ) / ( 18 gm./mole ) ≈ 56 moles. Thus, it contains
56 moles · 6.02 · 1023 molecules/mole ≈ 3 · 1025 molecules of water ,
of which ( 3 · 10−4 ) · ( 3 · 1025 ) ~ 1 · 1022 have been drunk by someone else at some
Qtime in the past. (If we were to include water put to use by humans for any purpose,
this number could easily be 100 times larger.)
2) In dimensional analysis, we can use the units associated with physical quantities (or
the “dimensions” of length, time, mass, etc. measured by those units) to establish
possible relationships among the quantities. Keep in mind that dimensional analysis
tells us nothing about any physical model connecting those quantities, so any
relationship found by this method isn’t necessarily a real one. Further, because there is
no particular physical model under consideration, dimensional analysis cannot tell us
what “dimensionless” numerical constants belong in the relationship we’ve determined.
A number which would appear due to geometry or a simple numerical relationship, such
as 2 or π , will not turn up in our dimensional result. Therefore, dimensional analysis
tends to give us proportionality relationships, rather than strict equations.
For the relationship concerning the flow of a fluid through a tube, we want to
express the “volume flow rate”, Q = dV/dt , to measured quantities for the tube or the
fluid: the radius r and length L of the tube; the difference in pressure between the
ends of the tube, ΔP ; and the viscosity of the fluid, η , which is a measure of the
resistance of the fluid to flowing motion. We need first to set out the units and
dimensions of these five physical quantities:
Q
volume/time
m3/sec
[L3]/[T]
r
length
m
[L]
L
length
m
[L]
ΔP
pressure
Pa = N/m2
= kg/(m · sec2)
[M]/[L · T2]
Pa·sec = (N · sec)/m2
= kg/(m · sec)
[M]/[L · T]
η
viscosity = force · time / area
We are seeking an expression for Q in terms of the other four quantities; this means
that the units or dimensions on the left-hand side of the proportionality must equal
those on the right-hand side:
Q ∝ r α ⋅ L β ⋅ (ΔP)γ ⋅ ηδ
→
kg.
kg. δ
m.3
α
β
γ
sec. = m. ⋅ m. ⋅ ( m. ⋅ sec.2 ) ⋅ ( m. ⋅ sec. )
or
→
[L]3
[M ] γ
[M ] δ
= [L]α ⋅ [L] β ⋅ (
) ⋅ (
)
2
[T ]
[L] ⋅ [T ]
[ L] ⋅ [T ]
.
€
€
We can now “gather up” all the factors of various fundamental units or of
dimensions to determine the required equations for each dimension:
€
m. or [L] :
3 =
α + β + (−γ) + (−δ)
kg. or [M] :
0 =
γ + δ
sec. or [T] :
−1 = (−2γ) + (−δ)
.
We can find two of these unknown exponents immediately:
0 = γ + δ
⇒ γ = −δ ;
−1 = − 2γ + (−δ ) ⇒ − 1 = − 2γ + γ = − γ
⇒ γ = 1 ⇒ δ = −1 .
€
€
This leaves us with one dimension equation still to be resolved:
3 = α + β + (−1) + (−[−1]) = α + β
€
,
which unfortunately contains two unknowns; the best we can do with this is to obtain a
relation between the unknowns, α = 3 – β . We are given an additional piece of
information
about the physical phenomenon itself, however: doubling the length of the
€
tube causes the rate of volume flow to decrease by a factor of one-half. This tells us
that Q ∝ 1/L or L−1 ; therefore, β = −1 and α = 4 . We can now conclude that
4
Q ∝ r ⋅ L
−1
1
−1
⋅ (ΔP) ⋅ η
or
r 4 (ΔP)
ηL
.
It requires a more detailed physical theory to arrive at the complete expression of
Poiseuille’s Law, Q =
€
€
π r 4 (ΔP)
.
8η L
3) Since all of the activity we are following takes place on a single railroad track, we
may use one-dimensional kinematics. We will call the moment when the engineer in the
speeding train applies its brakes t = 0 , and x = 0 is the position of the front end of the
km. 1000 m.
1 hr.
m.
train then. The initial velocity of the train is v 0 = 135
,
⋅
⋅
≈ 37.5
hr.
1 km.
3600 sec.
sec.
with the positive x-direction being the direction to the railway station. The rear end of
the stopped train lies 3000 meters ahead. In order to avoid a collision in this situation,
the moving train must reach zero velocity when it reaches that position or sooner. The
€
train’s brakes must then provide a minimum deceleration given by the “velocitysquared” equation
m.
v 2f = v 02 + 2 a ( x f − x 0 ) → 0 2 = (37.5 sec. ) 2 + 2 a ( 3000 m. − 0 )
⇒ a =
€
−(37.5
m. 2
)
sec.
2 ⋅ 3000 m.
≈ − 0.234
m.
sec.2
If the moving train is unable to decelerate at least this rapidly, the other train will
have to be moved out of the way in order to avoid a collision. For the actual capability of
its brakes,
the front end of the moving train under maximum deceleration will be at
€
x = x0 + v0 · t + ½ a1 · t2 = 0 + 37.5 t + ½ ( −0.16 ) · t2 = 37.5 t − 0.08 · t2 .
The engineer of the parked train is able to start it moving at t = 30 seconds. This delay
must be included in the kinematic equations for this train, so the position of its rear end
is given by x’ = x0’ + v0’ · ( t − 30 ) + ½ a2 · ( t − 30 )2
= 3000 + 0 · ( t − 30 ) + ½ ( 0.12 ) · ( t − 30 )2 = 3000 + 0.06 · ( t2 – 60 t + 900 )
= 3000 + 0.06 t2 – 3.6 t + 54 = 3054 – 3.6 t + 0.06 t2 m. , for t ≥ 30 seconds.
At the moment this parked train begins to accelerate, the front end of the
approaching train is at x = 37.5 · 30 − 0.08 · 302 ≈ 1125 – 72 = 1053 m. A collison
hasn’t occurred yet, so we may continue with our analysis.
The quantity of genuine interest here is the separation between the ends of the
two trains, which is given by
x’ – x = ( 3054 – 3.6 t + 0.06 t2 ) − ( 37.5 t − 0.08 · t2 )
= 3054 – 41.1 t + 0.14 t2 , for t ≥ 30 seconds.
A collision occurs if this separation function is ever zero. While we could use the
quadratic formula to solve for the zeroes of this polynomial, we might first check to see
whether there are any such solutions to be found. The discriminant for this quadratic
polynomial is ( −41.1 )2 − 4 · ( 0.14 ) · ( 3054 ) ≈ −21.0 < 0 ; since it is negative, there
are no solutions in real numbers where x’ – x = 0 . This means that a collision between
the trains has been averted.
Since no collision takes place, we are interested in finding out how close a call
this was, what the speed of both trains was at the moment of closest approach, and
where they were then. If we differentiate the separation function with respect to time,
we obtain a new function for the relative speed of the two trains,
€
d
( x' − x ) = v' − v .
dt
The moment of closest approach between the trains occurs when the relative speed is
zero (or when v’ = v ), hence
d
d
( x' − x ) =
( 3054 − 41.1t + 0.14t 2 ) = − 41.1 + 0.28t = 0
dt
dt
m.
⇒ t =
€
41.1 sec.
0.28
m.
= 147 sec.
sec.2
The two ends of the trains that would have collided made their closest approach 147
seconds after the engineer in the moving train first applied the brakes. (This could also
be worked out
€ by setting up the velocity equations for the individual trains and
proceeding from there.) At that moment, the separation between the trains was
( x' − x )
t = 147
= 3054 − 41.1 ⋅147 + 0.14 ⋅147 2 ≈ 38 m.
At that time, the front end of the incoming train is at
x = 37.5 · 147 − 0.08 · 1472 ≈ 3784 meters , while the rear end of the formerly parked
€ train is now at x’ = 3054 – 3.6 t + 0.06 t2 ≈ 3821 meters, confirming the minimum
separation found above (allowing for round-off), and indicating that the parked train has
moved forward 820 meters in the interval of acceleration. By setting up the velocity
functions for each train (or differentiating the two position functions), we can use either
to find that the velocity of the trains at the time of closest approach is
v = 37.5 − 0.16 · 147 = v’ = – 3.6 + 0.12 · 147 = 14.0 m./sec.
4)
first section –
Because we are following the motion of an elevator car along its shaft, we may
use one-dimensional kinematics. The travel of the car proceeds in three phases, the first
being under uniform linear acceleration, the middle portion being a “cruise” at constant
speed, and the final part being uniform deceleration to a stop.
For a run sufficiently long for the elevator car to reach “cruising speed”, the
model for the velocity and position functions of the car during the acceleration is
v1 = vi1 + a1 · t = 0 + a1 · t = a1 · t
;
y1 = yi1 + vi1 · t + ½ a1 · t2 = 0 + 0 · t + ½ a1 · t2 = ½ a1 · t2 , 0 ≤ t ≤ 5 .
At the end of this five-second interval, the elevator will be moving at a speed vf1 = 5a1
and will have reached a height of yf1 = ½ a1 · 52 = (25/2) a1 . These will be the starting
conditions for the “crusing phase”.
During the interval of constant velocity, the elevator car’s speed remains at
v2 = 5a1 . Its vertical position is given by
y2 = yi2 + vi2 · t + ½ a2 · t2 = yf1 + vf1 · t + ½ a2 · t2
= (25/2) a1 + ( 5a1 ) · t + ½ · 0 · t2 = (25/2) a1 + ( 5a1 ) · t , 5 ≤ t ≤ T .
There is one piece of information we are given which will be helpful in resolving things
later: when the elevator car travels the full distance along the shaft, it reaches the halfway point, a height of 39 meters , at a time 0.475 · τ , where τ is the total time for the
trip. We may assume that this point is reaches during the “cruising phase”, which
allows us to write
y = 39 = (25/2) a1 + ( 5a1 ) · ( 0.475 ·
τ − 5)
.
When the elevator is approaching the end of its travel, it begins braking to a
stop; for the full motion along the elevator shaft, this requires eight seconds. The
velocity function during this portion is
v3 = vi3 + a3 · t = vf2 + a3 · t = 5a1 + a3 · t , T ≤ t ≤ T + 8 .
Since the elevator car has come to a stop, we have
vf3 = 0 = 5a1 + a3 · 8
⇒
a3 = −(5/8) · a1 .
The car starts to decelerate at time t = τ – 8 ; since it began “cruising” at t = 5 , the car
spent ( τ – 8 ) – 5 = τ – 13 seconds at constant velocity. So, when it begins its
deceleration, it is at a height of yf2 = (25/2) a1 + ( 5a1 ) · (
position of the car during this last portion is
τ – 13 ) = yi3 . The vertical
y3 = yi3 + vi3 · t + ½ a3 · t2 = yf2 + vf2 · t + ½ a3 · t2
= [ (25/2) a1 + ( 5a1 ) · (
τ – 13 ) ] + ( 5a1 ) · t + ½ · [ −(5/8) · a1 ] · t2 .
At the end of the eight seconds, the elevator car will have arrived at the top of the shaft,
so
yf3 = 78 = [ (25/2) a1 + ( 5a1 ) · ( τ – 13 ) ] + ( 5a1 ) · 8 + ½ · [ −(5/8) · a1 ] · 82 .
We now have equations for two specific positions along the elevator shaft,
containing two unknowns which we can now solve for:
y = 39 = 12.5 a1 + 2.375 a1τ − 25 a1 = 2.375 a1τ − 12.5 a1
,
yf3 = 78 = 12.5 a1 + 5 a1τ − 65 a1 + 40 a1 − 20 a1 = 5 a1τ − 32.5 a1 .
We can most easily solve these two equations simultaneously by using the fact that 78
is twice 39 , permitting us to write
2 · ( 2.375 a1τ − 12.5 a1 ) = 5 a1τ − 32.5 a1
⇒
4.75 a1τ − 25 a1 = 5 a1τ − 32.5 a1
⇒
0.25 a1τ = 7.5 a1 .
Since a1 is certainly not zero, we can divide this last equation through by it to find
0.25 τ = 7.5 ⇒ τ = 30 seconds, the time for the entire elevator ride. We can put this
result into either of the position equations above to solve for a1 :
78 = 5 a1 · 30 − 32.5 a1 = 117.5 a1
⇒
a1 ≈ 0.664 m./sec.2 ,
and thus, a3 = −(5/8) · 0.664 ≈ −0.415 m./sec.2 . The “cruising” speed of the elevator
is therefore v = 5 · 0.664 = 3.32 m./sec.
second section –
A passenger riding in the elevator car is subjected to two forces, gravity and the
normal force from the floor of the car. If this person is standing on a weight scale, the
normal force N is applied to them through the body and mechanism of the scale; the
scale displays the magnitude of this force N , which we read as that person’s “weight”.
We have been referring to the upward direction along the axis of the elevator
shaft as “positive”, so we write the net force on the passenger as N – Mg = Ma , with a
being the acceleration of the elevator as seen by a stationary (inertial) observer. The
normal force on the passenger, and thus the reading on the scale, is then
N = M·(g+a).
For the purpose of calculation, we can express the scale reading in terms of the
a
person’s weight Mg in ordinary Earth gravity as N = Mg (1 + g ) . For a passenger of
ordinary Earth-gravity weight Mg = 800 N , during the acceleration phase for the
a
0.664
) ≈ 854 N . In
elevator car, the scale would read N1 = Mg (1 + g1 ) ≈ 800 N ⋅ (1 +
9.81
€
other words, the passenger would “feel heavier”
than usual; in order to accelerate them
upward, the elevator must apply a force larger than Mg against the person’s feet.
€ reaches its “cruising phase”, it maintains a constant
Once the elevator car
velocity v . Thus a = 0 , and the scale reads N2 = Mg ; the passenger experiences their
usual weight, as if they were standing still on the Earth’s surface. During the car’s
a
−0.415
) ≈ 766 N .
deceleration, the scale would read N 3 = Mg (1 + g3 ) ≈ 800 N ⋅ (1 +
9.81
The passenger “feels lighter” than usual, since the floor of the elevator car can apply a
€
force less than Mg against the person’s feet, in order to provide an “upward
deceleration”. (In effect, the elevator is actually permitting the passenger to fall under
gravity, but subject to restraint, so that their acceleration is toward the Earth’s surface,
but with a magnitude much less than g .)
5) It will be helpful to establish first a two-dimensional coordinate system for the
situation. If we call place the origin at the location of the adult ( xA = 0 , yA = 0 ) , the
positive horizontal direction pointing toward the child, and the positive vertical
direction pointing upward, then the child is located at ( xC = 20 cos 15º ≈ 19.32 m. ,
yC = 20 sin 15º ≈ 5.18 m. ).
In the first part of the problem, we wish to find how fast each person needs to
throw the ball to reach the other person exactly. (We ignore the difference in the
heights of adult and child, as far as how that affects where they each catch the ball.)
The adult throws at a 30º angle to the 15º upward slope, so their throw starts at an
angle of θA = 45º to the horizontal. The child is throwing downhill at the same angle
relative to the slope, so their throw starts at θC = 15º to the horizontal. We will
neglect air resistance, so the only force acting on the ball in flight is vertically
downward-directed gravity.
We will solve in parallel for the speeds at which adult and child each must throw
the ball for a “perfect catch”:
adult to child
child to adult
1
⋅ 0 ⋅ t2
2
⇒ 20 cos 15o = 0 + ( v A cos 45o ) t €
1
⋅ 0 ⋅ t2
2
⇒ 0 = 20 cos 15o − ( vC cos 15o ) t
x f = x A = xC − ( vC cos θ C ) t +
x f = xC = x A + ( v A cos θ A ) t +
the sign of the velocity is reversed, since
the child is throwing in the opposite x-direction
€
€
20 cos 15o
⇒ t =
v A cos 45o
€
;
y f = yC = y A + ( v A sin θ A ) t −
1
⋅ g ⋅ t2
2
⇒ t =
20 cos 15o
20
= v
o
vC cos 15
C
y f = y A = yC + ( vC sin θ C ) t −
;
1
⋅ g ⋅ t2
2
1 2
1 2
o
o
o
o
€ ⇒ 20 sin 15 = 0 + ( v A sin 45 ) t − 2 gt € ⇒ 0 = 20 sin 15 + ( vC sin 15 ) t − 2 gt
€
€
€
€
We can substitute our result for t into these equations to find
2
o
⎛ 20 cos 15o ⎞
1 ⎛ 20 cos 15 ⎞
20 sin 15o = ( v A sin 45o ) ⎜
−
g
⎜
⎟
⎟
o
2 ⎝ v cos 45o ⎠ ⎝ v A cos 45 ⎠
A
2
⎛ 20 ⎞
1 ⎛ 20 ⎞
0 = 20 sin 15o + ( vC sin 15o ) ⎜ v ⎟ − g ⎜ v ⎟
2 ⎝ C ⎠ ⎝ C ⎠
€
⇒ ( 20 cos 15o tan 45o − 20 sin 15o ) v A2 =
o 2
1 ⎛ 20 cos 15 ⎞ €
g ⎜
⎟
2 ⎝ cos 45o ⎠ ⇒ 2 ⋅ 20 sin 15o ⋅ vC2 =
1
⋅ g ⋅ 20 2
2
multiplying through by the velocity squared, since it is not zero
€
⇒
€
⇒
v A2 ≈
9.81 ⋅ 20 2 ⋅ 0.966 2
⎛ 2 ⎞2
2 ⋅ ⎜ 2 ⎟ ⋅ 20 ⋅ ( 0.966 ⋅1 − 0.259)
⎝ ⎠
m.
v A ≈ 16.1 sec.
≈ 259.0
€2
m.
sec.2
€
⇒ vC2 ≈
9.81 ⋅ 20 2
m.2
≈ 189.4
sec.2 2 ⋅ 2 ⋅ 20 ⋅ 0.259
m.
⇒ vC ≈ 13.8 sec.
For the second part of the problem, the physical arrangement for the two
throwers is the same, but they will now both throw
€ the ball at vA = vC = 20 m./sec. ; we
want to know where the ball would land in each case. The hillside has a slope of 15º, so
we can idealize it by describing it by a line of slope m = tan 15º . Since the origin of our
coordinate system is on the hillside, the equation for this line is y = (tan 15º) x . For
each thrower, we need to work out the trajectory of the ball thrown by them and find
where it intersects the line representing the hillside.
€
We are not asked for the time the ball spends in flight, so we can take an
approach in which we calculate the equation for the parabola of the ball’s trajectory for
each thrower, and then find where the parabola intersects the line of the hillside. The
position equations now yield
adult to child
x = ( 20 cos 45o ) t ⇒ t =
child to adult
x
20 cos 45o
;
x = 20 cos 15o − ( 20 cos 15o ) t
⇒ t = 1 −
€
;
€
⎛
⎞
⎞2
x
1 ⎛
x
y = ( 20 sin 45o ) ⎜
−
g
⎟
2 ⎜⎝ 20 cos 45o ⎟⎠
⎝ 20 cos 45o ⎠
⎛
⎞
x
y = 20 sin 15o + ( 20 sin 15o ) ⎜1 −
o ⎟
20
cos
15
⎝
⎠
€
−
€
x
20 cos 15o
€
⎛
⎞ 2
g
⇒ y = ( tan 45o ) x − ⎜
⎟ x
2
2
o
⎝ 2 ⋅ 20 cos 45 ⎠
⎞2
1 ⎛
x
g ⎜1 −
⎟
2 ⎝
20 cos 15o ⎠
⇒
y = ( 2 ⋅ 20 sin 15o −
€
€
⎛
⎞
1
g
g ) + ⎜
− tan 15o ⎟ x
2
⎝ 20 cos 15o
⎠ ⎛
⎞ 2
g
− ⎜
⎟ x
2
2
o
⎝ 2 ⋅ 20 cos 15 ⎠
€
We set each of these functions for the parabolic trajectories
of each ball equal to the
€
function for the line of the idealized hillside to find
⎛
⎞ 2
g
( tan 15o ) x = ( tan 45o ) x − ⎜
⎟ x
2
2
o
⎝ 2 ⋅ 20 cos 45 ⎠
( tan 15o ) x = ( 2 ⋅ 20 sin 15o −
€
⎛
⎞ 2
g
− ⎜
⎟ x
2
2
o
⎝ 2 ⋅ 20 cos 15 ⎠
€
⎛
⎞ 2
g
⇒ ( tan 45o − tan 15o ) x − ⎜
⎟ x = 0
⎝ 2 ⋅ 20 2 cos 2 45o ⎠
( 2 ⋅ 20 sin 15o −
⇒
€
( 2 ⋅ 20 ⋅ 0.259 −
⇒
€
⇒
either x = 0 (which is xA ) or
x =
€
⎛ 9.81
⎞
1
⋅ 9.81) + ⎜
− 2 ⋅ 0.268⎟ x
2
⎝ 20 ⋅ 0.966
⎠
⎛
⎞ 2
9.81
− ⎜
⎟ x = 0
⎝ 2 ⋅ 400 ⋅ 0.933 ⎠
this requires the quadratic formula
in order to solve for x
€
⎡
⎛
⎞ ⎤
g
⇒ x ⋅ ⎢ ( tan 45o − tan 15o ) − ⎜
⎟ x ⎥ = 0
2
2
o
⎝ 2 ⋅ 20 cos 45 ⎠ ⎦
⎣
⎛
⎞
1
g
g ) + ⎜
− 2 ⋅ tan 15o ⎟ x
2
⎝ 20 cos 15o
⎠
⎛
⎞ 2
g
− ⎜
⎟ x = 0
⎝ 2 ⋅ 20 2 cos 2 15o ⎠
€
€
⎛
⎞
1
g
g ) + ⎜
− tan 15o ⎟ x
2
⎝ 20 cos 15o
⎠
€
⇒ x ≈
−0.02810 ±
0.02810 2 − 4 ⋅ 0.01314 ⋅ (−5.4478)
2 ⋅ 0.01314
⇒ either x ≈ 19.29 m. (which is xC )
€
( tan 45o − tan 15o ) ( 2 ⋅ 20 2 cos 2 45o )
g
or x ≈ −21.43 m.
€
≈
(1 − 0.268) ( 2 ⋅ 400 ⋅
9.81
1
)
2
≈ 29.85 m.
(If we solved this problem by working out the time of flight, t , first, we find that
t ≈ 2.11 seconds in both directions – perhaps a surprising result!)
€
The y-coordinate of the position where the ball will land on the hillside is
y = ( tan 15o ) x ≈ 0.268 x ≈ 8.00 m.
y ≈ 0.268 x ≈ − 5.74 m.
The distance along the hillside from each thrower to the point where the ball lands is
given by the “distance formula”:
€
€
d ≈
€
( 29.85 − 0) 2 + (8.00 − 0) 2 ≈ 30.9 m.
d ≈
(− 21.43 − 19.32) 2 + (− 5.74 − 5.18) 2 ≈ 42.2 m.
6) For the purpose of analyzing the forces
and torques acting on the ladder, we will
€
divide it into two halves, with the person standing on its left half. We will call the
distance along the left leg, measuring upward from the floor, at which the person is
located, x . The person’s weight is Mg and the weight of the ladder is mg . The
reaction force with which the two legs of the ladder push against one another is R , the
tension force in the support rod halfway up the ladder is T , and the normal forces
from the floor against the feet of the ladder are NL and NR .
Since this is intended to be a static configuration, the sum of the horizontal
forces, the sum of the vertical forces, and the sum of the torques on the ladder are all
zero. The net horizontal force on each leg is T – R = 0 ⇒ T = R . The net vertical
force on the entire ladder is NL + NR – Mg – mg = 0 ⇒ NL + NR = Mg + mg . To
analyze the torques on each leg, we will chose the reference point to be at the top of the
ladder. For that choice, the reaction forces R produce no torque, since the moment
arm for those forces is zero.
The sum of the torques on each leg must be zero, so we have:
left leg -1
+( 2.8 − x ) ⋅ Mg ⋅ sin θ + 1.4 ⋅ ( 2 mg ) ⋅ sin θ + 1.4 ⋅ T ⋅ sin (θ + 90 o ) − 2.8 ⋅ N L ⋅ sin (180 o − θ )
⇒ + ( 2.8 − x ) Mg sin θ + 0.7 mg sin θ + 1.4 T cos θ − 2.8 N L sin θ = 0
sin(
€
θ + 90º ) = cos θ
sin(
θ – 180º ) = sin θ
right leg --
€
1
− 1.4 ⋅ ( 2 mg ) ⋅ sin θ − 1.4 ⋅ T ⋅ sin (θ + 90 o ) + 2.8 ⋅ N R ⋅ sin (180 o − θ )
⇒ − 0.7 mg sin θ − 1.4 T cos θ + 2.8 N R sin θ = 0
€
If we now use the measurements from the statement of the problem, along with
the diagram in green, at the start of the solution, showing the right triangle representing
€ a leg of the ladder, these equations become
left leg --
⎛ 0.45 ⎞
⎛
⎞
⎛
⎞
⎟ + 0.7 (12 ⋅ 9.81) ⎜ 0.45 ⎟ + 1.4 T ⎛⎜1.326 ⎞⎟ − 2.8 N L ⎜ 0.45 ⎟ = 0 ( 2.8 − 2.3) ( 900) ⎜
⎝ 1.4 ⎠
⎝ 1.4 ⎠
⎝ 1.4 ⎠
⎝ 1.4 ⎠
⇒ 144.6 + 26.5 + 1.33T − 0.9 N L = 0 ⇒ 0.9 N L − 1.33T ≈ 171.1 N
€
right leg --
⎛ 0.45 ⎞
⎟ − 1.4 T
− 0.7 (12 ⋅ 9.81) ⎜
⎝ 1.4 ⎠
€
⎛ 0.45 ⎞
⎛⎜1.326 ⎞⎟
⎟ = 0
+ 2.8 N R ⎜
⎝ 1.4 ⎠
⎝ 1.4 ⎠
⇒ − 26.5 − 1.33T + 0.9 N R = 0 ⇒ 0.9 N R − 1.33T ≈ 26.5 N
€ At the moment, it appears that we have two equations with three unknowns. However,
we have a way to eliminate two of these unknowns: if we add the equations for the left
and right
€ legs together, and use the result from the vertical forces equation, we obtain
0.9 N L + 0.9 N R − 2.65T ≈ 197.6 N ⇒ 0.9 ( N L + N R ) − 2.65T ≈ 197.6 N
⇒ 0.9 ( Mg + mg ) − 2.65 T ≈ 197.6 N ⇒ 0.9 ( 900 N + 117.7 N ) − 2.65 T ≈ 197.6 N
€
€
⇒ 2.65 T ≈ 718.3 N ⇒ T ≈ 271 N
;
the tension forces at each end of the support rod are directed toward the center of the
rod, so it is said to be “in compression”. We can now use the results from the torque
€ equations to solve for the two normal forces on the ladder’s feet:
€
€
0.9 N L − 1.33 ( 271 N ) ≈ 171.1 N ⇒ N L ≈ 589 N ; 0.9 N R − 1.33 ( 271 N ) ≈ 26.5 N ⇒ N R ≈ 429 N . For the second part of this problem, the physical situation is the same, but we
will leave the location of the person using the ladder unspecified; hence, we will use the
distance x , rather than the specific value 2.3 meters. The derivation will be as it was
before, except that the torque equation for the left ladder leg is now
( 2.8 − x ) ⋅ 289.3 + 26.5 + 1.33T − 0.9 N L = 0
⇒ 0.9 N L − 1.33T ≈ (836.5 − 289.3 x ) N
.
€ When we add this new equation to the equation for the right leg, we find
0.9 N€
L + 0.9 N R − 2.65T ≈ (836.5 − 289.3 x ) + 26.5 N = (863.0 − 289.3 x ) N
⇒ 2.65 T ≈ [ 0.9 ( 900 + 117.7 ) − (863.0 − 289.3 x ) ] N
€
⇒ T ≈
€
( 52.9 + 289.3 x ) N
= ( 20.0 + 109.2 x ) N
2.65
.
The sign of T is always positive, meaning that the support rod is always in
compression, and increasingly so as the person climbs the ladder. If the substituted
€ support rod will undergo a tension of T < 200 N before it crumples, the climber is
limited to locations above the floor given by
T ≈ ( 20.0 + 109.2 x ) N < 200 N ⇒
x <
200 − 20.0
≈ 1.65 m. ,
109.2
as measured along the ladder leg upward from the foot, or about 59% of the way to the
top of the ladder.
€
7)
a) This is a problem best analyzed in stages. The general principle that is
followed here is that the center of mass of a stack of blocks must be supported by the
block below it; otherwise the net torque due to gravity will cause the stack to tip over.
Since every block is taken to have uniform density, its center of mass is at the center of
the block.
If we start by looking a single block, it must be placed so that no more than half
its length overhangs the edge of the table; that is, the center of mass of the single block
must be no further out than just at the edge of the table (which would make the block
just marginally stable). The same would be true if this block were placed atop a second
block: its center of mass must lie no further out than the edge of the lower block.
Thus, the overhang distance of the first block is a1 = ½ L .
With a stack of two blocks, its center of mass is midway between the centers of
mass of the invididual blocks, or one-quarter of the length of the bottom block from the
edge under the first block. This stack can then only be placed as far over on the
tabletop as to have this total center of mass at the table’s edge. So the maximum
overhang of the second block is a2 = ¼ L .
Once we get to a stack of three blocks, it’s a bit less obvious how to proceed, so
we’ll look explicitly at the torque applied to the stack due to gravity. The amount of
overhang for the third block, a3 , is unknown. This puts the center of mass for the
upper two blocks beyond the edge of the table by a distance a3 . Since the entire stack
will pivot at that edge, the torque about that point due to the weight of the upper two
blocks is τupper = −2Mg · a3 sin 90º . The bottom block overhangs the table edge by a
distance a3 , so its center of mass is ½L – a3 from the edge, and thus from the pivot
point. The torque about that point due to the weight of the bottom block is then
τlower = +Mg · ( ½L – a3 ) sin 90º . In order for the stack to remain static, then, the total
torque on it must be
Mg · ( ½L – a3 ) − 2Mga3 = 0
3a3 = ½L
⇒
1
a3 = 6 L .
⇒
Now that we have established a method of analysis, it’s straightforward to extend it
to a stack of four blocks. The torque due to gravity on the upper three blocks is
€ is τlower = +Mg · ( ½L – a4 ) sin 90º ,
τupper = −3Mg · a4 sin 90º and that on the bottom block
so the sum of the torques for the stationary stack must be
Mg · ( ½L – a4 ) − 3Mga4 = 0
4a4 = ½L
⇒
1
a4 = 8 L .
⇒
The total distance by which the leading edge of the top block in this stack
extends beyond the edge of the table is then
d = a1 + a 2 + a 3 + a 4 =
1
2
L +
1
4
€
1
25
L + L + 8 L = 24 L .
1
6
This tells us that the length of the uppermost block in fact lies completely beyond the
edge of the table!
€
We could go on to find the result for a stack of any number of identical blocks
arranged according to this rule. If we use n blocks, the total torque on the stack would
be Mg · ( ½L – an ) − ( n – 1 )Mgan = 0
⇒
nan = ½L
an =
⇒
amount by which the bottommost block extends over the edge of the table diminishes
toward zero as the number of blocks in the stack increases. What about the distance that
the top block extends past that edge? The leading edge of the top block reaches past the
1
1
1
1
€
table’s edge by a distance d = a1 + a 2 + a 3 + K + a n = 2 + 4 + 6 + K + 2n L .
If we let the number of blocks increase without end, this total distance becomes
L
1
1
1
1
1 1 1
1
d = 2 + 4 + 6 + K + 2n + K L = 1 + 2 + 3 + K + n + K
. If you have learned
2
€ you’ll recognize this unending sum in parentheses as the harmonic
about infinite series,
series, which also grows without limit. This means that we can make the top block of the
stack reach beyond the end of the table by any distance we wish, provided that we use a
large enough number of blocks!
(
(
€
1
L . So the
2n
)
(
)
)
b) Because this arrangment of the four blocks is exactly symmetrical, its center
of mass is at the exact geometrical center of the stack. (Note also that in this
configuration, the center of mass is located where there is no actual mass.) This stack
will then be in static balance (but will be only marginally stable) if the center of mass is
just above the edge of the table. So the bottom block overhangs the table’s edge by
exactly half the block’s length, hence b2 = ½L .
We can just look at one of the blocks in the middle layer, again because of the
stack’s symmetry. The forces acting on this block are its own weight, Mg ; half the
weight of the top block, ½Mg ; and the normal force upward from the bottom block, N .
The weight force for a block acts effectively as if it were concentrated at its center of
mass. The partial weight force from the upper block and the normal force from the
bottom block will act effectively at points midway between the inner edge of the middle
block and the end of the bottom block. The bottom block’s edge will serve as the pivot
point for any tipping of the middle block.
We will now look at the torques acting on the middle block. Its own weight force
acts at a distance b1 – ½L from the pivot point, so that will produce a torque
τW = −Mg · ( b1 – ½L ) · sin 90º . The partial weight force from the top block acts at a
distance ½ ( L – b1 ) from the pivot point, which provides a torque of
τu = + ( ½ Mg ) · ½ ( L – b1 ) · sin 90º . Since we want to find the arrangement for which
these blocks just remain in static balance, we consider the situation where the middle
block is just about to tip. In this case, the middle block is on the verge of breaking
contact with the bottom block, at which point the normal force is N = 0 , thereby
contributing no torque to the middle block. The net torque on the middle block in
marginal static balance is then
1
1
1
+ ( 2 Mg ) ⋅ 2 ( L − b 1 ) − Mg ⋅ (b 1 − 2 L ) = 0 ⇒
⇒ b1 +
€
€
1
b
4 1
=
1
2
L +
1
4
L ⇒
5
b
4 1
=
1
4
( L − b1 ) = b1 − 2 L
1
3
4
L ⇒ b1 =
3
L
5
.
The outer end of the middle block extends beyond the edge of the table by a
3
1
11
distance h = b 1 + b 2 = 5 L + 2 L = 10 L . So, as with part (a) of this problem, no
portion of this block is actually over the surface of the table. (The situation for the
middle blocks is an example of the concept of the architectural cantilever – albeit a
marginally stable one here.)
€
8)
a) When the bicycle wheel comes into contact with the step, there are four forces
acting on it at that moment: its own weight, Mg ; the normal force upward from the
ground, N ; the applied force, F ; and the reaction force from the point of contact with
the step, R . If a large enough force is applied (and in the right place, as we shall see),
the wheel will pivot about the edge of the step and lift up over it. For the minimal
amount of force that would just make this happen, the wheel would be very nearly in
static equilibrium. It would just begin to lift off the ground, causing the normal force N
to drop to zero. The reaction force R acts at the contact point where the wheel is
pivoting, so the moment arm along which it acts is zero, hence the torque it produces
on the wheel is also zero.
So only two of the aforementioned forces produce a torque on the wheel and
only one of these is unknown, so the torque equation alone will suffice to determine the
value of F . The magnitude of a torque
r
"
is given by
r
r r
" = r # F = r F sin $ ,
which is equivalent to the magnitude of the force F times the perpendicular (or closest)
distance, r sin θ , from the pivot point to the “line of action” of the force. Thus we find:
!
!
With the wheel just about to pivot around the step, the net torque is exactly zero, so the
minimum applied force needed to lift the wheel over the step is given by
+ Mg "
% 2r h # h 2 (
* . 2r h # h 2 # F " ( r # h ) = 0 $ F = Mg ''
*
r
#
h
&
)
The form of this result appear reasonable. When there is no step ( h = 0 ) , the
horizontal force required to lift the wheel becomes F = 0 . Somewhat less trivially, we
! also find that it is impossible to lift the wheel, with a horizontally applied force, over a
step with a height equal to or greater than the wheel’s radius ( F → ∞ as r → h ) .
Notice that the moment of inertia, I , of the wheel does not enter into the
calculation for F . This indicates that the applied force required would be the same if
the wheel were a uniform solid disk of the same mass and radius.
b) There is more than one way to analyze the motion of an object rolling down
an incline. We will use the most direct approach here.
The forces parallel to the incline are the parallel component of the weight force
and the rolling friction force, fr . If we call the “downhill” direction positive, then we
have Mg sin θ − fr = Ma|| . Rather than attempt to evaluate fr in terms of the normal
force, N , it will be convenient to solve this equation for fr and apply that result; thus,
fr = M ( g sin θ − a|| ) . [ A ]
To analyze the torques on the rolling object, we will place the reference point at its
center-of-mass. The weight force on the object then produces zero torque, since the moment
arm for this force is zero. The normal force points directly toward the object’s center of
mass along a radius (if the object is uniformly dense, or has a density distribution which is
radially symmetrical), so the torque it produces is τN = R · N · sin 180º = 0 .
Since the rolling friction force acts along the incline, it is perpendicular to the moment
arm, which is on a radius of the object’s cross-section; the torque produced by this
friction is then τf = R · fr · sin 90º = fr R .
For an object that rotates rigidly (all parts rotate at the same angular velocity),
the magnitude of the net torque is τnet = I α . When the object is rolling, the angular
acceleration is related to the linear acceleration by
α =
a
R
. For each of the objects
we are considering, the moment of inertia about an axis parallel to its length and
through its center-of-mass is given by ICM = C · MR2 , where C = 1 for the tube (the
same value as for a thin ring or hoop), C = ½ for the solid cylinder (same value as for a
uniform disk), and
C =
2
€ solid sphere. If we now combine the
for the uniform
5
equations related to torque with Equation A , we obtain
τ net = τ W + τ N + τ f = 0 + 0 + f r R = I CM α
€
⎛ a ⎞
⇒ M ( g sin θ − a|| ) ⋅ R = C M R 2 ⋅ ⎜ || ⎟ = C M R a||
⎝ R ⎠
€
⇒ g sin θ − a|| = C a||
⇒ a|| =
€
g sin θ
1 + C
.
Notice that the mass, radius, and length of the rolling object don’t matter at all (at least
in this idealization of rolling); only the distribution of mass about the center-of-mass
affects the acceleration of the object going down the slope.
€
As all three of our objects are rolling downhill through the same distance, they
will reach the bottom of the slope in order of decreasing linear acceleration. The
uniform solid sphere arrives there first, since its acceleration along the incline is
a|| sph =
g sin 22º
5
m.
=
g sin 22º ≈ 2.62
2
7
sec.2 . The uniform solid cylinder will arrive
1+
5
next, as its linear acceleration is
a|| cyl =
g sin 22º
2
m.
= g sin 22º ≈ 2.45
1
2 .
3
sec.
1+
2
€
Coming in third and last is the thin-walled tube, which accelerates down the slope at
g sin 22º
1
m.
=
g sin 22º ≈ 1.84
. Note that for a sliding block, for which
1+1
2
sec.2
€
g sin 22º
m.
= g sin 22º ≈ 3.67
C is effectively zero, we have a|| block =
, as expected.
1+ 0
sec.2
a|| tube =
€
€
Now that we have worked out the linear accelerations down the ramp, we can use
the kinematic equation for position to find the time each object requires to reach the
bottom of the ramp. We will call the starting point at the top of the incline x0 = 0 ; all
three objects start from rest ( v0 = 0 ) , so we can write
x f = x0 + v0 t +
1
1
a t 2 → L = 0 + 0 ⋅ t + a|| t 2
2 ||
2
⎛ 2L ⎞1/ 2
⇒ t = ⎜ ⎟ =
⎝ a|| ⎠
€
⎛
⎞1/ 2
2L
⎜
⎟ =
⎝ g sin θ [ 1 + C ] ⎠
⎛ 2 [ 1 + C ] L ⎞1/ 2
.
⎜
⎟
⎝ g sin θ ⎠
For this situation in this Problem, this becomes
€
⎛
⎞1/ 2
⎜ 2 [ 1 + C ] ⋅ 1.5 m. ⎟
t = ⎜
≈ 0.904 (1 + C )1/ 2 seconds .
⎟
m.
⎜ 9.81 2 ⋅ sin 22º ⎟
⎝
⎠
sec.
The three objects will thus reach the foot of the ramp after
€ solid sphere:
solid cylinder:
€
thin-walled tube:
2
t ≈ 0.904 (1 + 5 )1/ 2 ≈ 1.070 seconds ,
1
t ≈ 0.904 (1 + 2 )1/ 2 ≈ 1.107 seconds , and
t ≈ 0.904 (1 + 1)1/ 2 ≈ 1.278 seconds .
€ block would reach the bottom in 0.904 seconds.
By contrast, the sliding
9)
€
a) In this situation, due to the absence of friction, the two masses are in a
marginally stable static equilibrium.
We will call the downhill motion of m1 positive, which automatically makes the uphill
motion of m2 positive in turn. The net force on m1 acting parallel to the incline it is on
is m1g sin 40º − T = 0 , while the net force on m2 acting parallel to its incline is
T − m2g sin
θ2 = 0 . When we solve each of these equations for the tension T in the
connecting cord and then equate the results, we find m1g sin 40º = T = m2g sin θ2 .
Since both of the masses of the blocks are known, we can solve for the angle θ2 that is
required in order for this static equilibrium to exist:
# 1.8 kg. &
#m &
o
sin " 2 = % 1 ( sin 40 o ) %
( * 0.643 ) 0.386 + " 2 ) 22.7
m
$ 2'
$ 3.0 kg. '
.
b) With the inclusion of friction, the static equilibrium of the two masses on
their inclines becomes far more stable. In this portion of the Problem, the angle for the
! incline m2 rests upon is chosen to be θ2 = 32º . There are now two cases to consider
for the limits of static equilibrium: one in which the mass m1 is just prevented from
sliding downhill, the other in which it is just prevented from being drawn uphill.
The forces on each mass in the case where m1 resists sliding downhill are
forces perpendicular
to incline
forces parallel to
incline
m1
m2
o
N1 " m1g cos 40 = m1a# = 0
N 2 " m2 g cos 32 o = m 2 a# = 0
m1g sin 40 o " T " f s1
T " m2 g sin 32 o " f s 2
max
= m1a || = 0
max
= m2 a || = 0
From the equations for the perpendicular (normal) forces on the blocks, we obtain
!
!
N1 = m1g cos
! 40º and N2 = m2g cos 32º . Since
! the limits of static friction on each
f
=
µ
N
and
f
=
µ
N
block are s 1
s 1
s2
s 2 , the equations for the forces parallel to
max
max
each incline can be solved for the tension T in the cord, yielding
!
m1g sin 40º −
µsm1g cos 40º = T = m2g sin 32º + µsm2g cos 32º
.
If we now solve this last equation for m2 , we have
# sin 40 o " µ cos 40 o &
# 0.643 " 0.38 ) 0.766 &
s
m2 = %
)
m
*
(
%
( )1.8 kg. * 0.74 kg. ,
1
o
o
$ 0.530 + 0.38 ) 0.848 '
$ sin 32 + µ s cos 32 '
which is the smallest mass m2 may have in order to keep m1 from sliding downhill.
!
If the block m2 is massive enough, it will instead apply enough tension in the
cord to pull block m1 uphill. The limit of static friction on m1 in this direction is
found by changing the equations for the forces parallel to each incline to
m1g sin 40 o " T ' + f s1
#
max
= m1a || = 0 and
T ' " m2 g sin 32 o + f s 2
#
max
= m2 a || = 0 ;
the equations for the normal forces are not affected. Repeating the rest of the
calculations as we did before, we now have
# sin 40 o + µ cos 40 o & !
# 0.643 + 0.38 ) 0.766 &
s
m2 = %
)
m
*
(
%
( )1.8 kg. * 8.09 kg. ,
1
o
o
$ 0.530 " 0.38 ) 0.848 '
$ sin 32 " µ s cos 32 '
!
the greatest mass this block may have without overcoming static friction and pulling the
other block along after it.
!
c) We continue to use the Atwood machine described in part (b), but now with a
mass m2 = 9.2 kg. We have seen that this will be more than sufficient mass to
overcome static friction and set both blocks sliding along their inclines. The normal
forces acting on the blocks will still be as they were in part (b). However, with the
blocks in motion, their accelerations will no longer be zero; with kinetic friction acting
on the blocks, the equations for the forces parallel to the inclines become
m1g sin 40 o " T ' ' + f k 1 = m1a ||
and
T ' ' " m2 g sin 32 o + f k 2 = m2 a || ,
with f k = µkN . We now solve each equation for the new cord tension T’’ and equate
the results to find
!
!
m1g sin 40 + µ k m1g cos 40 " m1a || = T ' ' = m 2 g sin 32 o " µ k m2 g cos 32 o + m 2 a ||
o
o
" ( m1 + m2 ) # a || = m1g sin 40 o + µ k m1g cos 40 o $ m 2 g sin 32 o + µ k m2 g cos 32 o
!
!
$ m (sin 40 o + µ cos 40 o ) # m (sin 32 o # µ cos 32 o ) '
k
2
k
" a || = & 1
)*g
m
+
m
%
(
1
2
% 1.8 kg. # ( 0.643 + 0.26 # 0.766) $ 9.2 kg. # ( 0.530 $ 0.26 # 0.848) (
m.
" '
)
* # ( 9.81
sec.2
1.8 + 9.2 kg.
&
)
!
" # 0.121 g or #1.19
m.
sec.2
.
!
!
The negative sign for this acceleration agrees with our expectation that m1 is being
pulled uphill and it is m2 that is sliding downhill.
-- G.Ruffa
original notes developed during 2006-08
revised: August-September 2010, February 2011