Download P. LeClair

PH 125 / LeClair
Spring 2014
Problem Set 2
Daily problem due 24 Jan 2014
1. A car is traveling at a constant velocity of 35 m/s and passes a stopped police cruiser. Exactly
2.5 s after passing, the cruiser begins pursuit, with a constant acceleration of 2.5 m/s2 . How long
does it take for the cruiser to overtake the car (from the moment the cop car starts moving)?
Solution: Given: The constant velocity and initial position of a speeding car, the constant acceleration and initial position of a pursuing police cruiser 2.5 s later.
Find: How long it takes the police cruiser to overtake the car. This means we need the position
versus time for each vehicle, from which we can find their intersection point.
Sketch: Let the car’s and cruiser’s initial position be x = 0, with the direction of travel being
the +x axis. Let time t = 0 be when the cruiser begins accelerating. Thus, the car starts off with
~
vi = 35 m/sı̂ at t = −2.5 s from x = 0, and the police cruiser starts off with ~
ai = 2.5 m/s2 ı̂ at t = 0 ,
also from x = 0.
bandit
|!vi |
x
t = −2 s
O
smokey
bandit
|!vi |
x
O
A car chase. Upper: at time t =
−2.5 s the car passes the stationary
police cruiser. Lower at time t = 0 s,
the police cruiser begins chase with
constant acceleration.
t=0
smokey
!ai
Relevant equations: We need the general equation for position in one dimension under the
conditions of constant (or zero) acceleration:
1
x(t) = xi + vix t + ax t2
2
(1)
Symbolic solution: For the car, we have xi = 0 and ax = 0. Let the car’s initial velocity be vix .
The car has been moving at constant velocity for two seconds before the cruiser starts, so we need
to shift the time coordinate from t 7→ (t + 2.5).
xcar = vix (t + 2.5 s)
(2)
You can verify that this correctly gives xcar (−2.5 s) = 0. In order to keep things more general,
however, let us say that the cruiser starts δt seconds later (we can later set δt = 2.5 s):
xcar = vix (t + δt)
(3)
For the police cruiser, we have xi = 0 and vix = 0. Let the cruiser’s initial acceleration be ax . The
cruiser starts out at t = 0, so things are simple:
1
xcop = ax t2
2
(4)
We need to find the time t at which xcar = xcop , from which we can get the time it takes for the
police cruiser to overtake the car, viz. t − δt.
xcar = xcop
1
vix (t + δt) = ax t2
2
1
vix t + δt vix = ax t2
2
0 = ax t2 − 2vix t − 2δt vix
=⇒
t=
2vix ±
(5)
(6)
(7)
(8)
q
2 + 8δt v a
4vix
ix x
(9)
2ax
It is the “+” solution in the ± we want, as our numerical solution below will verify.
Numeric solution: Using the numbers we are given:
t=
2vix ±
q
2 + 8δt v a
4vix
ix x
=
2 (35 m/s) ±
q
4 (35 m/s)2 + 8 (2.5 s) (35 m/s) (2.5 m/s2 )
2ax
2 (2.5 m/s2 )
(70 ± 81.5) m/s
70 m/s ± 4900 m2 /s2 + 1750 m2 /s2
=
=
= {30, −2.3} s
2
5 m/s
5 m/s2
p
(10)
The negative solution we can reject as unphysical, so it takes approximately 30 s for the cruiser to
overtake the car.
Double check: We carried units throughout the numerical phase of the calculation, and our end
result comes out in seconds as required. Another check we can make is to calculate the position of
both the car and cruiser at the time we found above to be sure they are the same:
sign.
xcar = vix (t + δt) = (35 m/s) (30.3 + 2.5 s) ≈ 1148 m −−−→ 330 m
(11)
1
1
sign.
xcop = ax t2 =
2.5 m/s2 (30.3 s)2 ≈ 1148 m −−−→ 330 m
digits
2
2
(12)
digits
Within the precision implied by the number of significant digits, the two positions are the same;
our answer is reasonable.
Daily problem due 27 Jan 2014:
2. HRW 5.31 A block is projected up a frictionless inclined plane with an initial speed of vo =
2.50 m/s. The angle of incline is θ = 17.0◦ . (a) How far up the plane does the block go? (b) How
long does it take to get there? (c) What is its speed when it gets back to the bottom?
Solution: The acceleration of the block, pointing down the ramp, is −g sin θ. If it is to travel a
distance d up the ramp given an initial velocity vi , and reach final velocity vf = 0, then:
vf2 − vi2 = 2ad = −2gd sin θ
(13)
vi2
≈ 1.09 m
2g sin θ
(14)
d=
Since we know the acceleration and initial velocity, we can find the time readily.
v(t) = vi + at = vi − gt sin θ = vf = 0
vi
≈ 0.87 s
t=
g sin θ
(15)
(16)
What is the speed at the bottom? Same as it was on the way up (since we have no friction). We
can verify that, noting that moving down the ramp the acceleration is now a = +g sin θ, and the
mass moves through distance d starting from rest:
vf2 − vi2 = 2gd sin θ
vf2 = 2gd sin θ = 2g
(17)
vi2
2g sin θ
!
sin θ = vi2
vf = |vi |
(18)
(19)
The problems below are due by the end of the day on 29 Jan 2014.
3. A projectile is launched with initial velocity ~
vi from the start of a ramp, with the ramp making
an angle ϕ with respect to the horizontal. The projectile is launched with an angle θ > ϕ with
respect to the horizontal. (a) At what position along the ramp does the projectile land? (b) What
angle θ maximizes the distance the particle makes it along the ramp (your answer will be in terms
of the angle ϕ? Note no numeric solution is required.
⃗vi
θ ϕ
Figure 1: A projectile is launched onto a ramp.
Solution: Find: The point at which a projectile launched from the base of a ramp inclined at
angle ϕ hits the ramp.
Given: The projectile’s launch speed and angle, the ramp angle.
Sketch: Let the origin be at the projectile’s launch position, with the x and y axes of a cartesian
coordinate system aligned as shown below.
⃗vi
tan ϕ =
θ ϕ
dy
= slope
dx
Figure 2: Where does the projectile hit the ramp?
Thus ramp begins at position (0, 0), and the projectile is launched from (0, 0). We seek the intersection of the projectile’s trajectory with the surface of the ramp at position (xhit , yhit ), subject to
the condition that yhit ≥ 0, i.e., the projectile actually reaches the ramp.
Relevant equations: We have already derived the trajectory y(x) for a projectile launched from
the origin:
yp = x tan θ −
gx2
2|~
vi |2 cos2 θ
(20)
The ramp itself can be described by a simple line. We know the slope m of the ramp is m =
∆x/∆y = tan ϕ, and we know it intersects the point (xo , yo ) = (0, 0). This is sufficient to derive an
equation of the line describing the ramp’s surface, yr (x), using point-slope form:
yr − yo = m (x − xo )
yr = (tan ϕ) x
The distance l the projectile goes along the ramp surface is found simply from (xhit , yhit ) or xhit
and ϕ:
l=
q
2
x2hit + yhit
or
l=
x
cos ϕ
(21)
We also need a sanity condition to check that the projectile actually hits the ramp, which means
we require yhit > 0 and xhit > 0. Finally, the point of intersection must occur when yr = yp ≡ yhit .
Symbolic solution: We need only impose the condition yr = yp to begin our solution. The
resulting x value is the xhit we desire.
yr = x tan ϕ = yp = x tan θ −
gx2
2|~
vi |2 cos2 θ
note x = l cos ϕ
gl2 cos2 ϕ
2|~
vi |2 cos2 θ
gl cos ϕ
tan ϕ = tan θ −
2|~
vi |2 cos2 θ
2|~
vi |2 cos θ
2|~
vi |2 cos2 θ
(tan θ − tan ϕ) =
(sin θ − tan ϕ cos θ)
l=
g cos ϕ
g cos ϕ
2|~
vi |2 cos θ
2|~
vi |2 cos θ sin (θ − ϕ)
l=
(sin
θ
cos
ϕ
−
sin
ϕ
cos
θ)
=
g cos2 ϕ
g cos2 ϕ
l cos ϕ tan ϕ = l cos ϕ tan θ −
(22)
(23)
(24)
(25)
(26)
(27)
As we should expect, the distance up the ramp depends on the relative angle between the ramp
and launch, θ−ϕ. We could have found this result a bit quicker if we had noted the identity (which
we basically just derived).
tan θ − tan ϕ =
sin (θ − ϕ)
cos θ cos ϕ
(28)
In order to maximize the distance up the ramp with respect to launch angle, we require dl/dθ = 0.
This is simple enough:
vi |2 cos θ sin (θ − ϕ)
d 2|~
2|~
vi |2
dl
=
=
cos (ϕ − 2θ) = 0
dθ
dθ
g cos2 ϕ
g cos2 ϕ
(29)
If we restrict ourselves to sensible ramp angles, ϕ ∈ [0−π/2], this implies ϕ−2θ = π2 . In the limiting
case of no ramp (ϕ = 0, launch over flat ground), this predicts θ = π/4 for maximum range, which
we know to be correct.
Incidentally, if you solve the more general problem of launching the projectile a distance d before
the start of the ramp (see picture below), this is the result:
"
|~
vi |2 cos2 θ
l=
g cos ϕ
"
|~
vi |2 cos θ
=
g cos2 ϕ
# "
s
tan θ − tan ϕ +
s
#"
(30)
#
sin2 (θ
sin (θ − ϕ) +
#
2d tan ϕ
d
(tan ϕ − tan θ) +
−
2
2
|~
vi | cos θ
cos ϕ
2
2d sin ϕ cos ϕ
d
− ϕ) +
−
2
2
|~
vi | cos θ
cos ϕ
(31)
You can verify that for d = 0 you recover our previous result. This presumes you have a launch
velocity sufficient to reach the ramp in the first place. That requirement you can find by setting
the projectile’s range equal to d, and you find
|~
vi |2 sin 2θ ≥ gd
(32)
I’ll leave it as an exercise to you to find the optimum launch angle in this case . . .
!vi
θ
ϕ
d
Figure 3: Launching the projectile before the start of the ramp.
Numerical solution:
Nothing much we can do here but make some nice plots. Here’s the distance along the incline as a
function of launch angle for various ramp angles, with the distance in units of 2vi2 /g for convenience:
Since the maximum distance for a given ϕ depends on ϕ − 2θ, if we instead plot the distance l
versus θ − ϕ/2 all the curves should line up symmetrically about θ−ϕ/2 = π/4. This is an important
conclusion: just as the angle for maximum range over level ground was π/4, for maximum range
up the ramp the launch angle minus half the ramp angle should be π/4.
distance to impact along incline (2vo2 /g)
ramp inclination
0◦
15 ◦
30 ◦
45 ◦
60 ◦
0.5
0.4
0.3
0.2
0.1
0.00
10
20
30
40
50
60
launch angle ( ◦ )
70
80
90
distance to impact along incline (2vo2 /g)
ramp inclination
0◦
15 ◦
30 ◦
45 ◦
60 ◦
0.5
0.4
0.3
0.2
0.1
0.00
10
20
30
40
50
60
70
launch angle - 1/2 ramp angle ( ◦ )
80
90
I’ve included the python code for the first plot at the end of this solution set in case you are curious.
Double check:
We’ve already checked the limit ϕ = 0 and verified that it makes sense, and our numerical simulations give sensible results that scale with relative angle in the right way. We could also check
the case of a vertical “ramp” which should give us free-fall motion. For ϕ = π/2 we find θ = π/2 the best launch for maximum height straight up is to launch the projectile straight up. Makes sense.
4. A mass m is released from rest at height h on the top of a ramp of inclination θ. The coefficient
of kinetic friction between the ramp and mass is µk . The block slides down the ramp, up a second
identical ramp, back down again, and so forth.
(a) After one round trip (from the top of the first ramp and back again), how far away from the
mass from its starting point?
(b) How about after n round trips?
h
h
θ
θ
Figure 4: A block is let go from the top of a ramp sitting on a table.
(c) Does the mass ever stop?
Solution: First we need the acceleration of the block down the ramp, Fig. 5. Align the x axis with
the ramp, and the y axis perpendicularly to it. The weight of the block mg points downward, the
normal force perpendicular to the ramp along +y, and the frictional force fk opposes the motion
of the block in the −x direction.
θ
h
θ
Figure 5: Going down one ramp . . .
Balancing forces along y gets us the normal force:
X
Fy = n − mg cos θ
=⇒
n = mg cos θ
(33)
Balancing forces along x gets us the acceleration down the ramp. We know the frictional force is
fk = µk n, so we’ll need the equation above for n.
X
Fx = max = mg sin θ − fk = mg sin θ − µk n = mg (sin θ − µk cos θ)
(34)
ax = g (sin θ − µk cos θ)
(35)
Since we have a constant acceleration, we can find the speed at the bottom of the ramp vf . The
block starts with vi = 0, and travels a lateral distance down the ramp of ∆x = h/ sin θ. Thus,
vf2 = vi2 + 2ax ∆x = 2g (sin θ − µk cos θ)
h
sin θ
(36)
The block then starts up the second ramp. Presume it goes to a height h0 up the second ramp.
If we find the acceleration, we can find the distance ∆x0 = h0 / sin θ it travels up the second ramp.
First another quick free body diagram.
θ
θ
Figure 6: . . . and then up the next.
Proceeding as before,
X
Fy = n − mg cos θ
n = mg cos θ
=⇒
(37)
Balancing forces along x gets us the acceleration down the ramp. We know the frictional force is
fk = µk n, so we’ll need the equation above for n.
X
Fx = ma0x = −mg sin θ − fk = mg sin θ − µk n = −mg (sin θ − µk cos θ)
(38)
a0x = −g (sin θ − µk cos θ)
(39)
With the new acceleration, we can find the distance up the ramp. We start with velocity vi0 = vf as
found above, and end with velocity vf0 = 0.
vf0
2
= 0 = vi0
2
+ 2a0x ∆x0 = 2g (sin θ − µk cos θ)
h
h0
− 2g (sin θ − µk cos θ)
sin θ
sin θ
h0
sin θ − µk cos θ
∆x0
=
=
h
sin θ + µk cos θ
∆x
(40)
(41)
We also noted that the ratio of the heights h0 /h is the same as the ratio of the distances up the
ramp, ∆x0 /∆x. What happens for subsequent trips? The ratio of the starting and ending heights
for one trip depends only a dimensionless ratio involving θ and µk . This ratio will be exactly the
same every time we make a round trip. Thus, for one full round trip, we must have
h00
=
h
sin θ − µk cos θ
sin θ + µk cos θ
2
=
∆x00
∆x
(42)
For n round trips, we pick up this squared ratio every time:
h(n)
=
h
sin θ − µk cos θ
sin θ + µk cos θ
2n
=
∆x(n)
∆x
(43)
How do we check this? First, if µk = 0 we should find h(n) /h = 1, which is the case. If there is no
friction, the block just keeps coming back to the same height over and over again. The block only
stops if we have some dissipative interaction present like friction.
Does the block really stop? The ratio of the heights or distances is a geometric series that smoothly
converges to zero, which is enough to tell us that it does stop. We can also demonstrate that the net
distance covered considering the sum of all trips up and down the ramps is finite, so the block must
indeed stop in finite time. The net distance traveled along the ramps is the sum of the distances
for each of the n ’half trips’ (not round trips), starting with n = 0 to go down the ramp the first
time. We also have to remember that what we calculated is how far the block goes up the ramp for
a given trip, it has to go back down the same distance. The total distance is then twice the sum
of all trips up the ramp. For the nth trip, we know from equation 43 that the distance traveled
∆x(n) is just a fraction to the nth power. This makes the net distance a geometric series, which in
∞
P
rn = 1/ (1 − r) provided the ratio r is less than 1 (which it is in our case,
general has the sum
n=0
since 0 ≤ θ ≤ π/2). We can then find the total distance covered before the block stops.
net distance = 2∆x
∞ X
sin θ − µk cos θ n
n=0
sin θ + µk cos θ
=
2∆x
1−
2 (sin θ + µk cos θ)
sin θ + µk cos θ − (sin θ − µk cos θ)
1
= ∆x 1 +
tan θ
µk
net distance = ∆x
(44)
sin θ−µk cos θ
sin θ+µk cos θ
= ∆x
sin θ + µk cos θ
µk cos θ
(45)
(46)
This sums the initial trip plus all following ‘half round trips’ that bring it back up another ramp,
accounting for the fact the particle goes back down the ramp the same distance. It is a finite
distance for non-zero µk , which tells us indeed that the block does stop even though in principle
it makes an infinite number of transits. This is closely related to Zeno’s paradoxi , or if you like,
simply a result of the fact that an infinite series can indeed have a finite sum.
How to check this result? We can examine limiting cases: no friction, and infinite friction. If µk → 0,
the distance diverges because the particle never stops moving. If we let the frictional force become
arbitrarily large (µ → ∞), the net distance is just ∆x - the block makes it down the first ramp
and gets stuck right there. Of course, if friction were really infinite the block would never start
moving in the first place, but our model presupposes that the block did start moving somehow, but
correctly predicts it won’t go up the second ramp at all. Finally, the smaller we make µk , the more
distance the particle covers before stopping, which make sense.
5. HRW 5.57 A block of mass ma = 3.70 kg on a frictionless plane inclined at an angle θ = 30.0◦ is
connected by a cord over a massless, frictionless pulley to a second block of mass mb = 2.30 kg (figure
below). The coefficient of kinetic friction between block B and the horizontal plane is µk = 0.20.
What are (a) the magnitude of the acceleration of each block, (b) the direction of the acceleration
i
See http://en.wikipedia.org/wiki/Zeno’s_paradoxes.
of the hanging block, and (c) the tension in the cord?
mb
ma
µk
θ
Solution: I did not specify the coefficient of friction for block b in the original problem, so if you
didn’t catch the update and only solved this symbolically that is fine.
Find: The tension in a cord connecting two blocks and the system’s acceleration, with one block
on a frictionless incline and the second on a flat surface with coefficient of kinetic friction µk .
Given: The mass of both blocks, the coefficient of friction for the block on the flat surface, and
the angle of incline for the ramp.
Sketch: We need free-body diagrams for each mass. Note the axis definitions for each mass. The
system will obviously accelerate to the left, as mass A slides down the ramp and pulls mass B with
it, meaning the acceleration is in the −x direction according to the sketch below.
n
n
T
fk
T
θ
mb g
ma g
y
y
x
x
Only for mass B do we need to consider a friction force. Since the rope is presumably taut the
entire time of interest, the acceleration is the same for both blocks. For the same reason, the tension
applied to both blocks is the same.
Relevant equations: Newton’s second law and geometry will suffice. Along the y direction for
either mass, the forces must sum to zero, while along the x direction, the forces must give the
acceleration for each mass.
X
Fy = 0
X
Fx = max
Symbolic solution: First consider mass A. The free body diagram above yields the following,
noting that the acceleration will be purely along the −x direction:
X
Fy = n − ma g cos θ = 0
X
Fx = T − ma g sin θ = −ma ax
ax = g sin θ −
=⇒
T
ma
For mass B, we must include the frictional force, fk = µk n.
X
Fy = n − mb g = 0
X
Fx = fk − T = µk mb g − T = −mb ax
=⇒
n = mb g
Using the latter equation with our result for the acceleration above allows us to solve for the tension
in terms of known quantities:
T = µk mb g + mb ax = µk mb g + mb g sin θ −
mb
T 1+
ma
mb
T
ma
= µk mb g + mb g sin θ
µk mb g + mb g sin θ
ma mb
T =
= g (µk + sin θ)
1 + mb /ma
mb + ma
Finally, this gives an expression for the acceleration in terms of known quantities:
T
mb
ma sin θ − µk mb
ax = g sin θ −
= g sin θ − g (µk + sin θ)
=g
ma
mb + ma
mb + ma
Numeric solution: Given ma = 3.70 kg, mb = 2.30 kg, µk = 0.2, and θ = 30◦ ,
ma mb
T = g (µk + sin θ)
≈ 9.74 N
mb + ma
and
ax = g
ma sin θ − µk mb
≈ 2.27 m/s2
mb + ma
6. A particle of mass m is subjected to the following force:
F (v) = a − bv
(47)
where a and b are positive constants with appropriate units, and v is the magnitude of the velocity.
(a) Find an expression for the velocity of the particle as a function of time in terms of a, b, m,
and the initial velocity vo ≥ 0. Express the velocity in the limit t → ∞ in terms of these same
quantities.
(b) Find an expression for the position of the particle as a function of time in terms of a, b, m,
vo , and the initial position xo .
(c) If the particle starts at rest, how far does it travel before its velocity is one half its maximum
velocity?
Solution: This was actually a problem on our graduate qualifying exam this year. We start just
by using Newton’s second law, and noting a = dv/dt.
F = a − bv = ma = m
dv
dt
(48)
We don’t know the velocity yet, but we know it is proportional to its own derivative. Right off the
bat we can guess it will be some kind of exponential function. But how to solve this equation?
Now, if we make sure our math professors aren’t watching, we can manipulate the differentials as
fractions and separate this equation.ii That is, put all the v-related stuff on one side, and t-related
stuff on the other.
dt
dv
=
m
a − bv
(49)
Now we can just integrate both sides, from time 0 to time t where the particle goes from initial
velocity vo to final velocity v:
ii
See http://www.theshapeofmath.com/oxford/physics/year1/calc/sepdiff for why this is sketchy, though it
works, and not OK in more than one dimension at all.
Zt
dt
=
m
0
Zv
vo
dv
a − bv
(50)
t
v
1
t
+ C = − ln (a − bv) + C m
b
(51)
vo
0
t
1
1
1
a − bv
= − ln (a − bv) + ln (a − bvo ) = − ln
m
b
b
b
a + bvo
a
−
bv
e−bt/m =
additional algebra ensues . . .
a + bvo
a
1 − e−bt/m + vo e−bt/m
v=
b
(52)
(53)
(54)
Looking at a plot of the particle’s velocity versus time, we can see under the influence of a drag
force it asymptotically approaches a terminal value of vt = a/b. This is borne out by taking the
limit as t → ∞:
a
≡ vt
b
v(t) = vt 1 − e−bt/m + vo e−bt/m
lim v(t) =
t→∞
=⇒
(55)
(56)
Starting with velocity vo , the particle smoothly approaches a limiting velocity of vt , with a characteristic time scale m/b. We can integrate once more to find position:
Zx
x(t) =
v(t) dt = vt t + vt
b
b −bt/m
e
− 1 + vo
1 − e−bt/m
m
m
(57)
0
= vt t +
b
(vo − vt ) 1 − e−bt/m
m
(58)
Here are some plots of x(t) and v(t) for various drag coefficients b/m. The starting velocity is set
at vo = 2.0 m/s, with vt = 1.0 m/s. The larger the drag coefficient b/m, the more quickly the particle
reaches terminal velocity.
How far does the particle travel before its velocity is 12 vt ? We just have to set v(t) = 21 vt to find
how long that takes, and plug that time into the position equation. It is tedious, so I’ll just tell
you the time at which v(t) = 21 vt and the resulting position.
m
2 (vt − vo )
t=
ln
b
vt
vt m
2 (vt − vo )
b
3
x=
ln
+
vo − vt
b
vt
m
2
(59)
(60)
1.0
v (m/s)
0.8
0.6
0.4
0.2
0.00
5
x (m)
4
1
2
3
4
5
3
4
5
Drag coefficient b/m
3
2
0.1
0.25
0.5
0.75
1
00
1
2
time (s)
7. HRW 6.30 A toy chest and its contents have a combined weight of 180 N. The coefficient of static
friction between toy chest and floor is µs = 0.42. A child attempts to move the chest across the floor
by pulling on an attached rope. (a) If the rope makes an angle of θ = 42◦ with the horizontal, what
is the magnitude of the force ~
F that the child must exert on the rope to pull the chest on the verge
of moving? (b) Write an expression for the magnitude F required to pull the chest on the verge of
moving as a function of the angle θ. Determine the value of θ for which F is (c) a minimum and
(d) a maximum magnitude.
Solution: Here’s a quick free-body diagram:
n
θ
F!
y
mg
x
Along the y direction, the net force must be zero for the block to stay on the floor:
X
Fy = N + F sin θ − mg = 0
(61)
N = mg − F sin θ
(62)
Along the horizontal direction, we want the box to be on the verge of moving, so the point where
acceleration is still zero:
X
Fx = F cos θ − µs N = F cos θ − µs mg + µs F sin θ = 0
µs mg
F =
≈ 74 N
cos θ + µs sin θ
(63)
(64)
Here’s a plot of F (θ). It is clear that there is a minimum at θ 6= 0.
1.0
coefficient of friction µs
0.1
0.2
0.3
0.4
0.5
0.8
F/mg
0.6
0.4
0.2
0.00
10
20
30
40
50
pull angle ( ◦ )
60
70
80
90
Figure 7: Force required to initiate motion as a function of pulling angle.
The force will be minimum when dF/dθ = 0:
d
µs mg
dθ cos θ + µs sin θ
sin θ − µs cos θ
= µs mg
=0
(cos θ + µ sin θ)2
(65)
The pre-factor and denominator are irrelevant; the equation above will only be zero and the force
at a minimum if sin θ = µs cos θ, or when tan θ = µs . In this case, that implies θ ≈ 22.8◦ for minimum
force.
What about the maximum? That is easier: it takes the most force when you are at 90◦ . You don’t
even really need math to prove this, though an inspection of F (θ) and the plot above will confirm.
At 90◦ , no motion will be initiated until you pull upward with a force equal to the box’s weight
mg. But, at that point, you can pull vertically as hard as you want and the box will never move
horizontally! Really, our graph should diverge at θ = 90, since we know a vertical pull can’t move
the box horizontally.
8. A point mass m starts from rest from the top of a stationary sphere of radius R and slides down
the frictionless surface. (a) At what angle θ (measured with respect to the vertical) does the mass
fly off the sphere? (b) If there is friction between the mass and the sphere, does the mass fly off
at a greater or smaller angle as the one found in part (a)? Explain your reasoning. An explicit
calculation is not necessary if a good physical argument can be put forth.
Appendix: Python code for plots in problems 3, 7
import math
import numpy a s np
import m a t p l o t l i b . p y p l o t a s p l t
p = [ 0 , 15 , 30 , 45 , 60]
# j u s t s i m u l a t e f o r a few i n t e r e s t i n g p h i v a l u e s
l a b e l s = [ ’ 0 $ ^\ c i r c $ ’ , ’ 15 $ ^\ c i r c $ ’ , ’ 30 $ ^\ c i r c $ ’ , ’ 45 $ ^\ c i r c $ ’ , ’ 60 $ ^\ c i r c $ ’ ]
colors = [ ’ r ’ , ’g ’ , ’b ’ , ’k ’ , ’o ’ ]
def x ( t , p ) :
return ( np . c o s ( t ) ∗ np . s i n ( t−p ) ) / ( np . c o s ( p ) ) ∗ ∗ 2
#f o r t e c h n i c a l r e a s o n s , u s e t h e s i n ( ) and c o s ( ) d e f i n e d by numpy , n o t math
#t h e l a t t e r can ’ t h a n d l e a r r a y i n p u t s , t h e former can .
#s p e c i f y a ran ge o f l a u n c h a n g l e s f o r t h e p l o t
t = np . a r a n g e ( 0 , 9 0 , 0 . 1 )
plt . axis ([0.0 ,90 , 0.0 ,0.6])
ax = p l t . gca ( )
ax . s e t _ a u t o s c a l e _ o n ( F a l s e )
f o r i in r a n g e ( l e n ( p ) ) :
p l t . p l o t ( t , x ( np . r a d i a n s ( t ) , np . r a d i a n s ( p [ i ] ) ) , l a b e l=l a b e l s [ i ] )
p l t . x l a b e l ( ’ l a u n c h ␣ a n g l e ␣ ( $ ^\ c i r c $ ) ’ )
p l t . y l a b e l ( ’ d i s t a n c e ␣ t o ␣ impact ␣ a l o n g ␣ i n c l i n e ␣ ( $2v_o ^2/ g$ ) ’ )
p l t . l e g e n d ( t i t l e =" ramp␣ i n c l i n a t i o n " , frameon =0)
#p l t . show ( ) #w r i t e t o s c r e e n
p l t . s a v e f i g ( ’ p r o j e c t i l e −ramp−p l o t s . p d f ’ , bbox_inches= ’ t i g h t ’ ) #w r i t e t o f i l e
import math
import numpy a s np
import m a t p l o t l i b . p y p l o t a s p l t
c = [0.1 , 0.25 , 0.5 , 0.75]
# d r a g c o e f f v a l u e s , c=b /m
l a b e l s = [ ’ 0.1 ’ , ’ 0.25 ’ , ’ 0.5 ’ , ’ 0.75 ’ ]
colors = [ ’ r ’ , ’g ’ , ’b ’ , ’k ’ , ’o ’ ]
v t = 1 #t e r m i n a l v e l o c i t y
vo = 0 . 2 #i n i t i a l v e l o c i t y
def v ( t , c ) :
return v t ∗(1−np . exp(−c ∗ t ))+ vo ∗np . exp(−c ∗ t )
#f o r t e c h n i c a l r e a s o n s , u s e t h e exp and l o g d e f i n e d by numpy , n o t math
#t h e l a t t e r can ’ t h a n d l e a r r a y i n p u t s , t h e former can .
def x ( t , c ) :
return v t ∗ t + c ∗ ( vo−v t )∗(1 −np . exp(−c ∗ t ) )
#s p e c i f y a ran ge o f t i m e s
t = np . a r a n g e ( 0 , 5 , 0 . 1 )
# one sub−p l o t f o r v ( t ) . s t a c k s u b p l o t s v e r t i c a l l y
plt . subplot (211)
#2 rows , 1 c o l , t h i s i s p l o t 1
p l t . y l a b e l ( ’ v␣ (m/ s ) ’ )
# x axis label
f o r i in r a n g e ( l e n ( c ) ) :
p l t . p l o t ( t , v ( t , c [ i ] ) , l a b e l=l a b e l s [ i ] )
#s i m u l a t e f o r v a r i o u s b /m
# one s u b p l o t f o r v ( t )
plt . subplot (212)
p l t . y l a b e l ( ’ x␣ (m) ’ )
f o r i in r a n g e ( l e n ( c ) ) :
p l t . p l o t ( t , x ( t , c [ i ] ) , l a b e l=l a b e l s [ i ] )
p l t . x l a b e l ( ’ bt /m ’ )
#s h a r e d l a b e l on x a x i s
p l t . l e g e n d ( t i t l e =" Drag ␣ c o e f f i c i e n t ␣b/m" , frameon =0 , l o c =2)
p l t . show ( ) #w r i t e t o s c r e e n
#p l t . s a v e f i g ( ’ x v t −drag −1D−s c a l e d . p d f ’ , b b o x _ i n c h e s =’ t i g h t ’ ) #w r i t e t o f i l e