Download Physics 18 Spring 2011 Homework 10

Physics 18 Spring 2011
Homework 10 - Solutions
Wednesday March 30, 2011
Make sure your name is on your homework, and please box your final answer. Because
we will be giving partial credit, be sure to attempt all the problems, even if you don’t finish
them. The homework is due at the beginning of class on Wednesday, April 6th. Because
the solutions will be posted immediately after class, no late homeworks can be accepted! You
are welcome to ask questions during the discussion session or during office hours.
1191
Static
and Elasticity
1. By considering the torques about the centers of the ball joints
in Equilibrium
your shoulders,
estimate the force your deltoid muscles (those muscles on top of the shoulder) must
extended
arm Farm
shown
at an angleat
θ with
the horizontal.
deltoid is
exert on your upper arm to
keep your
held
outacting
and extended
shoulder
level. The weight
of your
armexert
is thewhen
gravitational
force
m g exerted
through the center
Then, estimate the force they
must
you hold
a 10Fglb= weight
outby
toEarth
the side
at arm’s length.
of gravity of your arm. We can use the condition for rotational equilibrium to
estimate the forces exerted by your deltoid muscles. Note that, because its moment
————————————————————————————————————
arm is zero, the torque due to Fshoulder about an axis through point P and
perpendicularSolution
to the page is zero.
Let’s model the arm as a rod. The
deltoid muscle doesn’t attach right
at the shoulder, but down a little
distance, as seen in the figure. In
order to hold the weight out, all
the forces and torques acting on
your arm have to cancel out.
r
Fdeltoid
P
r
Fshoulder
L
θ
r
r
r
Fg = m g
Let’s look at the torques. Apply
The ∑
force
the extended
shoulder doesn’tFdeltoid
contribute
anything
sin θ − 12 mgL
=0
(1)
τ P =from
0 to your
since it is radial to the pivot
point.
Now,
if
the
distance
from
the
shoulder
to
the
arm:
deltoid muscle is `, then the torque is τdelt = Fdelt ` sin θ, while the torque on the center
Solving
Fdeltoid
yields:
of the arm is τarm = − 21 mgL.
Tofor
just
hold
your arm out, these two torques
mgL have to
F
=
cancel. So, solving for the force from the deltoid muscles gives deltoid 2 sin θ
mgL
.
(10 lb)(60 cm) ≈ 86 lb
delt =
Assuming F
that
≈ 20
2` cm,
sin θL ≈ 60 cm,
Fdeltoid =
2(20 cm )sin 10°
≈ 10
lb, andabout
θ ≈ 10°,10substitute
Now, let’s assume that themg
arm
weighs
pounds (about 5 kg), has a length of
numerical
values and
evaluate20 cm, and the angle is about 10◦ ,
about 60 cm, the deltoid muscle
attaches
at about
Fdeltoid:
then
(5)(10)(.6)
= a 10-lb weight
≈ 430 N,
IfFyou
F'deltoid sin θ − 12 mgL − m'gL = 0
delt hold
◦ ) at the end
2(.2)
sin(10
of your arm, equation (1) becomes:
where m′ is the mass of the 10-lb
which is about 86 pounds of force!
weight.
Now, if you hold out a 10 pound weight this adds an extra torque M gL. So, the deltoid
Solving for Fdeltoid yields:
mgL + 2m'gL
force now becomes
F' deltoid =
2 sin θ
(m + 2M )gL
(5 + 10)(10)(.6)
Fdelt =
≈
≈
1300
N,
2` sin θ
2(.2) sin(10◦ )
Substitute numerical values and
(10 lb)(60 cm) + 2(10 lb)(60 cm)
Fdeltoid =
which is about 260 poundsevaluate
of force!
F ′deltoid:
2(20 cm )sin 10°
1
Conditions for Equilibrium
≈ 260 lb
Because the vertical component of
the tension is 50 N:
Fd,max =
(500 N )2 − (50 N )2
= 0.50 kN
2. Romeo takes a uniform 10 m ladder and leans it against the smooth (frictionless) wall
30 ••
Romeo
takesresidence.
a uniform
it bottom
against
theonsmooth
of the Capulet
The10-m
ladder’sladder
mass is and
22 kgleans
and the
rests
the
m from
the wall.residence.
When Romeo,
whose
mass is 70
kg, gets
90 percent
of the
(frictionless)ground
wall 2.8
of the
Capulet
The
ladder’s
mass
is 22
kg and
the way to the top, the ladder begins to slip. What is the coefficient of static friction
bottom rests between
on thethe
ground
m ladder?
from the wall. When Romeo, whose mass is
ground 2.8
and the
70 kg, gets 90
percent of the way to the top, the ladder begins to slip. What is the
————————————————————————————————————
coefficient of static friction between the Solution
ground and the ladder?
r
Fby wall
Picture the Start
Problem
The ladder and
by looking at the forces on the ladder.
There is on
the weight
the ladder,
the forces acting
it at ofthe
criticalitself,
mg, as well as Romeo’s weight, M g. Since
moment of the
slipping
are shown
inwall,
thethere
ladder is leaning
against the
is a normal
from the wall,
FW . There
diagram. Use
the force
coordinate
system
is also the normal force from the floor, FN ,
shown. Because
the force,
ladder
and the frictional
Ff . is in
the ladder to keep from slipping, the
equilibrium, For
we
can apply the
sum of all these forces has to be zero. In
sum of all the torques
conditions addition,
for the
translational
andalso
has to be zero. Applying Newton’s laws
rotational equilibrium.
gives
0.9r
y
0.5r
r
Fn
Using its
x
0
10 m
r
m g Mgr
Ff =
FW
FN = (M + m)g.
Now, let’s consider the torques in the system.
r
θ
r
f s,max
2.8 m
The two gravitational forces will tend to twist the ladder. The torque from the ladder’s
weight is τladder = 5mg cos θ, where we have taken the cosine since only the component
f s,maxit. Similarly Romeo creates a
definition,
express
µs: to the ladder will torque
of the weight
perpendicular
(1)
μ
=
s the ladder from twisting, the wall also
torque, τRomeo = 9M g cos θ. Finally, to keep
F
provides a torque, τW = FW (10) sin θ. But, FW = Ff n= µs FN = µs (m + M ) g, and so
τW = 10µs (m + M ) g sin θ. The sum of the torques is zero, so
10µs (m + M ) g sin θ = 5mg cos θ + 9M g cos θ.
Solving for the coefficient gives
µs =
5m + 9M
cot θ.
10(m + M )
Now, since the ladder is 10 meters long, and it’s 2.8 meters away from the wall, then
θ = cos−1 (2.8/10) = 74◦ , and so cot θ = 0.29. Thus, we find
5m + 9M
5(22) + 9(70)
µs =
cot θ =
0.29 = 0.23.
10(m + M )
10(22 + 70)
2
3. When at cruising altitude, a typical airplane cabin will have an air pressure equivalent
to an altitude of about 2400 m. During the flight, ears often equilibrate, so that the
air pressure inside the inner ear equalizes with the air pressure outside the plane. The
Eustachian tubes allow for this equalization, but can become clogged. If an Eustachian
tube is clogged, pressure equalization may not occur on descent and the air pressure
inside an inner ear may remain equal to the pressure at 2400 m. In that case, by the
time the plane lands and the cabin is repressurized to sea-level pressure, what is the
net force on one ear drum due to this pressure difference, assuming the ear drum has
an area of 0.50 cm2 ?
————————————————————————————————————
Solution
If the pressures were equal, then there would be no net force - the inside pressure would
be pushing just as hard as the outside pressure, leading to no net force. So, the net
force comes from the difference in pressure, ∆P . So, the net force will be F = (∆P )A,
where A is the area. Now, the difference in pressure coming from a change in height
is, of course, ∆P = ρgh, where ρ is the density of air. So,
F = ρgAh.
Plugging in the numbers gives
F = ρgAh = (1.29)(9.8)(.5 × 10−4 )(2400) = 1.5 N.
3
Because ΔL/L = 9:
T=
9π r02Y
10
General Problems
55
•
[SSM]
A standard bowling ball weighs 16 pounds. You wish to ho
4. A standard bowling ball weights 16
pounds. You wish to hold a bowling ball in front
bowling ball in front of you, with the elbow bent at a right angle. Assume that y
of you, with your elbow bent at abiceps
rightattaches
angle.to Assume
thatat your
your forearm
2.5 cmbiceps
out fromattaches
the elbowtojoint, and that you
your forearm at 2.5 cm out from biceps
the elbow
and that
your that
biceps
pulls
musclejoint,
pulls vertically
upward,
is, it muscle
acts at right
angles to the forear
Also
assume
that to
thethe
ball forearm.
is held 38 cm
out assume
from the elbow
joint. Let the mass o
vertically upward, that is, it acts at
right
angles
Also,
that the
your forearm
5.0the
kg and
assume
its center
of gravity
is located
ball is held 38 cm out from the elbow
joint. be
Let
mass
of your
forearm
be 5.0
kg 19 cm out from
the elbow joint. How much force must your biceps muscle apply to forearm in o
and assume its center of gravity is located 19 cm out from the elbow joint. How much
to hold out the bowling ball at the desired angle?
force must your biceps muscle apply to the forearm in order to hold out the bowling
bar at the desired angle?
Picture the Problem We can model the forearm as a cylinder of length L = 38
————————————————————————————————————
with the forces shown in the pictorial representation acting on it. Because
forearm is in both translational and rotational equilibrium under the influenc
Solution
these forces, the forces in the diagram must add (vectorially) to zero and the
torquethis
with respect to any axis must also be zero.
Since the arm bends at the elbow,
r
is where we’ll set the pivot point. We
Fbicep
can draw the forces in the diagram to the
1
r
L
2 L
right. In order to hold the ball still, both elbow 0
x
the sum of the forces, and the sum of the
r
r
mforearm g
Felbow joint
r
torques have to be zero. Let’s look at the
m ball g
torques about the pivot point.
The force from the elbow joint doesn’t give a torque. The biceps provide a force
up, leading to a torque τbiceps = Fbiceps `, where ` is the distance where the biceps
attaches. Next, the arm, itself, creates a torque, τarm = − 21 marm gL, while the ball
creates τball = −mball gL. The sum of these torques gives
1
Fbiceps ` − marm gL − mball gL = 0.
2
Solving for the force from the biceps gives
Fbiceps =
1
m gL
2 arm
+ mball gL
=
`
1
marm + mball
2
gL
.
`
Plugging in all the numbers gives, after recalling that one kilogram is 2.2 pounds,
1
gL
1
9.8 × 38
Fbiceps =
marm + mball
=
(5) + 16/2.2
= 1500 N.
2
`
2
2.5
4
5. Blood flows at 30 cm/s in an aorta of radius 9.0 mm.
(a) Calculate the volume flow rate in liters per minute.
(b) Although the cross-sectional area of a capillary is much smaller than that of
the aorta, there are many capillaries, so their total cross-sectional area is much
larger. If all the blood from the aorta flows into the capillaries and the speed of
flow through the capillaries is 1.0 mm/s, calculate the total cross-sectional area
of the capillaries. Assume laminar nonviscous steady-state flow.
————————————————————————————————————
Solution
(a) The volume flow rate is just Iv = vA, or Iv = πr2 v, for a circular cross section.
So, we have
Iv = πr2 v = π(9 × 10−3 )2 × .3 = 7.6 × 10−5 m3 /sec.
Now, one cubic meter is 1000 liters, and so we can convert to liters per minute
by multiplying by 1000 liters per cubic meter, and multiplying by 60 seconds per
minute to find Iv = 7.6 × 10−5 × 60 × 1000 = 4.6 liters per minute.
(b) The continuity equation reads v1 A1 = v2 A2 , where v1 is the velocity of the blood
through a tube of area A1 , and v2 is its velocity through an area A2 . Now, we
know the rate through the capillaries and we know that area, so we can find the
area of the arteries
Aarteries =
vcapillaries
Iv
Acapillaries =
.
varteries
varteries
So,
Aarteries =
Iv
varteries
=
5
7.6 × 10−5
= 7.6 × 10−2 m2 .
10−3