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MASSACHUSETTS INSTITUTE OF TECHNOLOGY
Department of Physics
Physics 8.01
Fall 2012
Problem Set 8 Rigid Body Fixed Axis Rotational Motion
Due: Thursday Nov 8 at 9 pm. Place your solutions in the appropriate section slot in
the box outside 26-152. Write your name, section, and table number on the upper
right side.
Week 9 and 10 Schedule and Reading Assignments:
Sections 1-4 (Monday-Wednesday)
Oct 31 W09D2 Two-Dimensional Rotational Kinematics
Reading Assignment: Suggested: Young and Freedman: 9.1-9.6
8.01 Course Notes: Rigid Body Kinematics: Fixed Axis Rotation
http://web.mit.edu/8.01t/www/materials/modules/CN_W09D1_rigid_body_kinematics_v
01.pdf
Suggested: Young and Freedman: 1.10 (Vector Product), 9.1-9.6
Tuesday Oct 30 Problem Set 7 Due 9 pm
Nov 2 W09D3 Two Dimensional Rotational Dynamics
Reading Assignment: 8.01 Course Notes: Rigid Body Dynamics: Fixed Axis Rotation
http://web.mit.edu/8.01t/www/materials/modules/CN_W09D1_rigid_body_dynamics_v0
1.pdf
Suggested: Young and Freedman: 1.10 (Vector Product), 10.1-10.2, 10.4, 11.1-11.3
Nov 5 W10D1 Two-Dimensional Rotational Dynamics Hour Two, Angular
Momentum
Reading Assignment Part One: 8.01 Course Notes: Rigid Body Dynamics: Fixed Axis
Rotation
http://web.mit.edu/8.01t/www/materials/modules/CN_W09D1_rigid_body_dynamics_v0
1.pdf
Suggested: Young and Freedman: 1.10 (Vector Product), 10.1-10.2, 10.4, 11.1-11.3
Reading Assignment Part Two: 8.01 Course Notes: W010D1 Angular Momentum
http://web.mit.edu/8.01t/www/materials/modules/CN_W10D1_angular_momentum.pdf
Suggested: Young and Freedman Chapter 10.5-10.6
Sections 5-7 (Tuesday-Thursday)
Oct 30 W09D2 Two-Dimensional Rotational Kinematics
8.01 Course Notes: Rigid Body Kinematics: Fixed Axis Rotation
http://web.mit.edu/8.01t/www/materials/modules/CN_W09D1_rigid_body_kinematics_v
01.pdf
Suggested: Young and Freedman: 1.10 (Vector Product), 9.1-9.6
Nov 1 W09D2 Two-Dimensional Rotational Dynamics
Reading Assignment: 8.01 Course Notes: Rigid Body Dynamics: Fixed Axis Rotation
http://web.mit.edu/8.01t/www/materials/modules/CN_W09D1_rigid_body_dynamics_v0
1.pdf
Suggested: Young and Freedman: 1.10 (Vector Product), 10.1-10.2, 10.4, 11.1-11.3
Nov 2 W09D3 Problem Solving 07: Two Dimensional Rotational Dynamics
Reading Assignment: Review 8.01 Course Notes: Rigid Body Dynamics: Fixed Axis
Rotation
http://web.mit.edu/8.01t/www/materials/modules/CN_W09D1_rigid_body_dynamics_v0
1.pdf
Suggested: Young and Freedman: 1.10 (Vector Product), 10.1-10.2, 10.4, 11.1-11.3
Nov 6 W10D1 Angular Momentum (Longer Presentation)
Reading Assignment Part Two: 8.01 Course Notes: W010D1 Angular Momentum
http://web.mit.edu/8.01t/www/materials/modules/CN_W10D1_angular_momentum.pdf
Suggested: Young and Freedman Chapter 10.5-10.6
All Sections
Nov 7/8 W10D2 Experiment 3: Rotational Dynamics; Experiment 4: Angular
Momentum
8.01 Course Notes:
Thursday Nov 8 Problem Set 9 Due 9 pm
Nov 9 W10D3 Problem Solving 08: Conservation of Angular Momentum
Reading Assignment: Suggested: Young and Freedman: 10.5-10.6;
8.01 Course Notes:
Problem 1: Physical Pendulum
A physical pendulum consists of a disc of radius R and mass md fixed at the end of a rod
of mass mr and length l .
a) How far from the pivot point is the center of mass?
b) What is the moment of inertia of the physical pendulum about an axis passing
through the pivot perpendicular to the plane of motion of the pendulum.
Solution:
The physical pendulum consists of two pieces. A uniform rod of length d and a disk
attached at the end of the rod. The moment of inertia about the pivot point P is the sum
of the moments of inertia of the two pieces,
I P = I P,rod + I P,disk
(1)
We calculated the moment of inertia of a rod about the end point P in class, and found
that
1
I P,rod = mr l 2
3
(2)
We can use the parallel axis theorem to calculate the moment of inertia of the disk about
the pivot point P ,
I P,disk = I cm,disk + md l 2
(3)
We calculated the moment of inertia of a disk about the center of mass in class, and found
that
I cm,disk =
1
md R 2
2
(4)
So the total moment of inertia is
1
1
I P = mr l 2 + md gl 2 + md R 2
3
2
(5)
Problem 2: Claw Hammer
A claw hammer is used to pull a nail out of a board. The nail is an angle ! = 60! to the
!
board. The nail exerts a force on the hammer F1 along the line of the nail. The hammer
head contacts the board at !point A , which is a distance d from where the nail enters the
board. A horizontal force F2 is applied to the hammer handle at a distance h above the
!
!
board with d < h . The forces F1 and F2 on the hammer are shown in the figure to the
right (the arrows are not in scale with the magnitudes of the forces). You may ignore the
! !
weight of the hammer. What is the ratio, F1 / F2 ?
Solution:
Let‘s calculate the torque about the point A . The torque static equilibrium condition is
that
! !
! !
!
!
! A = rA,1 " F1 + rA,2 " F2 = 0
In the diagram below we show the relevant vectors.
!
!
We first calculate the torque about A due to the pulling force F2 , where rA,2 = xî + hĵ .
!
!
!
! A,2 = rA,2 " F2 = (xî + hĵ) " (# F2 î) = hF2k̂ .
Note that the even though the quantity x was not specified it doesn’t enter into the
torque
to calculate the torque
! about A . In order
! on the hammer due to the force of the
!
nail F1 , we note that rA,1 is perpendicular to F1 and the direction of the toque about A is
in the !k̂ direction. From the diagram below we can determine that the length of
!
rA,2 = d 3 / 2 .
!
Therefore the toque about A due to F1 is
!
!
d 3
!
! !
! A,1 = rA,1 " F1 = rA,1 F1 (#k̂) = #
F k̂
2 1
Therefore the torque about A is
%
! " d 3
0 = $!
F1 + hF2 ' k̂
2
#
&
Setting the above z -component equal to zero yields
d 3
F = hF2 .
2 1
Therefore the ratio, F1 / F2 is
F1 2 3 h
.
=
F2
3 d
Check
!
From the diagram above, we can determine the vector rA,1 from the point A to the point
!
where the force F1 acts (head of the nail)
d 3
d3
!
rA,1 =
î +
ĵ
4
4
!
Therefore the toque about A due to F1 is
! #d 3
!
d3 &
!
! A,1 = rA,1 " F1 = %
î +
ĵ( " (F1 cos(60" ) î ) F1 sin(60" ) ĵ)
4 '
$ 4
=
d 3
d3
F1 sin(60" )()k̂) +
F cos(60" )()k̂)
4
4 1
# d 3
# d 3 &
d3 3 &
= %)
F1 )
F1 ( k̂ = % )
F ( k̂
8
8
2 1'
$
'
$
which is in agreement with our earlier calculation.
,
Problem 3: Rotating Wheel
A light flexible cable is wrapped around a cylinder of mass m1 , radius R , and moment of
inertia I cm about the center of mass. The cylinder rotates about its axis without friction.
The free end of the cable is attached to an object of mass m2 . The object is released from
rest at a height h above the floor. You may assume that the cable has negligible mass.
Let g be the gravitational constant.
a) Draw free body diagrams of the cylinder and the object, indicating all of the forces
acting on each. Also choose appropriate coordinates the linear motion of the object
and the angular motion of the cylinder, and indicate the relation between them.
b) Find the acceleration of the falling object.
c) Find the tension in the cable.
d) Find the speed of the falling object just before it hits the floor.
Problem 4: Spinning Sphere
A uniform sphere of mass M s and radius R rotates about a vertical axis. Assume there is
no frictional torque acting along the axis of rotation. The sphere has moment of inertia
I S = (2 / 5)M S R 2 for an axis through its center of mass. A light string passes around the
equator of the sphere, and over a massive pulley of radius b and moment of inertia I P
about its center of mass, and the string is attached to a small object of mass m1 . When the
small object is released from rest, the string does not slip around the sphere or pulley, and
the small object falls vertically downward. What is the magnitude of the acceleration of
the small object after it is released? The downward gravitational acceleration has
magnitude g . Express your answer in terms of M s , m1 , I P , g , b , and R as needed.
Problem 5: Simple Pendulum
A simple pendulum consists of a massless string of length l and a point-like object of
mass m attached to one end. Suppose the string is fixed at the other end and is initially
pulled out at a small angle ! 0 from the vertical and released from rest. You may assume
the small-angle approximation, sin ! 0 ! ! 0 . The gravitational acceleration is g .
a) Find a differential equation for d 2! / dt 2 in terms of l , ! , m and g as needed by
calculating the torque about the pivot point.
b) What is your differential equation if ! << 1 ?
Solution We shall apply the torque equation about the pivot point. With our choice of a
cylindrical coordinate system with origin at the pivot point, the angular acceleration is
given by
! d 2!
" = 2 kˆ .
(6)
dt
The force diagram on the pendulum is shown below.
The only forces on the pendulum are the gravitational force on the pendulum bob (the
point-like object) and the tension in the string.
The torque about the pivot point is given by
!
!
! P = rP, F
grav
!
! !
" mg + rP,T " T .
(7)
!
!
Because rP,T is anit-parallel to T and hence the tension force exerts no torque. The rod is
uniform, therefore the center of mass is a distance d / 2 from the pivot point. So the
torque about the pivot point is given by
!
!
! P = rP, F
grav
!
" mg=lr̂ " mg(# sin $ $̂ + cos r̂) = #lmg sin $ k̂ .
(8)
The rotational dynamical equation (torque equation) is
!
!
! P = I P" .
(9)
The moment inertial of a point-like object about the pivot point is
I P = ml 2 .
(10)
d 2"
!lmg sin " k̂ = ml
k̂ .
dt 2
(11)
d 2!
g
= " sin ! .
2
l
dt
(12)
Therefore
2
We can rewrite this equation as
When the angle of oscillation is small, then we can use the small angle approximation
sin ! " ! .
(13)
d 2!
g
=" !
2
l
dt
(14)
Then the torque equation becomes
which is the simple harmonic oscillator equation.
Problem 6: Torsional Oscillator
A disk with moment of inertia about the center of mass I cm rotates in a horizontal plane.
It is suspended by a thin, massless rod. If the disk is rotated away from its equilibrium
position by an angle ! , the rod exerts a restoring torque given by ! cm = $"# .
a) Determine a differential equation that the angular displacement ! (t ) satisfies.
b) Show that the function ! (t) = Acos(" 0 t) + Bsin(" 0 t) satisfies the differential
equation you found in part a), where the angular frequency of oscillation is given
by ! 0 = " / I cm
c) Determine an expression for the z -component of the angular velocity.
d) Suppose the disk is released from rest at t = 0 , at an angle ! (t = 0) = ! 0 . What
are the constants A and B ? Based on that information determine an expression
for ! (t ) .
Solution: Choose a coordinate system such that k̂ is pointing upwards, then the angular
acceleration is given by
! d 2!
" = 2 kˆ .
(15)
dt
The torque about the center of mass is given in the statement of the problem as a
restoring torque
!
! cm = $"# kˆ .
(16)
!
!
Therefore the k̂ -component of the torque equation ! cm = I cm" is
!"# = I cm
d 2#
.
dt 2
(17)
b) Show that the function
! (t) = Acos(" 0 t) + Bsin(" 0 t)
(18)
where the angular frequency of oscillation is given by
! 0 = " / I cm
(19)
satisfies the differential equation you found in part a).
e) Determine an expression for the z -component of the angular velocity.
d!
(t) = "# 0 Asin(# 0 t) + # 0 Bcos(# 0 t) .
dt
(20)
f) Suppose the disk is released from rest at t = 0 , at an angle ! (t = 0) = ! 0 . What
are the constant A and B ?
Solution: ! (t = 0) = A = ! 0 , and (d! / dt)(t = 0) = " 0 B = 0 . Therefore
! (t) = ! 0 cos( " / I cm t) .
(21)
Problem 7: Race to the Bottom
A uniform stick of length l and mass m is pivoted at one end. The stick initially forms
an angle ! 0 with a table such that cos! 0 " 2 / 3 . At the other end of the stick a small
block is balanced on top of the stick. The stick is released from rest and hits the table.
The block is in free fall and also hits the table.
a) Calculate the speed of a point at the end of stick the instant before the stick hits the
table.
b) Calculate the speed of the block the instant before the block hits the table.
c) Does the ball or stick hit the table first?
Solution:
We assume the pivot is frictionless and so that energy of the stick is constant.
Choose U (! = 0) = 0 . The initial energy is then
Ei = ms g(l / 2)sin ! 0 .
The moment of inertia of the stick about the pivot point is I P = (1 / 3)ms l 2 . The final
energy of the stick when it just hits the ground is all rotational kinetic energy about the
pivot point
E f = (1 / 2)I P! f 2 = (1 / 6)ms l 2! f 2 .
Because energy is constant
ms g(l / 2)sin ! 0 = (1 / 6)ms l 2" f 2 .
Thus the angular speed of the stick just before it hits the ground
!f =
3g sin " 0
.
l
If the block fall straight down it hits the ground at a point A , a distance d A = l cos! 0
from the pivot point P . Therefore the speed of the point A of the stick just before it hits
the ground is
v A, f = d A
3g sin ! 0
3g sin ! 0
= l cos! 0
= cos! 0 3gl sin ! 0 .
l
l
b) The energy of the block is also constant. The initial energy is
Ei = mb gl sin ! 0 .
The final energy of the block is
E f = (1 / 2)mbvb, f 2 .
Because energy is constant
mb gl sin ! 0 = (1 / 2)mbvb,2 f .
Hence
vb, f = 2gl sin ! 0 .
c)
Let’s consider the point A on the stick that hits the ground where the block
hits. The point A on the stick is traveling in a circular arc which is a shorter distance than
the distance traveled by the block.
From the figure above, the distance traveled by the block is db = l sin ! 0 . Then distance
traveled by the point A is d A = (l cos! 0 )! 0 . The ratio of these distance is greater than
one,
db
l sin ! 0
tan ! 0
=
=
> 1.
d A (l cos! 0 )! 0
!0
Therefore is the speed of point A is greater than the speed of the block for fixed ! , point
A will hit the ground before the block hits the stick.
Using energy arguments for the block we have that at the angle ! ,
mb gl sin ! 0 " mb gl sin ! = (1 / 2)mbvb2 (! ) ,
hence
vb (! ) = 2gl(sin ! 0 " sin ! ) .
Using energy arguments for the stick we have that at the angle ! ,
ms g(l / 2)sin ! 0 " ms g(l / 2)sin ! = (1 / 6)ms l 2# s2 (! ) .
Hence the angular speed of the stick is
! s (" ) =
1
3gl(sin " 0 # sin " ) .
l
The speed of point A is then
v A (! ) = (l cos! 0 )" s (! ) = cos! 0 3gl(sin ! 0 # sin ! )
= cos! 0
3
3
2gl(sin ! 0 # sin ! ) = cos! 0 vb (! )
2
2
.
Recall that the angle ! 0 satisfied the condition that
cos! 0 >
2
3
therefore v A (! ) > vb (! ) . So the stick hit s the ground before the block hits the stick.
We could also argue as follows. The magnitude of the z -component of the end of the
stick is given by
vend ,z (! ) = (l cos! )" s (! ) = cos! 0 3gl(sin ! 0 # sin ! ) = cos!
3
v (! ) .
2 b
Because cos! " cos! 0 > 2 / 3 , the end of the stick is always traveling faster than the
block, so vend ,z (! ) > vb (! ) .
Problem 8: Static Equilibrium: Back-Bending Exercise
The human vertebral column is shown in
the figure. The sacrum is rigidly attached
to the pelvis. The fifth lumbar vertebra is
separated from the sacrum by a disc. In
this problem, approximate the spine as a
rigid body hinged at the bottom of the
lumbo-sacral disc.
! The top of the sacrum
exerts a force Fdisc on the base of the
lumbo-sacral disc at an angle ! relative
to the axis of the vertebral column.
When a person of mass m = 50 kg bends
over or lifts an object, the main muscles
that lift the back are the erector spinae
(sacrospinal muscles).
These muscles! act approximately as a single cord (“rope”) on the vertebral column and
exert a force, Fmuscle , at a point that is (2 / 3) of the distance from base of the spine (where
the sacrum pushes on base of the lumbo-sacral disc) to the center of gravity of the head
and arms. The angle of insertion of this “rope” is about ! = 12! relative to the axis of the
vertebral column. Assume that mass of the head and arms is m2 = (1 / 5)m and that the
mass of the trunk is m1 = (1 / 5)m , where m is the mass of the person. The center of mass
of the trunk is (1 / 2) of the distance from the base of the spine to the center of gravity of
the head and arms.
Suppose the person bends over so that the spine makes an angle of ! = 30! with respect
to the horizontal.
!
a) Find the magnitude of the force, Fmuscle = Fmuscle , that the muscles exert on the
spine.
b) Find the angle, ! , of the disc force with respect to the spinal axis and the
!
magnitude, Fdisc = Fdisc , of the force on the base of the lumbo-sacral disc.
c) What is the ratio of the magnitude of the force Fdisc to the weight of the body?
d) Suppose you lift a mass equal to the mass of the head and arms with your arms
hanging straight down. Assume you bend bent at the same angle of 30! with the
horizontal. By how much does the magnitude of the reaction force increase?
e) Explain why you should bend your knees when you pick up a mass. Base your
analysis on the work done in parts a) through d).
Solution:
a) Choose coordinates and a sense for positive rotation as shown in the force
diagram shown below.
We first vector decompose each of the forces with respect to the unit vectors shown in the
figure above.
!
m1 g = !m1 g sin " î ! m1 g cos" ĵ
(1.1)
!
m2 g = !m2 g sin " î ! m2 g cos" ĵ
(1.2)
!
Fmuscle = ! Fmuscle cos " î + Fmuscle sin " ĵ
(1.3)
!
Fdisc = Fdisc cos ! î " " Fdisc sin ! ĵ
(1.4)
The two equations for force equilibrium are then
ˆi : F cos ! $ m g sin " $ m g sin " $ F
disc
1
2
muscle cos # = 0
ˆj : $ F sin ! $ m g cos" $ m g cos" + F
disc
1
2
muscle sin # = 0.
(1.5)
Choose the base of the lumbo-sacral disc as the point S about which to calculate torques.
Then torque equilibrium becomes
!
!
!
!
!
! S , total = ! S ,1 + ! S ,2 + ! S , muscle = 0 .
(1.6)
We begin with the gravitational force acting at the center of mass of the trunk. The vector
from S to the center of mass of the trunk is
!
rS ,1 = (1 / 2)l î
(1.7)
The torque about S due to the center of mass of the trunk is
!
!
!
! S ,1 = rS ,1 " m1 g = (1 / 2)l î " (#m1 g sin $ î # m1 g cos$ ĵ)
= #(1 / 2)l m1 g cos$ k̂.
(1.8)
The vector from S to the center of mass of the head and arms is
!
rS , 2 = l î
(1.9)
The torque about S due to the center of mass of the head and arms is
(
)
!
!
!
! S , 2 = rS , 2 " m2 g = l î " #m2 g sin $ î # m2 g cos$ ĵ
= #l m2 g cos$ k̂.
(1.10)
The vector from S to where the muscles act approximately as a single cord (“rope”) on
the vertebral column is
!
rS , muscle = (2 / 3)l î
(1.11)
The torque about S due to the muscular force is
!
!
!
! S , muscle = rS , muscle " Fmuscle = (2 / 3)l î " # Fmuscle cos $ î + Fmuscle sin $ ĵ
(
)
= (2 / 3)l Fmuscle sin $ k̂.
(1.12)
The total component of the torque in the k̂ direction must vanish,
!(1 / 2)lm1 g cos" ! l m2 g cos" + (2 / 3)l Fmuscle sin # = 0 .
(1.13)
We can solve the torque equation (1.13) for the magnitude of the muscular force,
Fmuscle =
((3 / 4)m1 + (3 / 2)m2 )g cos!
.
sin "
(1.14)
Substituting the given values for the masses of the trunk m1 = (1 / 5)m , and arms and
head, m2 = (1 / 5)m , into Eq. (1.14) yields
Fmuscle =
(9 / 4)m1 g cos! (9 / 20)cos(30! )
=
mg
.
sin "
sin(12! )
(1.15)
= 1.9 mg = 9.2 # 102 N
The force in the erector spine muscles is 1.9 times the body weight.
b), c) We can solve for the angle the force on the disc makes with respect to the spine, by
rewriting the force equations (1.5) as
Fdisc cos ! = m1 g sin " + m2 g sin " + Fmuscle cos #
Fdisc sin ! = $m1 g cos" $ m2 g cos" + Fmuscle sin # .
(1.16)
Dividing these equations yields
Fdisc cos !
m g sin " + m2 g sin " + Fmuscle cos #
.
= 1
Fdisc sin ! $m1 g cos" $ m2 g cos" + Fmuscle sin #
Substituting Eq. (1.14) into Eq. (1.17) yields
(1.17)
cotan! =
m1 g sin " + m2 g sin " + ((3 / 4)m1 + (3 / 2)m2 )g cos" cotan#
$m1 g cos" $ m2 g cos" + ((3 / 4)m1 + (3 / 2)m2 )g cos"
m g sin " + m2 g sin " + ((3 / 4)m1 + (3 / 2)m2 )g cos" cotan#
= 1
($(1 / 4)m1 + (1 / 2)m2 )g cos"
.
(1.18)
We now substitute m1 = m2 , ! = 30! , and ! = 12! , yielding
cotan! = 8 tan " + 9cotan# = 8 tan 30! + 9cotan12! = 47 .
(1.19)
Therefore The magnitude of the disc force cancels and we can take the inverse cotangent
to find the angle
( )
! = cot "1 47 = 1.2!
(1.20)
Finally, we can solve for the magnitude of the disc force by using the first equation in
(1.16), with ! = 1.2! and Fmuscle from the first equation in (1.15);
Fdisc =
m1 g sin ! + m2 g sin ! + Fmuscle cos "
cos #
= m1 g sin ! + m2 g sin ! + (9 / 4)m1 g cos! cotan"
cos #
=
=
(
(1 / 5)mg 2sin ! + (9 / 4)cos! cotan"
cos #
)
(
(1 / 5)(50 kg)(9.8m $ s %2 ) 2sin(30! )+(9 / 4)cos(30! )cot(12! )
(1.21)
)
!
cos(1.2 )
= (1.0 & 10 N) " 2.0mg.
3
d) Suppose you lift a mass equal to the mass of your head and arms with your arms
hanging straight down. Assume you bend bent at the same angle of 30! with the
horizontal. By how much does the magnitude of the reaction force increase?
Lifting an additional weight m2 introduces extra torque at the end of the spine. We
assume that lifting the extra weight just doubles the center of mass of the head and arms.
Therefore the extra torque,
!
!
! S , weight = ! S , 2 = "l m2 g cos# k̂ .
(1.22)
We adjust all the above calculations by introducing this extra factor of two,
Fmuscle =
=
(1 / 2)m1 g cos! + 2m2 g cos! m1 g((1 / 2)cos! + 2cos! ) (3 / 4)mg cos!
=
=
(2 / 3)sin "
(2 / 3)sin "
sin"
(3 / 4)(50 kg)(9.8m# s $2 )cos(30! )
sin(12! )
(1.23)
= 1.5 % 103 N " 3.1mg.
% m g sin # + 2m2 g sin # + (3 / 4)mg cos# cot $ (
! = cot "1 ' 1
*
& "m1 g cos# " 2m2 g cos# + (3 / 4)mg cos# )
(
)
% (3 / 5)sin # + (3 / 4)cos# cot $ (
= cot "1 '
*
cos# ("(3 / 5) + (3 / 4))
&
)
(
= cot "1 (16 / 3)tan(30! ) + 5cot(12! )
(1.24)
)
= 2.15!.
Similarly
Fdisc =
=
m1 g sin ! + 2m2 g sin ! + (3 / 4)mg cos! cot "
cos #
(
mg (3 / 5)sin ! + (3 / 4)cos! cot "
cos #
)
((3 / 5)sin(30 ) + (3 / 4)cos(30 )cot(12 ))
)
!
= (50 kg)(9.8m $ s
%2
!
(1.25)
!
cos(2.15! )
= (2.1 & 103 N) " 4.3 mg.
The force on the disc has increased by 1.0 ! 103 N when lifting a weight of
m2 g = (1 / 5)mg = (1 / 5)(50 kg)(9.8m ! s "2 ) = 9.8 # 101 N ,
(1.26)
about ten times the weight lifted.
e) The large force found in part d) is in part due to the horizontal forces needed for torque
equilibrium. By bending the knees while lifting, the forces are closer to vertical, and the
force on the disc will be approximately the weight of the upper body plus the weight
lifted.