rawashdeh (ar73422) – Homework 8, torque 21-22 – dowd – (VickeryRPHY1 2)
This print-out should have 25 questions.
Multiple-choice questions may continue on
the next column or page – find all choices
before answering.
001 (part 1 of 2) 10.0 points
The arm of a crane at a construction site is
15.0 m long, and it makes an angle of 16.6◦
with the horizontal. Assume that the maximum load the crane can handle is limited
by the amount of torque the load produces
around the base of the arm.
What maximum torque can the crane withstand if the maximum load the crane can
handle is 459 N?
Answer in units of N · m.
Correct answer: 6598.05 N · m.
Explanation:
Let : d = 15.0 m and
Wmax = 459 N .
θ = 90.0◦ − 16.6◦ = 73.4◦ ,
so
τmax = F d sin θ
= Wmax d sin θ
= (459 N)(15 m)(sin 73.4◦ )
= 6598.05 N · m .
003 10.0 points
If the torque required to loosen a nut that
is holding a flat tire in place on a car has a
magnitude of 46 N · m, what minimum force
must be exerted by the mechanic at the end
of a 24 cm-long wrench to loosen the nut?
Answer in units of N.
Correct answer: 191.667 N.
Explanation:
Let : τ = 46 N · m and
d = 24 cm = 0.24 m .
For a given torque, the minimum force must
be applied perpendicular to the lever arm so
that sin θ = sin 90◦ = 1, and
τ =Fd
τ
F =
d
46 N · m
=
0.24 m
= 191.667 N .
004 (part 1 of 2) 10.0 points
The figure shows a claw hammer as it pulls a
nail out of a horizontal board.
002 (part 2 of 2) 10.0 points
What is the maximum load for this crane at
an angle of 25.0◦ with the horizontal?
Answer in units of N.
Correct answer: 485.343 N.
F
31 cm
Explanation:
Let :
θ = 90.0◦ − 25.0◦ = 65.0◦
We have the same maximum torque, so
τmax
W=
d sin θ
6598.05 N · m
=
(15 m) sin 65◦
= 485.343 N .
1
Single point
of contact
34◦
4.94 cm
rawashdeh (ar73422) – Homework 8, torque 21-22 – dowd – (VickeryRPHY1 2)
If a force of magnitude 211 N is exerted
horizontally as shown, find the force exerted
by the hammer claws on the nail. (Assume
that the force the hammer exerts on the nail
is parallel to the nail).
Answer in units of N.
Correct answer: 1597.14 N.
X
2
Fx = f + F − R sin α = 0
Explanation:
f = R sin α − F
= (1597.14 N) sin 34◦
− 211 N
= 682.109 N
and vertically,
Let : h = 31 cm ,
ℓ = 4.94 cm ,
α = 34◦ , and
F = 211 N .
X
F @h
α
R @ ℓ cos α
Let R be the reaction force of the nail on
the hammer. Taking the sum of the torques
about the point of contact of the hammer with
the table,
X
τ = R ℓ cos θ − F h = 0
F =
=
p
q
Fy = n − R cos α = 0
n = R cos α
= (1597.14 N) cos 34◦
= 1324.09 N , so
f 2 + n2
(682.109 N)2 + (1324.09 N)2
= 1489.46 N .
006 (part 1 of 2) 10.0 points
A ladder rests against a vertical wall. There
is no friction between the wall and the ladder.
The coefficient of static friction between the
ladder and the ground is µ = 0.419 .
Fh
ℓ cos α
(211 N) (31 cm)
=
(4.94 cm) cos 34◦
R=
Fw
ℓ
= 1597.14 N .
005 (part 2 of 2) 10.0 points
Find the force exerted by the surface on the
point of contact with the hammer head. Assume that the force the hammer exerts on the
nail is parallel to the nail.
Answer in units of N.
Correct answer: 1489.46 N.
Explanation:
Let f be the horizontal force exerted by the
surface at the point of contact on the hammer.
From Newton’s second law, horizontally,
h
θ
N
W
b
µ = 0.419
f
b
Consider the following expressions:
A1: f = Fw
A2: f = Fw sin θ
rawashdeh (ar73422) – Homework 8, torque 21-22 – dowd – (VickeryRPHY1 2)
W
B1: N =
2
B2: N = W
C1: ℓ Fw sin θ = 2 Fw cos θ
C2: ℓ Fw sin θ = ℓ W cos θ
1
C3: ℓ Fw sin θ = ℓ W cos θ ,
2
where
f : force of friction between the ladder and
the ground,
Fw : normal force on the ladder due to the
wall,
θ: angle between the ladder and the ground,
N : normal force on the ladder due to the
ground,
W: weight of the ladder, and
ℓ: length of the ladder.
Identify the set of equations which is correct.
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
A2,
A2,
A1,
A2,
A1,
A1,
A1,
A2,
A1,
A1,
B2,
B1,
B1,
B1,
B1,
B2,
B1,
B1,
B2,
B2,
C1
C1
C3
C2
C1
C3 correct
C2
C3
C2
C1
Explanation:
From translational equilibrium,
X
Fx = f − Fw = 0
f = Fw
X
and
3
007 (part 2 of 2) 10.0 points
Determine the smallest angle θ for which the
ladder remains stationary.
Answer in units of ◦ .
Correct answer: 50.037◦ .
Explanation:
f = µs N = Fw ,
so
1
W cos θ
2
1
µs W sin θ = W cos θ
2
1
µs =
2 tan θ
1
tan θ =
2 µs
Fw sin θ =
1
θ = tan
2 µs
1
−1
= tan
2 (0.419)
−1
= 50.037◦ .
Fy = N − W = 0
N =W.
Applying rotational equilibrium about the
foot of the ladder,
X
ℓ
τ = W cos θ − Fw ℓ sin θ = 0
2
1
ℓ Fw sin θ = ℓ W cos θ
2
So the correct set of equations is
A1 , B2 , C3 .
008 10.0 points
An 18.8 kg person climbs up a uniform ladder with negligible mass. The upper end of
the ladder rests on a frictionless wall. The
bottom of the ladder rests on a floor with a
rough surface where the coefficient of static
friction is 0.33 . The angle between the horizontal and the ladder is θ . The person wants
to climb up the ladder a distance of 1.2 m
along the ladder from the ladder’s foot.
rawashdeh (ar73422) – Homework 8, torque 21-22 – dowd – (VickeryRPHY1 2)
X
Fy = Nf − m g = 0
3.3 m
µ=0
X
18.8 kg
b
τ◦ = m g d cos θ − Nw L sin θ = 0
Nw L sin θ = m g d cos θ
f L sin θ = m g d cos θ
mgd
f=
L tan θ
θ
1.2 m
4
µ = 0.33
The ladder may slip when f = fmax = Nw ,
so
What is the minimum angle θmin (between
the horizontal and the ladder) so that the
person can reach a distance of 1.2 m without
having the ladder slip? The acceleration of
gravity is 9.8 m/s2 .
Answer in units of ◦ .
Correct answer: 47.7763 ◦.
Explanation:
Let : g = 9.8 m/s2 ,
d = 1.2 m ,
L = 3.3 m ,
m = 18.8 kg , and
µ = 0.33 .
mgd
L tan θ
d
tan θ ≥
µL
d
θ ≥ arctan
µL
1.2 m
≥ arctan
(0.33) (3.3 m)
fmax ≡ µ m g ≥
≥ 47.7763 ◦ .
Nw
009 (part 1 of 2) 10.0 points
A square plate is produced by welding together four smaller square plates, each of side
a. The weight of each of the four plates is
shown in the figure.
y
(2a, 2a)
(0, 2a)
60 N
70 N
20 N
40 N
θ
f
mg
x
b
P ivot
Nf
X
Fx = f − Nw = 0
f = Nw
(2a, 0)
(0, 0)
Find the x-coordinate of the center of gravity (as a multiple of a).
Answer in units of a.
Correct answer: 1.07895 a.
Explanation:
rawashdeh (ar73422) – Homework 8, torque 21-22 – dowd – (VickeryRPHY1 2)
5
P
i Wi yi
ycg =
W1 + W2 + W3 + W4
1
3a
a
Let : x1 = x2 = a ,
(W1 + W4 ) + (W2 + W3 )
2
2
2
=
3
x3 = x4 = a ,
W
2
(20 N + 40 N) + 3 (60 N + 70 N)
W1 = 20 N ,
a
=
2 (190 N)
W2 = 60 N ,
= 1.18421 a .
W3 = 70 N , and
W4 = 40 N .
011 10.0 points
A uniform brick of length 30 m is placed
over the edge of a horizontal surface with a
maximum overhang of 15 m attained without
tipping.
y
3a
2
a
2
W2
W3
W1
W4
a
2
The total weight is
3a
2
30 m
x
W = W1 + W2 + W3 + W4 = 190 N .
Applying the definition of center of gravity,
P
i Wi xi
xcg =
W1 + W2 + W3 + W4
a
3a
(W1 + W2 ) + (W3 + W4 )
2
2
=
W
(20 N + 60 N) + 3 (70 N + 40 N)
a
=
2 (190 N)
= 1.07895 a .
010 (part 2 of 2) 10.0 points
Find the y-coordinate of the center of gravity
(as a multiple of a).
Answer in units of a.
Correct answer: 1.18421 a.
x
Now two identical uniform bricks of length
30 m are stacked over the edge of a horizontal
surface.
30 m
x
What maximum overhang is possible for
the two bricks (without tipping)?
Answer in units of m.
Correct answer: 22.5 m.
Explanation:
Let :
L = 30 m
n = 1.
and
The center of mass (CM) of a stack of n
bricks (of equal mass M ) is
Explanation:
1
Let : y1 = y4 = a and
2
3
y2 = y3 = a .
2
x≡
n
X
xi M
i=1
n
X
i=1
M
n
1X
=
xi ,
n
i=1
rawashdeh (ar73422) – Homework 8, torque 21-22 – dowd – (VickeryRPHY1 2)
where xi is the CM position of the ith brick.
One brick: The CM of a single brick is in
its middle, so the maximum overhang is
1
1
L = (30 m) = 15 m .
2
2
Two bricks: The bricks will just balance
when the CM of the stack is placed at the
fulcrum (the edge of the horizontal surface).
Place the left edge of the new brick below the
CM of the previous brick and measure from
the left edge of the top brick. The top brick
1
can extend L from the left edge of the brick
2
under it, so the position of the CM of the new
brick is
Let : m1 = 2 kg ,
x1 = 1 m , and
ℓ = 7 m.
x=
x2 = x
n=1
+
1
1
1
L = L+ L = L,
2
2
2
012 10.0 points
A 2 kg rock is suspended by a massless
string from one end of a 7 m measuring stick.
2
3
4
5
6
7
2 kg
What is the weight of the measuring stick
if it is balanced by a support force at the
1 m mark? The acceleration of gravity is
9.81 m/s2 .
Answer in units of N.
Correct answer: 7.848 N.
Explanation:
x1
0
ℓ
2
and
x2 =
ℓ
− 1.
2
4
5
x2
1
2
3
6
7
m2 g
m2 g x2 = m1 g x1
m1 g x1
m2 g =
x2
(2 kg) (9.81 m/s2 ) (1 m)
=
2.5 m
= 7.848 N .
x
1
xCM =
The static equation for torques gives us
30 m
0
Because the stick is a uniform, symmetric
body, we can consider all of its weight to be
concentrated at the center of mass, so
m1 g
and for the stack of 2 bricks,
1
1
3
=
x
1+
L= L
n=2
2
2
4
3
= (30 m) = 22.5 m .
4
6
013 10.0 points
An Atwood machine is constructed using two
wheels (with the masses concentrated at the
rims). The left wheel has a mass of 2.5 kg and
radius 24.03 cm. The right wheel has a mass
of 2.3 kg and radius 31.38 cm. The hanging
mass on the left is 1.64 kg and on the right
1.27 kg.
m1
m3
m2
m4
What is the acceleration of the system?
The acceleration of gravity is 9.8 m/s2 .
Answer in units of m/s2 .
Correct answer: 0.470298 m/s2 .
rawashdeh (ar73422) – Homework 8, torque 21-22 – dowd – (VickeryRPHY1 2)
Explanation:
7
Adding these four equations gives
m1 a + m2 a + m3 a + m4 a = m3 g − m4 g
Let :
m1
r1
m2
r2
m3
m4
a
= 2.5 kg ,
= 24.03 cm ,
= 2.3 kg ,
= 31.38 cm ,
= 1.64 kg , and
= 1.27 kg .
T2
m1 r1
m2r2
(m3 − m4 ) g
m1 + m2 + m3 + m4
(1.64 kg − 1.27 kg) (9.8 m/s2 )
=
2.5 kg + 2.3 kg + 1.64 kg + 1.27 kg
a=
= 0.470298 m/s2 .
a
T3
T1
m4
m3
The net acceleration a = r α is in the direction of the heavier mass m3 . Each pulley’s mass is concentrated on the rim, so
I = mpulley r 2 and
2 a
= mra
τnet = Iα = m r
r
014 10.0 points
A horizontal 809 N merry-go-round of radius
1.23 m is started from rest by a constant
horizontal force of 74.2 N applied tangentially
to the merry-go-round.
Find the kinetic energy of the merry-goround after 2.46 s. The acceleration of gravity
is 9.8 m/s2 . Assume the merry-go-round is a
solid cylinder.
Answer in units of J.
Correct answer: 403.604 J.
Explanation:
so that for the leftmost pulley
T1 r1 − T2 r1 = m1 r1 a
m1 a = T1 − T2
Let : r = 1.23 m ,
F = 74.2 N ,
W = 809 N .
(1)
The moment of inertia of the cylinder is
and for the rightmost pulley
I=
T2 r2 − T3 r2 = m2 r2 a
m2 a = T2 − T3 .
(2)
T3
a
m3
m4
m3g
m4g
a
1
I ω2
2
2
2F gt
1 1W 2
r
=
2 2 g
Wr
K=
and
m4 a = T3 − m4 g .
and
2F gt
so the angular velocity is ω = α t =
,
Wr
and
the net forces are
m3 a = m3 g − T1
1
1W 2
m r2 =
r
2
2 g
τ = Iα
W r2
α
Fr=
2g
2gF
α=
Wr
Applying Newton’s Law to the hanging
masses,
T1
and
(3)
(4)
rawashdeh (ar73422) – Homework 8, torque 21-22 – dowd – (VickeryRPHY1 2)
F 2 t2 g
W
(74.2 N)2 (2.46 s)2 (9.8 m/s2 )
=
809 N
= 403.604 J .
=
015 10.0 points
A solid sphere rolls along a horizontal, smooth
surface at a constant linear speed without
slipping.
What is the ratio between the rotational
kinetic energy about the center of the sphere
and the sphere’s total kinetic energy?
Answer in units of s.
Correct answer: 1.31353 s.
3
5
3
2.
7
2
3.
5
7
4.
2
5
5.
3
6. None of these
2
7. correct
7
Explanation:
1.
Krot
1
=
2
2
v 2 1
2
mr
= m v2
5
r
5
and
Ktot = Ktrans + Krot
1
7
1
m v2 ,
= m v2 + m v2 =
2
5
10
Krot
Ktot
so
1
m v2
10
2
1
5
=
=
=
7
5
7
7
m v2
10
keywords:
016
10.0 points
Explanation:
8
rawashdeh (ar73422) – Homework 8, torque 21-22 – dowd – (VickeryRPHY1 2)
9
shopkeeper.
By what percentage is the true weight of the
goods being marked up by the shopkeeper?
Assume the balance has negligible mass.
Answer in units of %.
Correct answer: 4.79836 %.
Explanation:
Let :
L = 51.9 cm and
ℓ = 0.608 cm .
Let W be the standard weight and W ′ the
measured weight.
L
L
′
W
−ℓ =W
+ℓ
2
2
W
L+ 2ℓ
=
,
′
W
L− 2ℓ
W − W′
=
W′
so
L+2ℓ
− 1 × 100%
L−2ℓ
4ℓ
× 100%
=
L− 2ℓ
4 0.608 cm
=
× 100%
51.9 cm − 2 0.608 cm
= 4.79836 % .
018 10.0 points
The center of mass of a pitched baseball or
radius 5.53 cm moves at 21.3 m/s. The ball
spins about an axis through its center of mass
with an angular speed of 124 rad/s.
Calculate the ratio of the rotational energy
to the translational kinetic energy. Treat the
ball as a uniform sphere.
Correct answer: 0.0414567.
Explanation:
017 10.0 points
Two pans of a balance are 51.9 cm apart.
The fulcrum of the balance has been shifted
0.608 cm away from the center by a dishonest
Let : r = 5.53 cm = 0.0553 m ,
v = 21.3 m/s , and
ω = 124 rad/s .
For the sphere I =
2
m R2 ,
5
so
rawashdeh (ar73422) – Homework 8, torque 21-22 – dowd – (VickeryRPHY1 2)
Krot
Ktrans
I ω2
I ω2
= 22 =
m v2
mv
2
2 m r2
ω2
2 r2 ω 2
5
=
=
m v2
5 v2
2 (0.0553 m)2 (124 rad/s)2
=
5 (21.3 m/s)2
= 0.0414567 .
keywords:
019 (part 1 of 2) 10.0 points
A figure skater on ice spins on one foot. She
pulls in her arms and her rotational speed
increases.
Choose the best statement below:
1. Her angular speed increases because her
angular momentum increases.
2. Her angular speed increases due to a net
torque exerted by her surroundings.
3. Her angular speed increases because her
angular momentum is the same but her moment of inertia decreases. correct
4. Her angular speed increases because air
friction is reduced as her arms come in.
5. Her angular speed increases because by
pulling in her arms she creates a net torque in
the direction of rotation.
6. Her angular speed increases because her
potential energy increases as her arms come
in.
Explanation:
The initial angular momentum of the figure
skater is Ii ωi . After she pulls in her arms, the
angular momentum of her is If ωf . Note that
If < Ii because her arms now rotate closer to
the rotation axis and reduce the moment of
inertia. Since the net external torque is zero,
angular momentum remains unchanged, and
so Ii ωi = If ωf = L. Therefore, ωf > ωi .
020 (part 2 of 2) 10.0 points
And again, choose the best statement below:
10
1. When she pulls in her arms, her rotational potential energy increases as her arms
approach the center.
2. When she pulls in her arms, her rotational
kinetic energy is conserved and therefore stays
the same.
3. When she pulls in her arms, the work she
performs on them turns into increased rotational kinetic energy. correct
4. When she pulls in her arms, her rotational
kinetic energy must decrease because of the
decrease in her moment of inertia.
5. When she pulls in her arms, her moment
of inertia is conserved.
6. When she pulls in her arms, her angular momentum decreases so as to conserve
energy.
Explanation:
The kinetic energy of the figure skater,
E=
1
1
I ω2 = L ω .
2
2
Since ω increases after she pulls in her arms
as mentioned above, the total kinetic energy
increases. This additional energy comes from
the figure skater, namely she has to perform
some work to achieve this.
021 (part 1 of 2) 10.0 points
A student sits on a rotating stool holding two
2 kg objects. When his arms are extended
horizontally, the objects are 1 m from the axis
of rotation, and he rotates with angular speed
of 0.73 rad/sec. The moment of inertia of the
student plus the stool is 8 kg m2 and is assumed to be constant. The student then pulls
the objects horizontally to a radius 0.28 m
from the rotation axis.
ωi
(a)
ωf
(b)
rawashdeh (ar73422) – Homework 8, torque 21-22 – dowd – (VickeryRPHY1 2)
Calculate the final angular speed of the
student.
Answer in units of rad/s.
Correct answer: 1.0537 rad/s.
Explanation:
Let : m = 2 kg ,
R = 1 m,
r = 0.28 m ,
ω = 0.73 rad/sec ,
Is = 8 kg m2 .
X
1
I ω2
2
The initial moment of inertia of the system is
Krot =
Ii = Is + 2 m R 2
= (8 kg m2 ) + 2 (2 kg) (1 m)2
= 12 kg m2 .
023 (part 1 of 2) 10.0 points
A merry-go-round rotates at the rate of
0.12 rev/s with an 92 kg man standing at
a point 2.8 m from the axis of rotation.
What is the new angular speed when the
man walks to a point 0 m from the center?
Consider the merry-go-round is a solid 13 kg
cylinder of radius of 2.8 m.
Answer in units of rad/s.
Correct answer: 11.4257 rad/s.
Explanation:
The final moment of inertia of the system is
If = Is + 2 m r 2
= 8 kg m2 + 2 (2 kg) (0.28 m)2
= 8.3136 kg m2 .
From conservation of the angular momentum
it follows that
I i ωi = I f ωf
Ii
ωf = ωi
If
12 kg m2
= (0.73 rad/sec)
8.3136 kg m2
∆K = Kf − Ki
1
1
= If ωf2 − Ii ωi2
2
2
1
8.3136 kg m2 (1.0537 rad/s)2
=
2
1
− 12 kg m2 (0.73 rad/sec)2
2
= (4.61518 J) − (3.1974 J)
= 1.41778 J .
and
~ = Constant.
L
11
= 1.0537 rad/s .
Given :
ωi = 0.12 rev/s ,
m = 92 kg ,
ri = 2.8 m ,
rf = 0 m ,
M = 13 kg , and
R = 2.8 m .
The merry-go-round can be modeled as a
solid disk with angular momentum
1
2
Ld = Id ω =
MR ω
2
and the man as a point mass with angular
momentum
Lm = Im ω = m r 2 ω .
Angular momentum is conserved, so
022 (part 2 of 2) 10.0 points
Calculate the change in kinetic energy of the
system.
Answer in units of J.
Correct answer: 1.41778 J.
Explanation:
Lf
Lm,f + Ld,f
Im,f ωf + Id ωf
1
2
2
m r f + M R ωf
2
= Li
= Lm,i + Ld,i
= Im,i ωi + Id ωi
1
2
2
= m r i + M R ωi
2
rawashdeh (ar73422) – Homework 8, torque 21-22 – dowd – (VickeryRPHY1 2)
ωf =
=
1
2
2
2 M R + m ri
ω
1
2 i
2
2 M R + m rf
1
2
2
2 (13 kg) (2.8 m) + (92 kg) (2.8 m)
1
2
2
2 (13 kg) (2.8 m) + (92 kg) (0 m)
× (0.12 rev/s) ·
2π
rev
= 11.4257 rad/s .
024 (part 2 of 2) 10.0 points
What is the change in kinetic energy due to
this movement?
Answer in units of J.
Correct answer: 3106.84 J.
Explanation:
The moment of inertia of the system about
the axis of rotation at any moment is
I = Id + Im =
1
M R2 + m r 2 ,
2
and the kinetic energy at any time is
1
I ω2
2
1 1
2
2
=
M R + m r ω2
2 2
1
1
= M R2 ω 2 + m r 2 ω 2
4
2
Krot + Ktrans =
Thus the change in kinetic energy of the system is
∆K = Kf − Ki
1
1
2
2
=
M R + m rf ωf2
4
2
1
1
2
2
−
M R + m ri ωi2
4
2
(13 kg)(2.8 m)2 (92 kg)(0 m)2
=
+
4
2
× (11.4257 rad/s)2
(13 kg)(2.8 m)2 (92 kg)(2.8 m)2
−
+
4
2
2
× (0.12 rev/s) (2 π rad/rev)2
= 3106.84 J .
12
025 10.0 points
A 109 kg man sits on the stern of a 3.9 m long
boat. The prow of the boat touches the pier,
but the boat isn’t tied. The man notices his
mistake, stands up and walks to the boat’s
prow, but by the time he reaches the prow,
it’s moved 2.91 m away from the pier.
Assuming no water resistance to the boat’s
motion, calculate the boat’s mass (not counting the man).
Answer in units of kg.
Correct answer: 37.0825 kg.
Explanation:
In the absence of external forces, the center
of mass of the man–boat system remains at
rest. So if the man moves distance ∆Xman
and the boat moves distance ∆Xboat , then we
must have
Mman Xman + Mboat Xboat
∆XCM = ∆
Mman + Mboat
Mman ∆Xman + Mboat ∆Xboat
=0
=
Mman + Mboat
and therefore
Mman ∆Xman + Mboat ∆Xboat = 0 .
Solving this equation for the boat’s mass, we
find
∆Xman
Mboat = Mman ×
.
−∆Xboat
Now, let’s be careful about the displacements. Taking the stern-to-prow direction to
be positive, we have the boat moving backward, so
∆Xboat = −2.91 m < 0.
As to the man, his displacement relative to the
boat is the boat’s full length (stern to prow),
so
∆Xrel = +Lboat = +3.9 m,
but relative to the pier his displacement is only
∆Xman = ∆Xrel + ∆Xboat
= +3.9 m − 2.91 m = +0.99 m .
rawashdeh (ar73422) – Homework 8, torque 21-22 – dowd – (VickeryRPHY1 2)
Consequently,
∆Xman
−∆Xboat
+0.99 m
= 109 kg ×
+2.91 m
= 37.0825 kg.
Mboat = Mman ×
13