These so
appear in
however.
1. Three
of the wa
downwar
rope and
The horiz
(a) We so
obtain
(b) We so
obtain
N
F
2. The ob
causing a
problem
accelerat
string and
Therefor
lutions use t
n your person
.
forces act on
all
N
F
r
(actin
rd). Since th
the vertical.
zontal comp
olve the first
mg
T
=
olve the seco
22
Tr
Lr
=
=
+
bject exerts a
a “kink” sim
in the text).
tion is zero)
d its “relaxe
e,
T
=
F
/(2si
Phys 210
the paramete
nal homewo
n the sphere
ng horizontal
he sphere is i
. Then, the v
onent is
F
N
t equation fo
(
L
+
=
ond equation
2
mg L
r
L
+
a downward
ilar to that s
By analyzin
2
T
sin
θ
=
F
,
d” position (
in
) = 7.92
01
Home
er values fro
ork assignme
: the tension
lly away from
n equilibrium
vertical comp
–
T
sin
= 0
or the tension
0.85 kg)(9.8
n for the nor
2
r
=
+
force of ma
shown for pr
ng the forces
, where
is
(when the tw
tan
−
=
× 10
3
N.
work 9 Sol
m the proble
ent. The logic
n force
r
T
of
m the wall),
m they sum
ponent of Ne
0.
n:
T
=
mg
/co
2
8 m/s ) (0.0
0.080 m
rmal force:
F
(0.8
mgr
L
=
agnitude
F
=
roblem 10 (se
s at the “kink
the angle (ta
wo segments
1
0.35m
1.72m
⎛⎞
=
⎜⎟
⎝⎠
lution Spr
ems printed i
c used to get
the rope (ac
and the forc
to zero. Let
ewton’s seco
os
. We sub
2
080 m)
(0.
m
+
.
sin
N
T
=
.
5 kg)(9.8 m/
(0.080
3160 N at th
ee the figure
k” where
r
F
aken positive
s are collinea
11.5 .
°
ring `11
in the book,
t to the answ
cting along th
ce of gravity
be the ang
ond law is
T
bstitute cos
2
042 m)
9
=
. Usingsin
2
/s )(0.042 m
0 m)
he midpoint
that accom
is exerted, w
e) between e
ar). In this pr
not those th
wer is the sam
he rope), the
y
mg
r
(acting
gle between
cos
–
mg
=
+
LL
/
2
.4 N
=+
rL
r
/
2
)
4.4 N.
=
of the rope,
panies that
we find (sinc
each segmen
roblem, we h
hat
me,
e force
g
the
= 0.
r
2
to
2
, we
,
ce the
nt of the
have
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View Full Document3. P. 120
diving bo
the weigh
forces eq
(a) The s
which sh
(b) Since
(c) The f
which sh
(d) The r
(e) The f
pedestal
(f) The fo
compress
4. (a) An
(b) Equil
(c) Comp
07. We take
oard. We tak
ht of the div
qual to zero (
econd equat
hould be roun
e
F
1
is negati
first equation
hould be roun
result is posi
force of the d
on the divin
orce of the d
sed.
nalyzing the
librium of ve
puting torque
v
F d
the force of
ke the force o
er, located a
(with upward
tion gives
1
F
= −
nded off to
F
ive, indicatin
n gives
2
F
=
nded off to
F
tive, indicat
diving board
ng board), so
diving board
horizontal fo
ertical forces
es about poi
23
d
Fb Fa
=+
f the left ped
of the right p
at
x
=
L
. The
ds positive),
1
F
Fd
Ld
W
d
−
−
= −
1
1.2 10
=−
×
ng that this f
1
58
WF
−=
3
2
1.7 10
=×
ing that this
d on the left p
this pedesta
on the right
orces (which
s leads to
F
v
nt
O
, we obt
(
10
d
⇒=
estal to be
F
pedestal to b
following tw
and the sum
12
()
FW
WL d
+
−
=
+
−
=
3.0m
(5
1.5m
⎛⎞
−
⎜⎟
⎝⎠
3
0 N . Thus,
force is down
0 N+1160 N
N . Thus, 
force is upw
pedestal is u
al is being st
t pedestal is d
h add to zero
=
F
1
+
F
2
=
tain
)( )
0
N3
.
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 Spring '07
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 Force, Friction, Work, FN

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