Problem 3.6 A zoologist estimates that the jaw of a
predator, Martes, is subjected to a force P as large as
800 N. What forces T and M must be exerted by the
temporalis and masseter muscles to support this value
of P?
'-'-,--,--J~36°
'- '-
Solution: Resolve the forces into scalar components, and solve the
equilibIium equations ... Express the forces in terms of horizontal and
vertical unit vectors:
T = ITI(icos 22° + j sin 22°) = ITI(0.927i + 0.375j)
p=
800(i cos 270° + j sin 270°) = Oi - 800j
= IMj(icos 144° + j sin 144°) = IMI( -0.809i
Iy the equilibrium
+ 0.588j)
conditions,
=O=T+M+P=O
like terms:
LFx
= (0.927ITI-
0.8091Ml)i = 0
= (0.375ITI + 0.5881MI - 800)j = 0
tbe first
'tute
tho equation,. ITI = (0.809)
0.927
(I)
(2)
IMI = 0.8731MI
~s value IIlto the second equation,
soLve: IMj = 874 N, ITI = 763.3 NJ
reduce algebraically,
Problem 3.17 In Problem 3.16, determine the magnitude of the total friction force exerted on the tow truck's
tires. (This is the friction force the truck's tires must
exert to prevent the truck and car from sliding down
the slope.)
Solution:
Use the value of the tension from the previous problem
L
Solving: I F
F'\ : F - 20.6 kN sin 10° - T cos 8°
= 9.43
20.6 kN
=0
kN I
100
F
---------------.
T
Problem 3.22 A construction worker holds a 180-kg
crate in the position shown. What force must she exert
on the cable?
Solution:
L
F)'
Eqns. of Equilibrium:
= T I COS 5° -
{ LFxmg =
= T2cos30°
(180)(9.81)
T 2 sin 30° - mg
- NTI sinS" = 0
=0
Solving, we get
T1
=
1867 N
T2
=
188 N
5°
-y
x
mg
= ( 180)
(9.81) N
Problem 3.30 An astronaut candidate conducts experiments on an airbearing platform. While he carries out
calibrations, the platform is held in place by the horizontal tethers AB, AC, and AD. The forces exerted by
the tethers are the only horizontal forces acting on the
platform. If the tension in tether AC is 2 N, what are the
tensions in the other two tethers?
Solution:
Isolate the platform. The angles a and
tan a
=
-3.5
(1.5
=
)
fJ
B
are
-I~
0.429,
3rm~D
1.5m
Also,
tan
fJ
= -3.5 = 0.857,
(3.0)
fJ
=
40.6°.
The angle between the tether AB and the positive x axis is (180° hence
TAB
= ITABI(i cos(l80°
TAB
=
-
fJ)
+ j sin( 180° -
...L
;
I
I
I
1~3rii)-I-t?-
fJ),
B
fJ»
D
ITABI(-icosfJ+jsinfJ).
The angle between the tether AC and the positive x axis is (180° + a).
The tension is
TAC
= ITAcl (i cos(l80°
=
+ a) + j sin( 180° + a»
Solve:
ITAcl(-icosa-jsina).
The tether AD is aligned with the positive x axis, TAD = ITADli +
The equilibrium
OJ.
sin(a
fJ
+ fJ») .
ITADI = (ITAcl Sin
condition:
For ITACI = 2 N,
Substitute
L
Fy
fJ
-
ITAcl sin a)j
23.2° and
fJ
=
ITABI = 1.21 N, ITADI = 2.76 N
and collect like terms,
= (ITABIsin
a=
= O.
40.6°,
Problem 3.51 The cable AB is 0.5 m in length. The
unstretched length of the spring is 0.4 m. When the
50-kg mass is suspended at B, the length of the spring
increases to 0.45 m. What is the spring constant k?
Solution:
The Geomelry
0.7 m
Law of Cosines and Law of Sines
F
1 : TAB sin
e + F sin
</> -
490.5 N
=0
0
490.5 N
Problem 3.SS The mass of each pulley of the system
is m and the mass of the suspended object A is mA.
Determine the force T necessary for the system to be in
equilibrium.
Solution: Draw free body diagrams of each pulley and the object
A. Each pulley and the object A must be in equilibrium. The weights
of the pulleys and object A are W = mg and W A = mAg. The equilibrium equations for the weight A, the lower pulley, second pulley, third
pulley, and the top pulley are, respectively, B - W A = 0, 2C - B W = 0, 2D - C - W = 0, 2T - D - W = 0, and Fs - 2T - W = o.
Begin with the first equation and solve for B, substitute for B in the
second equation and solve for C, substitute for C in the third equation
and solve for D, and substitute for D in the fourth equation and solve
for T, to get T in terms of Wand W A. The result is
WA
0= -
4
+ -3W
4 '
and T
= -WA + -7W
or in terms of the masses,
T
=
~(mA
+ 7m).
8
8'
I
IT
Problem 3.88 The cable AB is 0.5 m in length and
the ul1stretched length of the spring Be is 0.4 m. The
spring constant k is 5200 N/m. When the 50-kg mass
is suspended at B, what is the resulting length of the
stretched spring?
Solution:
Introduce
the distances
0.5 In
LBe
=
F
=
and equilibrium
equations
= J b2 + h2
J (0.7
In -
W + h2
B
(5200 N/m)(LBe - 0.4 m)
'\'
~Fx:
'\'
~
h.
(F, TAB, b, h, LBc).
Then we have 4 unknowns
We have the constraint
band
+
---TAB
b
0.5 m
h
h
F y : -0.5 m TAB + -LBe F - 490.5 N
Solving
=
0
we find
h
= 0.335
F
= 364
GOBe
F
0.7 m - b
---F=O
LBe
m, b
N,
= 0.37\
m, LBe
= 0.470
m,
TAB = 343 N
= 0.470
m I
490.5 N
Problem 3.92* The cable AB keeps the 8-kg collar
A in place on the smooth bar CD. The y axis points
upward. Determine the distance s from C to the collar
A for which the tension in the cable is 150 N.
Y
0.15 m
x
Z
Solution:
From the figure, the coordinates of the points
meters) are B(O, 0.5, 0.15), C(O.4, 0.3, 0), and D(0.2, 0, 0.25).
Y
(in
I
0.15 m
~I
The first unit vector is of the form,
1_0.4
B(.).
e'K
=
«XI
-
J (XI
xK)i
-
XK)2
(YI
+ (YI
-
YK)j
-
-
(ZI -
+ (ZI
YK)2
where I K takes on the value CD.
given by
Ax
1 ·
+
)k)
---;==============
=
Cx
+ seCDx,
Ay
=
Cy
ZK
-
ZK
!
)2 '
The coordinates
/'
of point A are
IF
220
of equilibrium
200
T
>,,\
N
~ '.
~
s
-
z/--0.2J~/
,
//"~
0.3 m
1:---.-,'D -c:/
0.25-,rm
,_ -L.-
""j ,./,/
where we do not know the value of s. The equations
for this problem are:
'r AB
'I~C"
0.5 m
+ secoy,
m--t
J_x
-'--=::::\\\-----"----
180
•...•..•
~
"-
I
n 160
LFy
=
TABeABy
+ FNy
-
W
=
140
0,
B
_120
N
100
where TAB
=
80
150 N.
.25
W
= mg,
or W
The condition
= (8)(9.81) = 78.48
= 80.7i -
that the force FN is perpendicular
47.7j
+ 7.29k
N.
.3
.325
.35
.375
.4
.425
.45
.475
.\
s (m)
N.
to CD is
We have three equilibrium equations plus the dot product equation
in the four unknowns, s and the three components of F N. Several
methods of solution are open to us. Any iterative algebraic solution
method should give the result s = 0.3046 m and that
FN
.275
Distance.
The weight of the collar is given by
Alternalive Solulion: The complication in the algebra in the solution
is because we do not know the location of point A. We can assume
the location of A is known (assume that we know the distance .1') and
solve for the value of the tension in cable AB which corresponds 10
that location for A. We can plot the value of the tension versus the
distance s and find the value of s at which the tension is 150 N. If we
do this, we get the plot shown. From the plot, s ~ 0.305 N.