Problem 2.10 The forces acting on the sailplane are
represented by three vectors. The lift L and drag Dare
perpendicular, the magnitude of the weight W is 3500 N,
and W + L + D = O. What are the magnitudes of the lift
and drag?
W
Solution:
Draw the force triangle and then use the geometry plus
cos25°
= l!1
sin25°
=~
IWI
W
IWI
IWI
= 3500
ILl = 3500
IDI
N
cos 25°
= 3500 sin 25°
3170N
----
-----_ _--..•.
Problem 2.23
A fish exerts a 40-N force on the line
that is represented by the vector F. Express F in terms
of components using the coordinate system shown.
Solution:
Fx
=
IFIcos 60°
Fy
=
-lFlsin60°
F
x
= 20i -
=
(40)(0.5)
=
= 20
-(40)(0.866)
34.6j (N)
(N)
=
-34.6 (N)
Problem 2.40 The hydraulic actuator BC in Problem
2.39 exerts a 1.2-kN force F on the joint at C that is
parallel to the actuator and points from B toward C.
Determine the components of F.
Solution:
eBC
= -0.78
From the solution to Problem 2.39,
Ii
+ 0.625j
The vector F is given by F
= IFleBc
F
= (1.2)(-0.78li + 0.625j)
F
= -9371 + 750j
(N)
(k· N)
Problem 2.46 Four groups engage in a tug-of-war. The
magnitudes of the forces exerted by groups B, C, and D
are IFBI = 800 lb, IFel = 1000 lb, IFDI = 900 lb. If the
vector sum of the four forces equals zero, what are the
magnitude of FA and the angle ot?
y
x
Solution: The strategy is to use the angles and magnitudes to
detennine the force vector components. to solve ~orthe unknown force
FA and then take its magnitude. The force vectors are
FB = 800(lcos 110° +jsin 110°) = -273.6i+751.75j
Fe
= 1000(1cos 30° + j sin 30°) = 866i + 500j
FD = 9OO(lcos(_20°) + jsin(-200))
FA = iFAI(icos(l80
= iFAI(-icosa
= 845.72i -
307.8j
+ a) + j sin(I80 + a))
-
j sina)
The sum vanishes:
FA + FB + Fe
+ FD = 1(1438.1 + j(944
IFAIcosa)
-IFAI sin a)
=0
From which FA = 1438.li + 944j. The magnitude is
IIFAI
=
)(1438)2
The angle is: tana
+ (944)2 =
=
1720 Ib
I
194~8= 0.6565. or I a
=
:t3.3° I
Problem 2.83 The distance from point 0 to point A is
20 ft. The straight line AD is parallel to the y axis, and
point B is in the x-z plane. Express the vector rOA in
terms of scalar components.
y
Strategy:
You can resolve rOA into a vector from 0 to
B and a vector from B to A. You can then resolve the
vector form 0 to B into vector components parallel to
the x and z axes. See Example 2.9.
z
Solution:
See Example 2.10. The length BA is, from the right
The vector
rOA
is given by
rOA
= rOB
+
triangle OAB,
I rOA = 15i +
IrABI= IroAlsin30° = 20(0.5) = 10 ft.
Similarly, the length OB is
IrOBI
=
IrOAlcos30°
I
A
= 20(0.866) = 17.32 ft
The vector rOB can be resolved into components along the axes by the
right triangles OBP and OBQ and the condition that it lies in the x-z
plane.
Hence,
rOB
= IrOBI(icos 30° + j cos90°
rOB
= 15i +
+ kcos 60°) or
OJ+ 8.66k.
The vector rBA can be resolved into components from the condition
that it is parallel to the y-axis. This vector is
rBA
!OJ+ 8.66k (ft)
= IrBAI(icos90° +jcosO°
+kcos900)
= Oi+
10j + Ok.
Q
rBA,
from which
Problem 2.96 The cable AB exerts a 32-lb force T on
the collar at A. Express T in terms of scalar components.
Solution:
The coordinates
position of B is rOB
y
of point B are B (0, 7, 4). The vector
= Oi + 7j + 4k.
x
The vector from point A to point B is given by
From Problem 2.95, rOA = 2.67i + 2.33j + 2.67k. Thus
z
rAB = (0 - 2.67)i + (7 - 2.33)j + (4 - 2.67)j
rAB = -2.67i
+ 4.67j + 1.33k.
The magnitude is
IrABI = -/2.672
+ 4.672 + 1.332 = 5.54 ft.
The unit vector pointing from A to B is
UAB=
rAB =-0.4819i+0.8429j+0.240Ik
IrABI
The force T is given by
I TAB = ITABluAB
= 32uAB = -15.4i
+ 27.0j + 7.7k (Ib)
I
problem 2.105
20.
, (Ii)
:'
(b)
The magnitudes
IVI
=
to and IVI
=
y
Vse the definition of the dot product to determine
V·V.
Vse Eq. (2.23) to obtain V· V.
x
]'he definition of the dot product (Eq. (2.18» is
·V=
IUIIVlcosO. Thus
'V = (10)(20) cos(45° - 30°) = 193.2
components of U and V are
O(icos45°
+ jsin45°)
= 7.07i + 7.07j
(icos30°
+ jsin300)
= 17.32i + IOj
. (2.23)
I U·
V=(7.07)(l7.32)+
(7.07)(10)
= 193.21
Problem 2.120 In Problem 2.119, what is the vector
component of F parallel to the surface?
Solution:
F
= -O.123i
FNORMAL
Thus
From the solution to Problem 2.119,
+ O.123j - O.0984k (lb) and
= Oi + O.0927j + O.0232k
The component
= F NORMAL + F parallel).
=F-
Substituting,
(Ib)
parallel to the surface and the component
the surface add to give F(F
Fparallcl
normal to
Fparallel
FNORMAI.
we get
= -O.123Ii+O.0304j
-O.1216k
Ib
r
---------------
Problem 2.127
Determine the cross product
the position vector
= 16i -
r = 4i -
22j - 10k (N).
rx
F of
Solution:
12j + 3k (m) and the force
rxF=
I16~ -22
-I~
~I
-10
r x F = (120 - (-66»i + (48 - (-40»j
+ (-88
- (-19:2»k (N-m)
r x F = 186i + 88j
+
104k (N-m)