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?
Solution: Draw the force triangle and then use the geometry plus cos25°
= l!1
IWI sin25°
= ~
IWI
IWI
=
3500 N
ILl
=
3500 cos 25°
IDI
=
3500 sin 25°
3170N
W
W

----
..•.
_---
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° = (40)(0.5) = 20 (N)
Fy
= -lFlsin60° = -(40)(0.866) = -34.6 (N)
F
=
20i - 34.6j (N) x

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: From the solution to Problem 2.39, eBC
=
-0.78 Ii
+
0.625j
The vector F is given by F
=
IFleBc
F
=
(1.2)(-0.78li
+
0.625j) (k· N)
F
=
-9371 + 750j (N)

y
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?
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))
=
845.72i - 307.8j
FA
= iFAI(icos(l80
+ a)
+ j sin
(I
80
+ a))
= iFAI(-icosa j sina)
The sum vanishes:
FA
+
FB
+
Fe
+
FD
=
1(1438.1 - IFAIcosa)
+ j(944 -IFAI sin a)
=
0
From which FA
=
1438.li + 944j. The magnitude is
IIFAI = )(1438)2
+
(944)2 = 1720 Ib
I
The angle is: tana = 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.
Strategy:
B
You can resolve rOA into a vector from
0 to 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.
y
Solution: See Example 2.10. The length BA is, from the right triangle OAB,
IrABI = IroAlsin30° = 20(0.5) = 10 ft.
Similarly, the length OB is
IrOBI
=
IrOAlcos30°
= 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° + kcos 60°) or rOB
=
15i + 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
=
IrBAI(icos90° +jcosO° +kcos900)
=
Oi+ 10j + Ok.
z
The vector rOA is given by rOA
= rOB
+ rBA, from which
I rOA
=
15i + !OJ+ 8.66k (ft) I
A
Q

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 of point B are B (0, 7, 4). The vector position of B is rOB
=
Oi
+
7j
+
4k.
The vector from point A to point B is given by
From Problem 2.95, rOA = 2.67i + 2.33j + 2.67k. Thus 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
IrABI
=-0.4819i+0.8429j+0.240Ik
The force T is given by
I
TAB
=
ITABluAB = 32uAB = -15.4i
+ 27.0j + 7.7k (Ib)
I z y x

problem 2.105
The magnitudes IVI
20.
= to and IVI = y
, (Ii)
:'
Vse the definition of the dot product to determine
V·V.
(b) Vse Eq. (2.23) to obtain
V.
]'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
x

Problem 2.120
In Problem 2.119, what is the vector component of F parallel to the surface?
Solution: From the solution to Problem 2.119,
F
=
-O.123i
+ O.123j - O.0984k (lb) and
FNORMAL
=
Oi
+
O.0927j
+
O.0232k (Ib)
The component parallel to the surface and the component normal to the surface add to give F(F
=
F NORMAL
+
F parallel).
Thus
Fparallcl
= F -
FNORMAI.
Substituting, we get
Fparallel
=
-O.123Ii+O.0304j
-O.1216k
Ib r

Problem 2.127
Determine the cross product r x F of the position vector r =
4i - 12j
+
3k (m) and the force
=
16i - 22j - 10k (N).
Solution: rxF=
I ~
-I~ -10
~I r x F = (120 - (-66»i
+
(48 - (-40»j
+
(-88 - (-19:2»k (N-m) r x F = 186i
+
88j
+
104k (N-m)