PROBLEM 2.18
PROBLEM 2.5
Solve Problem 2.3 using trigonometry
Two control rods are attached at A to lever AB. Using trigonometry and
knowing that the force in the left-hand rod is F1
120 N, determine
(a) the required force F2 in the right-hand rod if the resultant R of the
forces exerted by the rods on the lever is to be vertical, (b) the
corresponding magnitude of R.
SOLUTION
Problem 2.3: Two forces P and Q are applied as shown at point A of a
hook support. Knowing that P
60 N and Q
100 N, determine
graphically the magnitude and direction of their resultant using (a) the
parallelogram law, (b) the triangle rule.
SOLUTION
Using the force triangle and the Laws of Cosines and Sines
We have:
60 N
180q 15q 30q J
135q
Graphically, by the triangle law
Then:
F2 # 108 N
We measure:
= 22085.28 N2
R # 77 N
or
By trigonometry: Law of Sines
100 N
F2
sin D
D
90q 28q
R
sin 38q
62q, E
and
120
sin E
180q 62q 38q
R2 = (60 N)2 + (100 N)2 – 2 (60 N) (100 N) cos 135°
80q
R = 148.6 N
100 N
148.6 N
=
sin b
sin 135∞
sin b =
Then:
F2
sin 62q
R
sin 38q
FG 100 N IJ sin 135°
H 148.6 N K
0.4758
120 N
sin 80q
E
or (a) F2
(b)
R
107.6 N W
75.0 N W
Then:
28.41q
D E 75q
D
180q
76.59q
R = 148.6 N
76.6° J
PROBLEM 2.25
PROBLEM 2.34
While emptying a wheelbarrow, a gardener exerts on each handle AB a
force P directed along line CD. Knowing that P must have a 135-N
horizontal component, determine (a) the magnitude of the force P, (b) its
vertical component.
Determine the resultant of the three forces of Problem 2.23.
1.2 m
Problem 2.23: Determine the x and y components of each of the forces
shown.
1.4 m
2.25 m
900 N
950 N
108 kN
SOLUTION
1.5 m
2m
SOLUTION
The components of the forces were
determined in Problem 2.23.
P
(a)
F900 = – (417 N)i + (799.2 N) j
Px
cos 40q
F950 = (494.4 N)i + (812.3 N)j
135 N
cos 40q
Thus
or P
(b)
Py
Px tan 40q
F1800 = – (1440 N)i – (1080 N)j
R = Rx + Ry
176.2 N W
P sin 40q
R = – (528.6)i + (531.5)j
135 N tan 40q
Now:
or Py
113.3 N W
tan a =
R
Ry = 531.5 N
a = tan–1
a
Rx = –528.6 N
and
R=
5315
. N
528.6 N
5315
.
= 45.16°
528.6
b-528.6 Ng + b5315. Ng
2
2
= 749.6 N
R = 750 N
45.2° J
PROBLEM 2.43
PROBLEM 2.54
Two cables are tied together at C and are loaded as shown. Determine the
tension (a) in cable AC, (b) in cable BC.
500 mm
300 mm
Two cables tied together at C are loaded as shown. Determine the range
of values of W for which the tension will not exceed 1050 N in either
cable.
750 mm
B
A
300 mm
400 mm
C
15
2.7 kN
8
690 N
400 mm
W
525 mm
SOLUTION
SOLUTION
Free-Body Diagram
Free-Body Diagram
From geometry:
The sides of the triangle with hypotenuse CB are in the ratio 8:15:17.
The sides of the triangle with hypotenuse CA are in the ratio 3:4:5.
Thus:
3
15
15
0: TCA TCB 680 N 5
17
17
6Fx
0
or
2700 N
1
5
TCA TCB
5
17
From the geometry, we calculate the distances:
AC =
BC =
b400 mmg + b300 mmg
b500 mmg + b525 mmg
2
2
2
2
6Fy
or
Hence:
(a)
(b)
0:
= 725 mm
4
8
8
TCA TCB 680 N W
5
17
17
0
or
Â Fx = 0: -
and
(1)
and
= 500 mm
Then, from the Free Body Diagram of point C:
or
200 N
T BC =
Â Fy = 0:
1
2
TCA TCB
5
17
400
525
TAC +
TBC = 0
500
725
300 T + 500 T – 2700 N = 0
AC
BC
500
725
F
H
1
W
4
(2)
Then, from Equations (1) and (2)
725 ¥ 4 T
AC
525 5
300
500 725 ¥ 4 T
T +
AC
500 AC 725 525 5
80 N I – 2700 N = 0
K
TCB
680 N TCA
25
W
28
17
W
28
Now, with T d 1050 N
TCA : TCA
TAC = 1982.5 N
TAC = 1982.5 N J
or
TBC = 2190.2 N J
and
W
TCB : TCB
or
W
1050 N
1176 N
1050 N
609 N
25
W
28
680 N 17
W
28
? 0 d W d 609 N W
PROBLEM 2.61
PROBLEM 2.71
Two cables tied together at C are loaded as shown. Knowing that the
maximum allowable tension in each cable is 900 N, determine (a) the
magnitude of the largest force P which may be applied at C, (b) the
corresponding value of D.
A load Q is applied to the pulley C, which can roll on the cable ACB. The
pulley is held in the position shown by a second cable CAD, which passes
800 N,
over the pulley A and supports a load P. Knowing that P
determine (a) the tension in cable ACB, (b) the magnitude of load Q.
SOLUTION
SOLUTION
Free-Body Diagram: Pulley C
Free-Body Diagram: C
Force Triangle
= 900 N
900 N
900 N
2E
P
(a)
0
2303.5 N
TACB
47.5q
2 900 N cos 47.5 q
180q 55q 47.5q
77.5q
6Fy
(b)
1216 N
P
Since P ! 0, the solution is correct.
D
0: TACB cos 30q cos 50q 800 N cos 50q
TACB
Hence
180q 85q
E
(b)
6Fx
(a)
Force triangle is isoceles with
1216 N W
D
77.5q W
0: TACB sin 30q sin 50q 800 N sin 50q Q
2303.5 N sin 30q sin 50q 800 N sin 50q Q
or
Q
3529.2 N
2.30 kN W
0
0
Q
3.53 kN W