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PROBLEM 4.37
Determine the value of a for which the magnitude of the reaction at B
is equal to 800 N.
SOLUTION
Free-Body Diagram:
Force triangle
80 mm 80 mm
tan tan
a
a
ββ
= =
(1)
From force triangle:
320 N
tan 0.4
800 N
β
= =
From Eq. (1):
80 mm
0.4
a=
200 mma=
consent of McGraw-Hill Education.
PROBLEM 4.38
For the frame and loading shown, determine the reactions at C and D.
SOLUTION
Since BD is a two-force member, the reaction at D must pass through Points B and D.
Free-Body Diagram:
(Three-force body)
Triangle CEF:
4.5 ft
tan 56.310
3 ft
ββ
= = °
Triangle ABE:
1
tan 26.565
2
γγ
= = °
consent of McGraw-Hill Education.
SOLUTION Continued
Force Triangle
Law of sines:
150 lb
sin 29.745 sin116.565 sin33.690
CD
= =
° °°
270.42 lb,
167.704 lb
C
D
=
=
270 lb=C
56.3 ;°
167.7 lb=D
26.6°
consent of McGraw-Hill Education.
PROBLEM 4.39
For the boom and loading shown, determine (a) the tension in cord
BD, (b) the reaction at C.
SOLUTION
Three-force body: 3-kip load and T intersect at E.
Geometry:
32
; 12 48
8 in.
BF CF BF
AG CG
BF
= =
=
32 32 8 24 in.
48
; ; 36 in.
24 32
36 32 4 in.
BH BF
JE DJ JE JE
BH DH
EG JE JG
= − = −=
= = =
= − =−=
4 in.
tan 48 in.
4.7636
24 in.
tan 32 in.
36.870
a
a
β
β
=
= °
=
= °
Force Triangle
Law of sines:
3 kips
sin94.764 sin32.106 sin53.13
BD
TC
= =
°°
(a)
5.63 kips
BD
T=
(b)
4.52 kips=
C
4.76°
Free-Body Diagram:
consent of McGraw-Hill Education.
PROBLEM 4.40
A slender rod BC of length L and weight W is held by two cables as
cable CD forms with the horizontal, (b) the tension in each cable.
SOLUTION
Free-Body Diagram:
(Three-force body)
1
2
tan
sin 40
cos40
2tan 40
59.210
CF
EF
L
L
θ
=
°
=°
= °
= °
59.2
θ
= °
(b) Force Triangle
tan30.790
0.59588
AB
TW
W
= °
=
0.596
AB
TW=
cos30.790
1.16408
CD
W
T
W
=°
=
1.164
CD
TW=
consent of McGraw-Hill Education.
PROBLEM 4.41
Knowing that
θ
= 30°, determine the reaction (a) at B, (b) at C.
SOLUTION
In ∆ CDE:
( 3 1)
tan
31
36.2
R
R
β
β
−
=
= −
= °
Force Triangle
Law of sines:
sin 23.8 sin126.2 sin30
2.00
1.239
P BC
BP
CP
= =
° °°
=
=
(a)
2P=B
60.0°
(b)
1.239P=
C
36.2°
Free-Body Diagram:
(Three-force body)
consent of McGraw-Hill Education.
PROBLEM 4.42
A slender rod of length L is attached to collars that can slide freely along
the guides shown. Knowing that the rod is in equilibrium, derive an
expression for the angle
θ
in terms of the angle
β
.
SOLUTION
As shown in the free-body diagram of the slender rod AB, the three forces intersect at C. From the force
geometry:
Free-Body Diagram:
tan GB
AB
x
y
β
=
where
cos
AB
yL
θ
=
and
1sin
2
GB
xL
θ
=
1
2sin
tan cos
1tan
2
L
L
θ
βθ
θ
=
=
or
tan 2tan
θβ
=
consent of McGraw-Hill Education.
PROBLEM 4.43
An 8-kg slender rod of length L is attached to collars that can slide freely
along the guides shown. Knowing that the rod is in equilibrium and that
β
= 30°, determine (a) the angle
θ
that the rod forms with the vertical,
(b) the reactions at A and B.
SOLUTION
geometry of the forces:
Free-Body Diagram:
tan
CB
BC
x
y
β
=
where
1sin
2
CB
xL
θ
=
and
cos
1
tan tan
2
BC
yL
θ
βθ
=
=
or
tan 2tan
θβ
=
For
30
β
= °
tan 2tan30
1.15470
49.107
θ
θ
= °
=
= °
or
49.1
θ
= °
(b)
2
(8 kg)(9.81m/s ) 78.480 NW mg= = =
From force triangle:
tan
(78.480 N)tan30
AW
β
=
= °
45.310 N=
or
45.3 N=A
and
78.480 N 90.621 N
cos cos30
W
B
β
= = =
°
or
90.6 N=B
60.0°
consent of McGraw-Hill Education.
PROBLEM 4.44
Rod AB is supported by a pin and bracket at A and rests against a frictionless
peg at C. Determine the reactions at A and C when a 170-N vertical force is
applied at B.
SOLUTION
The reaction at A must pass through D where C and the 170-N force intersect.
160 mm
tan 300 mm
28.07
a
a
=
= °
28.07CAD
a
= = °�
Also, since
,CD CB⊥
reaction C forms angle
28.07
a
= °
with the horizontal axis.
Force triangle
180 (90 ) 2 90
aa a
°− °− − = °−
Force triangle is isosceles, and we have
170 N
2(170 N)sin
160.0 N
A
C
a
=
=
=
170.0 N=A
33.9°;
160.0 N
=C
28.1°
Free-Body Diagram:
(Three-force body)
consent of McGraw-Hill Education.
PROBLEM 4.37
Determine the value of a for which the magnitude of the reaction at B
is equal to 800 N.
SOLUTION
Free-Body Diagram:
Force triangle
80 mm 80 mm
tan tan
a
a
ββ
= =
(1)
From force triangle:
320 N
tan 0.4
800 N
β
= =
From Eq. (1):
80 mm
0.4
a=
200 mma=
consent of McGraw-Hill Education.
PROBLEM 4.38
For the frame and loading shown, determine the reactions at C and D.
SOLUTION
Since BD is a two-force member, the reaction at D must pass through Points B and D.
Free-Body Diagram:
(Three-force body)
Triangle CEF:
4.5 ft
tan 56.310
3 ft
ββ
= = °
Triangle ABE:
1
tan 26.565
2
γγ
= = °
consent of McGraw-Hill Education.
SOLUTION Continued
Force Triangle
Law of sines:
150 lb
sin 29.745 sin116.565 sin33.690
CD
= =
° °°
270.42 lb,
167.704 lb
C
D
=
=
270 lb=C
56.3 ;°
167.7 lb=D
26.6°
consent of McGraw-Hill Education.
PROBLEM 4.39
For the boom and loading shown, determine (a) the tension in cord
BD, (b) the reaction at C.
SOLUTION
Three-force body: 3-kip load and T intersect at E.
Geometry:
32
; 12 48
8 in.
BF CF BF
AG CG
BF
= =
=
32 32 8 24 in.
48
; ; 36 in.
24 32
36 32 4 in.
BH BF
JE DJ JE JE
BH DH
EG JE JG
= − = −=
= = =
= − =−=
4 in.
tan 48 in.
4.7636
24 in.
tan 32 in.
36.870
a
a
β
β
=
= °
=
= °
Force Triangle
Law of sines:
3 kips
sin94.764 sin32.106 sin53.13
BD
TC
= =
°°
(a)
5.63 kips
BD
T=
(b)
4.52 kips=
C
4.76°
Free-Body Diagram:
consent of McGraw-Hill Education.
PROBLEM 4.40
A slender rod BC of length L and weight W is held by two cables as
cable CD forms with the horizontal, (b) the tension in each cable.
SOLUTION
Free-Body Diagram:
(Three-force body)
1
2
tan
sin 40
cos40
2tan 40
59.210
CF
EF
L
L
θ
=
°
=°
= °
= °
59.2
θ
= °
(b) Force Triangle
tan30.790
0.59588
AB
TW
W
= °
=
0.596
AB
TW=
cos30.790
1.16408
CD
W
T
W
=°
=
1.164
CD
TW=
consent of McGraw-Hill Education.
PROBLEM 4.41
Knowing that
θ
= 30°, determine the reaction (a) at B, (b) at C.
SOLUTION
In ∆ CDE:
( 3 1)
tan
31
36.2
R
R
β
β
−
=
= −
= °
Force Triangle
Law of sines:
sin 23.8 sin126.2 sin30
2.00
1.239
P BC
BP
CP
= =
° °°
=
=
(a)
2P=B
60.0°
(b)
1.239P=
C
36.2°
Free-Body Diagram:
(Three-force body)
consent of McGraw-Hill Education.
PROBLEM 4.42
A slender rod of length L is attached to collars that can slide freely along
the guides shown. Knowing that the rod is in equilibrium, derive an
expression for the angle
θ
in terms of the angle
β
.
SOLUTION
As shown in the free-body diagram of the slender rod AB, the three forces intersect at C. From the force
geometry:
Free-Body Diagram:
tan GB
AB
x
y
β
=
where
cos
AB
yL
θ
=
and
1sin
2
GB
xL
θ
=
1
2sin
tan cos
1tan
2
L
L
θ
βθ
θ
=
=
or
tan 2tan
θβ
=
consent of McGraw-Hill Education.
PROBLEM 4.43
An 8-kg slender rod of length L is attached to collars that can slide freely
along the guides shown. Knowing that the rod is in equilibrium and that
β
= 30°, determine (a) the angle
θ
that the rod forms with the vertical,
(b) the reactions at A and B.
SOLUTION
geometry of the forces:
Free-Body Diagram:
tan
CB
BC
x
y
β
=
where
1sin
2
CB
xL
θ
=
and
cos
1
tan tan
2
BC
yL
θ
βθ
=
=
or
tan 2tan
θβ
=
For
30
β
= °
tan 2tan30
1.15470
49.107
θ
θ
= °
=
= °
or
49.1
θ
= °
(b)
2
(8 kg)(9.81m/s ) 78.480 NW mg= = =
From force triangle:
tan
(78.480 N)tan30
AW
β
=
= °
45.310 N=
or
45.3 N=A
and
78.480 N 90.621 N
cos cos30
W
B
β
= = =
°
or
90.6 N=B
60.0°
consent of McGraw-Hill Education.
PROBLEM 4.44
Rod AB is supported by a pin and bracket at A and rests against a frictionless
peg at C. Determine the reactions at A and C when a 170-N vertical force is
applied at B.
SOLUTION
The reaction at A must pass through D where C and the 170-N force intersect.
160 mm
tan 300 mm
28.07
a
a
=
= °
28.07CAD
a
= = °�
Also, since
,CD CB⊥
reaction C forms angle
28.07
a
= °
with the horizontal axis.
Force triangle
180 (90 ) 2 90
aa a
°− °− − = °−
Force triangle is isosceles, and we have
170 N
2(170 N)sin
160.0 N
A
C
a
=
=
=
170.0 N=A
33.9°;
160.0 N
=C
28.1°
Free-Body Diagram:
(Three-force body)
consent of McGraw-Hill Education.
