5–15.
Determine the horizontal and vertical components of
reaction at the pin at A and the reaction of the roller at B on
the lever.
14 in.
30⬚
F ⫽ 50 lb
A
SOLUTION
Equations of Equilibrium: From the free-body diagram, FB and A x can be obtained
by writing the moment equation of equilibrium about point A and the force
equation of equilibrium along the x axis, respectively.
a+ ©MA = 0;
50 cos 30°(20) + 50 sin 30°(14) - FB(18) = 0
FB = 67.56 lb = 67.6 lb
+ ©F = 0;
:
x
Ans.
A x - 50 sin 30° = 0
A x = 25 lb
Ans.
Using the result FB = 67.56 lb and writing the force equation of equilibrium along
the y axis, we have
+c ©Fy = 0;
A y - 50 cos 30° - 67.56 = 0
A y = 110.86 lb = 111 lb
Ans.
20 in.
B
18 in.
*5–20.
The pad footing is used to support the load of 12 000 lb.
Determine the intensities w1 and w2 of the distributed
loading acting on the base of the footing for the
equilibrium.
12 000 lb
5 in.
9 in.
9 in.
w2
w1
SOLUTION
35 in.
Equations of Equilibrium: The load intensity w2 can be determined directly by
summing moments about point A.
a + ©MA = 0;
w2 a
35
b 117.5 - 11.672 - 12114 - 11.672 = 0
12
w2 = 1.646 kip>ft = 1.65 kip>ft
+ c ©Fy = 0;
Ans.
35
35
1
1w - 1.6462a b + 1.646 a b - 12 = 0
2 1
12
12
w1 = 6.58 kip>ft
Ans.
5–23.
The smooth disks D and E have a weight of 200 lb and 100 lb,
respectively. Determine the largest horizontal force P that
can be applied to the center of disk E without causing the
disk D to move up the incline.
1.5 ft
5
4
1 ft
3
E
A
B
SOLUTION
For disk E:
224
≤ = 0
5
+ ©F = 0;
:
x
- P + N¿ ¢
+ c ©Fy = 0;
1
NC - 100 - N¿ a b = 0
5
For disk D:
+ ©F = 0;
:
x
4
224
NA a b - N¿ ¢
≤ = 0
5
5
+ c ©Fy = 0;
3
1
NA a b + NB - 200 + N¿ a b = 0
5
5
Require NB = 0 for Pmax. Solving,
N¿ = 214 lb
Pmax = 210 lb
NA = 262 lb
NC = 143 lb
Ans.
P
D
C
5–33.
The woman exercises on the rowing machine. If she exerts a
holding force of F = 200 N on handle ABC, determine the
horizontal and vertical components of reaction at pin C and
the force developed along the hydraulic cylinder BD on the
handle.
F ⫽ 200 N
30⬚
0.25 m
A
B
0.25 m
C
D
SOLUTION
0.75 m
Equations of Equilibrium: Since the hydraulic cylinder is pinned at both ends, it can
be considered as a two-force member and therefore exerts a force FBD directed
along its axis on the handle, as shown on the free-body diagram in Fig. a. From the
free-body diagram, FBD can be obtained by writing the moment equation of
equilibrium about point C.
a+ ©MC = 0;
FBD cos 15.52°(250) + FBD sin 15.52°(150) - 200 cos 30°(250 + 250)
-200 sin 30°(750 + 150) = 0
FBD = 628.42 N = 628 N
Ans.
Using the above result and writing the force equations of equilibrium along the x and
y axes, we have
+
: ©Fx = 0;
Cx + 200 cos 30° - 628.42 cos 15.52° = 0
Cx = 432.29 N = 432 N
+ c ©Fy = 0;
0.15 m
Ans.
200 sin 30° - 628.42 sin 15.52° + Cy = 0
Cy = 68.19 N = 68.2 N
Ans.
0.15 m
*5–56.
The disk B has a mass of 20 kg and is supported on the
smooth cylindrical surface by a spring having a stiffness of
k = 400 N>m and unstretched length of l0 = 1 m. The spring
remains in the horizontal position since its end A is attached
to the small roller guide which has negligible weight.
Determine the angle u for equilibrium of the roller.
0.2 m
A
r
u
SOLUTION
+ c ©Fy = 0;
R sin u - 20(9.81) = 0
+ ©F = 0;
:
x
R cos u - F = 0
tan u =
Since cos u =
1.0 +
20(9.81)
F
F
400
2.2
2.2 cos u = 1.0 +
20(9.81)
400 tan u
880 sin u = 400 tan u + 20(9.81)
Solving,
u = 27.1°
and u = 50.2°
B
k
Ans.
2m
*5–72.
z
The pole is subjected to the two forces shown. Determine
the components of reaction of A assuming it to be a balland-socket joint. Also, compute the tension in each of the
guy wires, BC and ED.
30°
20°
2 m F = 450 N
2
45°
E
D
F1 = 860 N
SOLUTION
B
3m
6m
Force Vector and Position Vectors:
6m
FA = Ax i + Ay j + Az k
C
F1 = 8605cos 45°i - sin 45°k6 N = 5608.11i - 608.11k6 N
4m
4.5 m
F2 = 4505-cos 20° cos 30°i + cos 20° sin 30°k - sin 20°k6 N
x
1-6 - 02i + 1-3 - 02j + 10 - 62k
21-6 - 022 + 1-3 - 022 + 10 - 622
R
2
1
2
= - FEDi - FEDj - FEDk
3
3
3
FBC = FBC B
=
16 - 02i + 1-4.5 - 02j + 10 - 42k
216 - 022 + 1-4.5 - 022 + 10 - 422
R
9
8
12
F i F j F k
17 BC
17 BC
17 BC
r1 = 54k6 m
r2 = 58k6 m
r3 = 56k6 m
Equations of Equilibrium: Force equilibrium requires
©F = 0;
FA + F1 + F2 + FED + FBC = 0
aAx + 608.11 - 366.21 -
2
12
+
F
F bi
3 ED
17 BC
+ aAy + 211.43 -
1
9
F
F bj
3 ED
17 BC
+ aAz - 608.11 - 153.91 -
2
8
F
F bk
3 ED
17 BC
0
Equating i, j and k components, we have
©Fx = 0;
Ax + 608.11 - 366.21 -
©Fy = 0;
Ay + 211.43 -
©Fz = 0;
Az - 608.11 - 153.91 -
2
12
F
+
F = 0
3 ED
17 BC
1
9
FED F = 0
3
17 BC
2
8
F
F = 0
3 ED
17 BC
6m
y
= 5-366.21i + 211.43j - 153.91k6 N
FED = FED B
A
(1)
(2)
(3)
*5–72. (continued)
Moment equilibrium requires
©MA = 0;
4k * a
r1 * FBC + r2 * 1F1 + F22 + r3 * FED = 0
9
8
12
F i F j F kb
17 BC
17 BC
17 BC
+ 8k * 1241.90i + 211.43j - 762.02k2
2
1
2
+ 6k * a - FEDi - FEDj - FEDkb = 0
3
3
3
Equating i, j and k components, we have
©Mx = 0;
36
F + 2FED - 1691.45 = 0
17 BC
(4)
©My = 0;
48
F - 4FED + 1935.22 = 0
17 BC
(5)
Solving Eqs. (4) and (5) yields
FBC = 205.09 N = 205 N
FED = 628.57 N = 629 N
Ans.
Substituting the results into Eqs. (1), (2) and (3) yields
Ax = 32.4 N
Ay = 107 N
Az = 1277.58 N = 1.28 kN
Ans.
5–75.
z
If the pulleys are fixed to the shaft, determine the magnitude
of tension T and the x, y, z components of reaction at the
smooth thrust bearing A and smooth journal bearing B.
1m
1m
A
0.2 m
1m
0.3 m B
x
{900 i} N
SOLUTION
Equations of Equilibrium: From the free-body diagram of the shaft, Fig. a, A y, T,
and Bx can be obtained by writing the force equation of equilibrium along the y axis
and the moment equations of equilibrium about the y and z axes, respectively.
©Fy = 0;
Ay = 0
Ans.
©My = 0; 400(0.2) - 900(0.2) - 900(0.3) + T(0.3) = 0
T = 1233.33 N = 1.23 kN
©Mz = 0;
y
{400 i} N
Ans.
-Bx(3) - 400(1) - 900(1) = 0
Bx = -433.33 N = -433N
Ans.
Using the above results and writing the moment equation of equilibrium about the
x axis and the force equation of equilibrium along the x axis, we have
©Mx = 0; Bz(3) - 900(2) - 1233.33(2) = 0
Bz = 1422.22 N = 1.42 kN
Ans.
©Fx = 0; 400 + 900 - 433.33 - A x = 0
A x = 866.67 N = 867 N
Ans.
Finally, writing the force equation of equilibrium along the z axis, yields
©Fz = 0; A z - 1233.33 - 900 + 1422.22 = 0
A z = 711.11 N = 711 N
Ans.
{⫺900 k} N
T
*5–84.
Both pulleys are fixed to the shaft and as the shaft turns
with constant angular velocity, the power of pulley A is
transmitted to pulley B. Determine the horizontal tension T
in the belt on pulley B and the x, y, z components of
reaction at the journal bearing C and thrust bearing D if
u = 45°. The bearings are in proper alignment and exert
only force reactions on the shaft.
SOLUTION
z
200 mm
©Fx = 0;
©Fy = 0;
B
y
80 mm
A
65 N
80 N
6510.082 - 8010.082 + T10.152 - 5010.152 = 0
Ans.
165 + 80210.452 - 50 sin 45°10.22 - Cz 10.752 = 0
Ans.
58.010.22 + 50 cos 45°10.22 - Cy 10.752 = 0
Cy = 24.89 N = 24.9 N
Ans.
Dx = 0
Ans.
Dy + 24.89 - 50 cos 45° - 58.0 = 0
Dy = 68.5 N
©Fz = 0;
150 mm
C
T
Cz = 77.57 N = 77.6 N
©Mz = 0;
θ
x
T = 58.0 N
©My = 0;
D
300 mm
Equations of Equilibrium:
©Mx = 0;
50 N
250 mm
Ans.
Dz + 77.57 + 50 sin 45° - 80 - 65 = 0
Dz = 32.1 N
Ans.