Problem 1:
Rod AB is subjected to the 200-N force.
Determine the reactions at the ball-and-socket
joint A and the tension in cables BD and BE.
z
Solution:
Equations of Equilibriu m
Az
FA  Ax i  Ay j  Az k
A
Ax
TE  TE i
y
Ay
x
rC
TD  TD j
F   200k N
C
Apply the force equation of equilibriu m
rB
200 N
F  0
 FA  TE  TD  F  0
TD
 ( Ax  TE )i  ( Ay  TD ) j  ( Az  200)k  0
F
F
F
x
 0 Ax  TE  0
(1)
y
 0 Ay  TD  0
( 2)
z
 0 Az  200  0
(3)
Summing moments about point A yields
M
A
 0 rC  F  rB  (TE  TD )  0
rB
, then
2
(0.5i  1 j  1k )  ( 200k )  (1i  2 j  2k )  (TE i  TD j )  0
Since rC 
TE
B
(0.5i  1 j  1k )  ( 200k )  (1i  2 j  2k )  (TE i  TD j )  0
Expanding and rearrangin g terms gives
( 2TD  200)i  ( 2TE  100) j  (TD  2TE )k  0
 M  0 2T  200  0
 M  0   2T  100  0
 M  0 T  2T  0
 F  0 A  T  0
 F  0 A  T  0
 F  0 A  200  0
 M  0 2T  200  0
 M  0   2T  100  0
 M  0 T  2T  0
x
D
y
( 4)
E
z
D
E
( 5)
( 6)
x
x
E
(1)
y
y
D
( 2)
z
z
x
D
y
E
z
D
E
(3)
( 4)
( 5)
( 6)
Solving Equations 1 through 6, we get
TD  100 N
TE  50 N
Ax  50 N
Ay  100 N
Az  200 N
Problem 2:
Determine the x, y, z
components of reaction acting
on the ball-and-socket at A, the
reaction at the roller B, and the
tension in the cord CD required
for equilibrium of the plate.
Solution:
The FBD is:
z
TDC
5 ft
2 ft
Ay
y
200 lb
100 lb.ft
2.5 ft
2.5 ft
Ax
x
M
x
Az
0
 100  Bz (5)  200( 2.5)  0
 Bz  80 lb
M
y
0
 200( 2)  Az ( 2)  80( 2)  0
 Az  120 lb
F
z
0
 80  120  200  TCD  0
 TCD  0 lb
F
x
0
 Ax  0
F
y
0
 Ay  0
Bz
Problem 3:
A skeletal diagram of the lower leg is
shown.
The quadriceps muscle attached to
the hip at A and to the patella bone at
B lift the leg. This bone slides freely
over cartilage at the knee joint.
The quadriceps is further extended
and attached to the tibia at C.
Determine the tension T in the
quadriceps at C and the magnitude of
the resultant force at the (femur) pin,
D, in order to hold the lower leg.
The lower leg has a mass of 3.2 kg
and a mass center at G1; the foot has
a mass of 1.6 kg and a mass center at
G2.
Solution:
The FBD is:
 25 
tan 1    18.43
 75 
90  75  15
75 mm
B
18.43
15
Dx
350 mm
300 mm
C
D
75
Dy
G1
G2
3.2(9.81) N
1.6(9.81) N
  M
D
0
T sin 18.43(75)  3.2(9.81)( 425 sin 75)  1.6(9.81)(725 sin 75)  0
 T  1006.82 N  1.01kN
   Fy  0
D y  1006.82 sin 33.43  3.2(9.81)  1.6(9.81)  0
 D y  507.66 N



 Fx  0
Dx  1006.82 cos 33.43  0
 Dx  840.20 N
FD  Dx2  D y2  (507.66) 2  (840.20) 2
 FD  982 N