PROBLEM 4.

156
A 2100-lb tractor is used to lift 900 lb of gravel. Determine the reaction at
each of the two (a) rear wheels A, (b) front wheels B.

SOLUTION

(a) From f.b.d. of tractor

ΣM B = 0: ( 2100 lb )( 40 in.) − ( 2 A)( 60 in.) − ( 900 lb )( 50 in.) = 0

∴ A = 325 lb or A = 325 lb

(b) From f.b.d. of tractor

ΣM A = 0: ( 2B )( 60 in.) − ( 2100 lb )( 20 in.) − ( 900 lb )(110 in.) = 0

∴ B = 1175 lb or B = 1175 lb
 PROBLEM 4.157
A tension of 5 lb is maintained in a tape as it passes the support system
shown. Knowing that the radius of each pulley is 0.4 in., determine the
reaction at C.

SOLUTION

From f.b.d. of system

ΣFx = 0: C x + ( 5 lb ) = 0

∴ Cx = −5 lb

ΣFy = 0: C y − ( 5 lb ) = 0

∴ C y = 5 lb

( Cx )2 + ( C y )
2
Then C = = ( 5 )2 + ( 5 )2 = 7.0711 lb

 +5 
and θ = tan −1   = −45°
 −5 

or C = 7.07 lb 45.0°

ΣM C = 0: M C + ( 5 lb )( 6.4 in.) + ( 5 lb )( 2.2 in.) = 0

∴ M C = −43.0 lb ⋅ in or M C = 43.0 lb ⋅ in.

 PROBLEM 4.158
Solve Problem 4.157 assuming that 0.6-in.-radius pulleys are used.
P4.157 A tension of 5 lb is maintained in a tape as it passes the support
system shown. Knowing that the radius of each pulley is 0.4 in., determine
the reaction at C.

SOLUTION
From f.b.d of system

ΣFx = 0: C x + ( 5 lb ) = 0

∴ Cx = −5 lb

ΣFy = 0: C y − ( 5 lb ) = 0

∴ C y = 5 lb

( C x )2 + ( C y )
2
Then C = = ( 5 )2 + ( 5 )2 = 7.0711 lb

 5 
and θ = tan −1   = −45.0°
 −5 

or C = 7.07 lb 45.0°

ΣM C = 0: M C + ( 5 lb )( 6.6 in.) + ( 5 lb )( 2.4 in.) = 0

∴ M C = −45.0 lb ⋅ in.

or M C = 45.0 lb ⋅ in.
 PROBLEM 4.159
The bent rod ABEF is supported by bearings at C and D and by wire AH.
Knowing that portion AB of the rod is 250 mm long, determine (a) the
tension in wire AH, (b) the reactions at C and D. Assume that the bearing
at D does not exert any axial thrust.

SOLUTION
(a) From f.b.d. of bent rod

( )
ΣM CD = 0: λ CD ⋅ rH /B × T + λ CD ⋅ rF /E × F = 0 ( )
where λ CD = i

rH /B = ( 0.25 m ) j

T = λ AH T

=
( y AH ) j − ( z AH ) k T
( y AH )2 + ( z AH )2
y AH = ( 0.25 m ) − ( 0.25 m ) sin 30°

= 0.125 m

z AH = ( 0.25 m ) cos 30°

= 0.21651 m

T
∴T= ( 0.125j − 0.21651k )
0.25
rF /E = ( 0.25 m ) k

F = −400 N j

1 0 0 1 0 0
 T 
∴ 0 1 0 ( )
0.25  0 0 1 ( 0.25 )( 400 N ) = 0
+
 0.25 
0 0.125 −0.21651 0 −1 0

−0.21651T + 0.25 ( 400 N ) = 0

∴ T = 461.88 N
or T = 462 N
PROBLEM 4.159 CONTINUED

(b) From f.b.d. of bent rod

ΣFx = 0: C x = 0

ΣM D( z -axis ) = 0: − ( 461.88 N ) sin 30°  ( 0.35 m ) − C y ( 0.3 m )

− ( 400 N )( 0.05 m ) = 0

∴ C y = −336.10 N

ΣM D( y -axis ) = 0: Cz ( 0.3 m ) − ( 461.88 N ) cos 30°  ( 0.35 m ) = 0

∴ Cz = 466.67 N

or C = − ( 336 N ) j + ( 467 N ) k

ΣFy = 0: Dy − 336.10 N + ( 461.88 N ) sin 30° − 400 N = 0

∴ Dy = 505.16 N

ΣFz = 0: Dz + 466.67 N − ( 461.88 N ) cos30° = 0

∴ Dz = −66.670 N

or D = ( 505 N ) j − ( 66.7 N ) k
 PROBLEM 4.160
For the beam and loading shown, determine (a) the reaction at A, (b) the
tension in cable BC.

SOLUTION
(a) From f.b.d of beam

ΣFx = 0: Ax = 0

ΣM B = 0: (15 lb )( 28 in.) + ( 20 lb )( 22 in.) + ( 35 lb )(14 in.)

+ ( 20 lb )( 6 in.) − Ay ( 6 in.) = 0

∴ Ay = 245 lb

or A = 245 lb
(b) From f.b.d of beam
ΣM A = 0: (15 lb )( 22 in.) + ( 20 lb )(16 in.) + ( 35 lb )(8 in.)
− (15 lb )( 6 in.) − TB ( 6 in.) = 0

∴ TB = 140.0 lb

or TB = 140.0 lb
Check:
ΣFy = 0: −15 lb − 20 lb − 35 lb − 20 lb

− 15 lb − 140 lb + 245 lb = 0?
245 lb − 245 lb = 0 ok
 PROBLEM 4.161
Frame ABCD is supported by a ball-and-socket joint at A and by three
cables. For a = 150 mm, determine the tension in each cable and the
reaction at A.

SOLUTION
− ( 0.48 m ) i + ( 0.14 m ) j
First note TDG = λ DGTDG = TDG
( 0.48)2 + ( 0.14 )2 m

−0.48i + 0.14 j
= TDG
0.50
TDG
= ( 24i + 7 j)
25
− ( 0.48 m ) i + ( 0.2 m ) k
TBE = λ BETBE = TBE
( 0.48)2 + ( 0.2 )2 m
−0.48i + 0.2k
= TBE
0.52
TBE
= ( −12 j + 5k )
13
From f.b.d. of frame ABCD
 7 
ΣM x = 0:  TDG  ( 0.3 m ) − ( 350 N )( 0.15 m ) = 0
 25 
or TDG = 625 N

 24   5 
ΣM y = 0:  × 625 N  ( 0.3 m ) −  TBE  ( 0.48 m ) = 0
 25   13 
or TBE = 975 N

 7 
ΣM z = 0: TCF ( 0.14 m ) +  × 625 N  ( 0.48 m )
 25 
− ( 350 N )( 0.48 m ) = 0

or TCF = 600 N
PROBLEM 4.161 CONTINUED

ΣFx = 0: Ax + TCF + (TBE ) x + (TDG ) x = 0

 12   24 
Ax − 600 N −  × 975 N  −  × 625 N  = 0
 13   25 
∴ Ax = 2100 N

ΣFy = 0: Ay + (TDG ) y − 350 N = 0

 7 
Ay +  × 625 N  − 350 N = 0
 25 
∴ Ay = 175.0 N

ΣFz = 0: Az + (TBE ) z = 0

5 
Az +  × 975 N  = 0
 13 
∴ Az = −375 N

Therefore A = ( 2100 N ) i + (175.0 N ) j − ( 375 N ) k

 PROBLEM 4.162
Frame ABCD is supported by a ball-and-socket joint at A and by three
cables. Knowing that the 350-N load is applied at D (a = 300 mm),
determine the tension in each cable and the reaction at A.

SOLUTION
− ( 0.48 m ) i + ( 0.14 m ) j
First note TDG = λ DGTDG = TDG
( 0.48)2 + ( 0.14 )2 m
−0.48i + 0.14 j
= TDG
0.50
TDG
= ( 24i + 7 j)
25
− ( 0.48 m ) i + ( 0.2 m ) k
TBE = λ BETBE = TBE
( 0.48)2 + ( 0.2 )2 m
−0.48i + 0.2k
= TBE
0.52
TBE
= ( −12i + 5k )
13
From f.b.d of frame ABCD
 7 
ΣM x = 0:  TDG  ( 0.3 m ) − ( 350 N )( 0.3 m ) = 0
 25 
or TDG = 1250 N

 24   5 
ΣM y = 0:  × 1250 N  ( 0.3 m ) −  TBE  ( 0.48 m ) = 0
 25   13 
or TBE = 1950 N

 7 
ΣM z = 0: TCF ( 0.14 m ) +  × 1250 N  ( 0.48 m )
 25 
− ( 350 N )( 0.48 m ) = 0

or TCF = 0
PROBLEM 4.162 CONTINUED

ΣFx = 0: Ax + TCF + (TBE ) x + (TDG ) x = 0

 12   24 
Ax + 0 −  × 1950 N  −  × 1250 N  = 0
 13   25 
∴ Ax = 3000 N

ΣFy = 0: Ay + (TDG ) y − 350 N = 0

 7 
Ay +  × 1250 N  − 350 N = 0
 25 
∴ Ay = 0

ΣFz = 0: Az + (TBE ) z = 0

5 
Az +  × 1950 N  = 0
 13 
∴ Az = −750 N

Therefore A = ( 3000 N ) i − ( 750 N ) k

 PROBLEM 4.163
In the problems listed below, the rigid bodies considered were completely
constrained and the reactions were statically determinate. For each of
these rigid bodies it is possible to create an improper set of constraints by
changing a dimension of the body. In each of the following problems
determine the value of a which results in improper constraints.
(a) Problem 4.81, (b) Problem 4.82.

SOLUTION

(a)
(a) ΣM B = 0: ( 300 lb )(16 in.) − T (16 in.) + T ( a ) = 0

or T =
( 300 lb )(16 in.)
(16 − a ) in.
∴ T becomes infinite when
16 − a = 0
or a = 16.00 in.

 8 
(b) ΣM C = 0: (T − 80 N )( 0.2 m ) −  T  ( 0.175 m )
(b)  17 

 15 
−  T  ( 0.4 m − a ) = 0
 17 
0.2T − 16.0 − 0.82353T − 0.35294T + 0.88235Ta = 0
16.0
or T =
0.88235a − 0.23529
∴ T becomes infinite when
0.88235a − 0.23529 = 0
a = 0.26666 m
or a = 267 mm