4138.
The loading on the bookshelf is distributed as shown.
Determine the magnitude of the equivalent resultant
location, measured from point O.
2 lb/ft
3.5 lb/ft
O
A
2.75 ft
4 ft
SOLUTION
+ TFR O = F;
c + MR O = MO ;
FRO = 8 + 5.25 = 13.25 = 13.2 lbT
Ans
511.
Determine the magnitude of the reactions on the beam at A
and B. Neglect the thickness of the beam.
600 N
3
5
4
B
A
4m
SOLUTION
a + MA = 0;
By (12) - (400 cos 15)(12) - 600(4) = 0
By = 586.37 = 586 N
+
: Fx = 0;
Ans.
Ax - 400 sin 15 = 0
Ax = 103.528
45.
Determine the moment about point B of each of the three
forces acting on the beam.
F2 = 500 lb
F1 = 375 lb
5
A
4
3
B
0.5 ft
8 ft
SOLUTION
F3 = 160 lb
Ans.
4
a +1MF22B = 500 a b 152
5
= 2000 lb # ft = 2.00 kip # ft (Counterclockwise)
Ans.
a +1MF32B = 1
28 CHAPTER 2 FORCE VECTORS
2—1. Determine the magnitude of the resultant force
FR = F1 + F2 and its direction, measured counterclockw1se
from the positive x axis.
r1 = 2501b
F2 = 3751b
Prob. 2—1 '
2—2. If 0 = 60° and F = 450 N, determine the magnitude
l
94 CHAPTER 3 EQUILIBRIUM OF A PARTICLE 3.3 COPLANAR FORCE SYSTEMS 95 1
I
I F NleMEN'FAL "PliaBL‘EMS:
All oblem solutions must include an FBD F3—4. The block has a mass of 5 kg and rests on the smooth All problem solutions must include an FBD. *3—4. Co
562.
The uniform load has a mass of 600 kg and is lifted using a
uniform 30-kg strongback beam and four wire ropes as
shown. Determine the tension in each segment of rope and
the force that must be applied to the sling at A.
F
1.25 m
A
B
2m
SOLUTION
Equat
450.
The chain AB exerts a force of 20 lb on the door at B.
Determine the magnitude of the moment of this force along
the hinged axis x of the door.
z
3 ft
2 ft
A
SOLUTION
O
Position Vector and Force Vector:
F = 20 lb
B
y
4 ft
rOA = 513 - 02i + 14 - 02k6
*628.
Determine the force in members CD, HI, and CJ of the truss,
and state if the members are in tension or compression.
J
K
I
H
G
4 ft
A
B
SOLUTION
3 ft
Method of Sections: The forces in members HI, CH, and CD are exposed by cutting
the truss into two p
62.
Determine the force on each member of the truss and state
if the members are in tension or compression. Set
P1 = 500 lb and P2 = 100 lb.
6 ft
8 ft
C
A
8 ft
SOLUTION
Method of Joints: In this case, the support reactions are not required for
determining
2—89.
If F : {350i — 2503‘ 45014} N and cabie AB is 9 m long.
determine the x. y, z coordinates of point A.
SOLUTiON
Position Vector: The position vector 1313. directed from point A to point B. is given by
but = [9 ﬂ (’Uii "* (0 _ xii + (U w «0k
2 xi yj :
45.
Determine the moment about point B of each of the three
forces acting on the beam.
F2 = 500 lb
F1 = 375 lb
5
A
8 ft
SOLUTION
a +1MF12B = 3751112
0.5 ft
5 ft
F3 = 160 lb
Ans.
= 2000 lb # ft = 2.00 kip # ft (Counterclockwise)
Ans.
= 40.0 lb # ft (Counte
343.
Determine the magnitude and direction of the force P
required to keep the concurrent force system in
equilibrium.
z
(1.5 m, 3 m, 3 m)
P
F2 = 0.75 kN
120
SOLUTION
F3 = 0.5 kN
45
Cartesian Vector Notation:
F1 = 2 kN
F1 = 25cos 45i + cos 60j + cos 120k6