Chapter 4: Equilibrium of Rigid Bodies
In 2D, there are 3 equations
of equilibrium:
Fx = 0
Fy = 0
Mz = 0
(in the xy plane, rotations
are about the z=axis)
Ch4 Page 1
Ch4 Page 2
Ch4 Page 3
Ch4 Page 4
Ch4 Page 5
Ch4 Page 6
Ch4 Page 7
Sample Problem 4.1
A fixed crane has a mass of 100 kg and is used to
lift a 2400 kg crate. It is held in place in a pin at A
and a rocker at B. The center of gravity of the
crane is located at G.
Determine the components of the reactions at A
and B.
Free-body diagram: We now know the types of reactions by each support (A is a pin
joint and B is a rocker).
Equations of equilibrium: Point A is a good choice to calculate moments about
because there are two unknown reactions at that point. Recall that if the line of
action of a force intersects a point or axis, it has no moment about that point or axis.
Rule of thumb: choose a point with the most number of unknown reactions when
calculating moments.
∑𝐹 = 0 = 𝐴 + 𝐵
∑𝐹 = 0 = 𝐴 − 9.81 kN − 23.5 kN
∑𝑀 = 0 = 𝐵(1.5 m) − (9.81 kN)(2 m) − (23.5 kN)(6 m)
Solving: Positive answer indicates that assumed direction in free-body diagram is
correct. Negative answer indicates that assumed direction is incorrect.
From Fy equation: 𝐴 = +33.3 kN
𝐴 = 33.3 kN ↑
From MA equation: 𝐵 = +107.1 kN
𝐵 = 107.1 kN →
From Fx equation: 𝐴 = −107.1 kN
𝐴 = 107.1 kN ←
Ch4 Page 8
Ch4 Page 9
Ch4 Page 10
Sample Problem 4.4
The frame supports par of the
roof of a small building. The
tension in the cable is 150 kN.
Determine the reactions at
fixed end E.
What are the reactions?
E is a fixed support, which exerts x and y components of force and a moment.
The cable exerts a tension, but where is the point of application of the tension?
Recall that a tension always pulls away from the body at the point of contact.
There are actually three points of contact:
Ch4 Page 11
Sample Problem 4.4 (cont'd)
The frame supports par of the
roof of a small building. The
tension in the cable is 150 kN.
Determine the reactions at
fixed end E.
The tension must be broken down into x and y components:
𝐷𝐹 =
⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯
4.5 + 6.0 = 7.5 m
All sides are divisible by 1.5, giving us
a 3-4-5 triangle.
𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡
3
cos 𝜃 = ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯= ⎯⎯
ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒 5
𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒
4
sin 𝜃 = ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯= ⎯⎯
ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒 5
3
𝑇 = ⎯⎯(150 kN) = 90 kN
5
4
𝑇 = ⎯⎯(150 kN) = 120 kN
5
Ch4 Page 12
Sample Problem 4.4 (cont'd)
The frame supports par of the
roof of a small building. The
tension in the cable is 150 kN.
Determine the reactions at
fixed end E.
Equilibrium equations:
∑𝐹 = 0 = 𝐸 + 90
∑𝐹 = 0 = 𝐸 − 120 − 4(20)
∑𝑀 = 0 = −90(6) + 20(4)(1.8) + 20(3)(1.8) + 20(2)(1.8) + 20(1.8) + 𝑀
Answers
𝐸 = 90 kN ←
𝐸 = 200 kN ↑
𝑀 = 180 kN m 
Ch4 Page 13
Ch4 Page 14
More 2D equilibrium examples:
4.15
Two links AB and DE are connected by a bell crank as
shown. Knowing that the tension in link AB is 720 N,
determine (a) the tension in link DE, (b) the reaction
at C.
3
𝛼 = atan ⎯⎯= 36.87°
4
𝛽 = 180° − 90° − 36.87° = 53.13°
∑𝐹 = 0 = −720 cos 53.13 + 𝐶
∑𝐹 = 0 = −720 sin 53.13 + 𝐶 − 𝑇
∑𝑀 = 0 = 720 cos 53.13 (60) + 720 sin 53.13 (80) − 𝑇 (120)
OR
∑𝑀 = 0 = 720(100) − 𝑇 (120)
𝑇
= 600 N
𝐶 = 432 N →
𝐶 = 1176 N ↑
Ch4 Page 15
4.21b
Determine the reactions at A and B when h = 200 mm.
∑𝐹 = 0 = −𝐵 cos 30 + 𝐴
∑𝐹 = 0 = 𝐵 sin 30 + 𝐴 − 150
∑𝑀 = 0 = −150(250) − 𝐵 cos 30 (200) + 𝐵 sin 30 (500)
OR (using cross product instead of right-hand rule)
∑𝑀 ⃗ = 0 = ∑ 𝑟⃗ × 𝐹⃗ = 𝐴𝐺⃗ × (−150𝚥̂) + 𝐴𝐵⃗ × (−𝐵 cos 30 𝚤̂ + 𝐵 sin 30 𝚥̂)
= 250𝚤̂ × (−150𝚥̂) + (500𝚤̂ − 200𝚥̂) × (−𝐵 cos 30 𝚤̂ + 𝐵 sin 30 𝚥̂)
∑𝑀 = −250(150) + 500(𝐵 sin 30) − 200(𝐵 cos 30)
𝐵 = 488.31 N
𝐴 = 422.89 N →
𝐴 = −94.157 N (↓)
Ch4 Page 16
3D Equilibrium
Ch4 Page 17
Ch4 Page 18
Ch4 Page 19
A fixed support is enough by itself to prevent all 6 motions because there are 6 reactions:
∑𝐹 = 0 = 𝐴 − 300
∑𝐹 = 0 = 𝐴 − 100
∑𝐹 = 0 = 𝐴 − 200
∑𝑀 = 0 = −200(1) + 𝑀
∑𝑀 = 0 = 200(1) + 𝑀
∑𝑀 = 0 = 300(1) − 100(1) + 𝑀
𝐴 = 300 N →
𝐴 = 100 N ↑
𝐴 = 200 N ↙
𝑀
= 200 N m
𝑀
= −200 N m
𝑀
= −200 N m
Ch4 Page 20
Ch4 Page 21
Ch4 Page 22
4.122 The assembly shown is welded to collar A that fits on the vertical pin shown. The
pin can exert couples [moments] about the x and z axes but does not prevent motion
about or along the y axis. For the loading shown, determine the tension in each cable
and the reaction at A.
Az
MRz
TCF
TDE
MRx Ax
Note that the two moment reactions at bearing A must be included because we
would have only 4 unknown reactions without them.
First, resolve the two tensions into their vector components. Both are in 2D
coordinate planes and form 3-4-5 triangles.
𝑇⃗
= −⎯⎯𝑇
𝚤̂ + ⎯⎯𝑇
𝚥̂
𝑇⃗
Ch4 Page 23
= ⎯⎯𝑇
𝚥̂ − ⎯⎯𝑇
𝑘
4.122 The assembly shown is welded to collar A that fits on the vertical pin shown. The
pin can exert couples [moments] about the x and z axes but does not prevent motion
about or along the y axis. For the loading shown, determine the tension in each cable
and the reaction at A.
Az
MRz
TCF
TDE
MRx Ax
Equilibrium equations:
∑𝐹 = 0 = −⎯⎯𝑇
+𝐴
∑𝐹 = 0 = ⎯⎯𝑇
+ ⎯⎯𝑇
∑𝐹 = 0 = −⎯⎯𝑇
+𝐴
− 480
Take moments about A, which has the most number of unknown reactions:
Using right-hand rule − ⎯⎯𝑇 (90) − ⎯⎯𝑇 (135) + 480(135)
∑𝑀
=0=𝑀
∑𝑀
= 0 = ⎯⎯𝑇 (80) − ⎯⎯𝑇 (135)
∑𝑀
=0=𝑀
+ ⎯⎯𝑇 (80) + ⎯⎯𝑇 (80) − 480(80)
Using cross products ∑𝑀⃗ = 0 = 𝑀
=𝑀
+
𝚤̂ + 𝑀
𝚤̂ + 𝑀
𝑘+
𝑘 + 𝐴𝐶⃗ × (−480 𝚥̂) + 𝐴𝐷⃗ × 𝑇⃗
+ (𝐴𝐶⃗ × 𝑇⃗ )
80 𝚤̂ + 135 𝑘 × (−480 𝚥̂)
80 𝚤̂ + 90 𝑘 × ⎯⎯𝑇
𝚥̂ − ⎯⎯𝑇
𝑘
+
Ch4 Page 24
80 𝚤̂ + 135 𝑘 × −⎯⎯𝑇
𝚤̂ + ⎯⎯𝑇
𝚥̂
4.122 The assembly shown is welded to collar A that fits on the vertical pin shown. The
pin can exert couples [moments] about the x and z axes but does not prevent motion
about or along the y axis. For the loading shown, determine the tension in each cable
and the reaction at A.
Az
MRz
TCF
Using cross products (cont'd) ∑𝑀⃗ = 0 = 𝑀
𝚤̂ + 𝑀
𝑘 + (−480)(80) 𝑘 − (−480)(135) 𝚤̂
+ ⎯⎯𝑇 (80) 𝑘 − −⎯⎯𝑇
(80) 𝚥̂ − ⎯⎯𝑇 (90) 𝚤̂
+ ⎯⎯𝑇 (80) 𝑘 + −⎯⎯𝑇
(135) 𝚥̂ − ⎯⎯𝑇
(135) 𝚤̂
Separating into x, y, and z equations − ⎯⎯𝑇 (90) − ⎯⎯𝑇 (135) + 480(135)
∑𝑀
=0=𝑀
∑𝑀
= 0 = ⎯⎯𝑇 (80) − ⎯⎯𝑇 (135)
∑𝑀
=0=𝑀
+ ⎯⎯𝑇 (80) + ⎯⎯𝑇 (80) − 480(80)
Ch4 Page 25
TDE
MRx Ax
Ch4 Page 26
Ch4 Page 27
Ch4 Page 28
Ch4 Page 29
Ch4 Page 30
Problem 1
The 5-ft long beam is fixed at A. What are the reactions at the support?
Fx = 0 = Ax - 200
Fy = 0 = Ay - 100
MA = 0 = -100(5) + MR
Ax = 200 lb
Ay = 100 lb
MR = 500 lb-ft
Problem 2
The 2-m long, 10-kg rod is pinned at A and rests on the ground at B. Find the normal reaction at B.
MA = 0 = -10(9.81)(1 cos30) + B(2 cos30)
B = 49.05 N
Ch4 Page 31
Problem 3
The lid of the box weighs 90 lb and is held by the cable CE so that the lid makes an angle of 60 degrees with
the horizontal xz plane. Hinge B supports axial thrust in the direction of the hinge axis AB while hinge A does
not support axial thrust. Find the reactions at the hinges and the tension in the cable. Hint: Analyze
equilibrium of the lid only.
CE = -4i - 3 sin60j + (3 + 3 cos60)k = -4i - 2.59808j + 4.5k
T = T (-4i - 2.59808j + 4.5k) / (42 + 2.598082 + 4.52)1/2 = T (-0.60999i - 0.3962j + 0.686244k)
T = -0.60999T i - 0.3962T j + 0.686244T k)
_____________________________________________________________________
Fx = 0 = -0.60999T + Bx -----------------------------------------------------------------> Bx = 34.641 lb
Fy = 0 = -0.3962T + Ay + By - 90 -------------------------------------------------------> By = 45 lb
Fz = 0 = 0.686244T + Az + Bz ----------------------------------------------------------> Bz = -25.981 lb
MBx = 0 = -0.3962T (3 cos60) + 0.686244T (3 sin60) - 90 (1.5 cos60) --------> T = 56.789 lb
MBy = 0 = 0.60999T (3 cos60) + Az (4) -----------------------------------------------> Az = -12.9904 lb
MBz = 0 = 0.60999T (3 sin60) - Ay (4) + 90 (2) --------------------------------------> Ay = 67.5 lb
_____________________________________________________________________
Fx = 0 = -0.60999T + Bx -----------------------------------------------------------------> Bx = 34.641 lb
Fy = 0 = -0.3962T + Ay + By - 90 -------------------------------------------------------> Ay = 67.5 lb
Fz = 0 = 0.686244T + Az + Bz ----------------------------------------------------------> Az = -25.981 lb
MAx = 0 = -0.3962T (3 cos60) + 0.686244T (3 sin60) - 90 (1.5 cos60) --------> T = 56.789 lb
MAy = 0 = 0.60999T (3 cos60) - 0.686244T (4) - Bz (4) ---------------------------> Bz = -25.981 lb
MAz = 0 = 0.60999T (3 sin60) - 0.3962T (4) + By (4) - 90 (2) --------------------> By = 45 lb
Ch4 Page 32
Problem 4
The mechanism shown consists of a vertical shaft rigidly connected to a horizontal shaft. The horizontal shaft
OA makes an angle of 60 degrees with the x axis. The bearing at B supports axial thrust. When the 40 N-m
torque is applied to the vertical shaft, cable AC prevents rotation of the mechanism about the y axis. Find the
tension in the cable and the reactions at bearing B.
AC = (180 - 160 cos60)i - 200j + 160 sin60k = 100i - 200j + 138.564k
T = T (100i - 200j + 138.564k) / (1002 + 2002 + 138.5642)1/2 = T (0.380143i - 0.76029j + 0.526742k)
T = 0.380143T i - 0.76029T j + 0.526742T k)
_____________________________________________________________________
Fx = 0 = 0.380143T + Bx -----------------------------------------------------------------> Bx = -160.375 N
Fy = 0 = -0.76029T + By ------------------------------------------------------------------> By = 320.75 N
Fz = 0 = 0.526742T + Bz -----------------------------------------------------------------> Bz = -222.222 N
MBx = 0 = -0.76029T (0.16 sin60) + 0.526742T (0.11) + MRx -------------------> MRx = 20 N-m
MBy = 0 = -0.380143T (0.16 sin60) - 0.526742T (0.16 cos60) + 40 -----------> T = 421.88 N
MBz = 0 = -0.380143T (0.11) - 0.76029T (0.16 cos60) + MRz -------------------> MRz = 43.301 N-m
Ch4 Page 33