ENS 161

(Statics of Rigid Bodies)

Lecture 3: Equilibrium of a Particle

Static Equilibrium
It is used to describe a particle at rest. This requires that the
resultant force acting on that particle to be equal to zero.
In mathematical expression:
For three-dimensional
Σ𝐅 = 0 force system:
For coplanar force system (two-dimensional):

Free-Body Diagram (FBD)

It is an outlined shape of the particle that shows all the forces
acting on it listed with their known or unknown magnitudes and
directions.
Procedure for Drawing an FBD:
Two Connections Commonly Encountered
in Equilibrium Problems:
1. Springs 2. Cables and Pulleys
Facts:
1. A cable can only support a tension or
“pulling” force, and this force always acts in
the direction of the cable.
2. A continuous cable passing over a
frictionless pulley has a constant
magnitude of tension force throughout its
length.

where: 𝑠 = 𝑙 − 𝑙𝑜
k = spring constant/stiffness
lo = initial/undeformed length
l = final/deformed length
Procedure
for Analysis
of 2-D
Force
System:
Procedure for Analysis of 3-D Force System:
Sample Problem 1
Determine the tension in cables BA and
BC necessary to support the 60-kg
cylinder in the figure. Notes:
1) From FBD of the cylinder (not
Solution: drawn), TBD = Weight of cylinder
2) Weight = mass x acceleration
(1) FBD: due to gravity
3) g = 9.81 m/s2 [acceleration
due to gravity]
4) 1 kg∙m/s2 = 1 N

(2) Eq’ns of Equilibrium: ▸ With 2 unknowns and 2 eq’ns,

Σ 𝐹𝑦 = 0 𝑇𝐶𝑦 + 𝑇𝐴𝑦 − 𝑇𝐵𝐷 = 0 unknowns can be solved by calculator.

𝑇𝐶 𝑠𝑖𝑛 45 + 𝑇𝐴 3 − 𝑇𝐵𝐷 = 0 𝑇𝐶 𝑠𝑖𝑛 45 + 𝑇𝐴 3 = 588.6 𝑁

5 5
𝑇𝐶 𝑐𝑜𝑠 45 − 𝑇𝐴 4 = 0
𝑇𝐶 𝑠𝑖𝑛 45 + 𝑇𝐴 3 = 588.6 𝑁 (eq’n 1)
5
5
Σ 𝐹𝑥 = 0 𝑇𝐶𝑥 − 𝑇𝐴𝑥 = 0 𝑇𝐴 = 420.43 𝑁 = 420 𝑁
𝑇𝐶 𝑐𝑜𝑠 45 − 𝑇𝐴 4 = 0 (eq’n 2) 𝑇𝐶 = 475.66 𝑁 = 476 𝑁
5
 (By geometry,
Solution: the two θ are
Sample Problem 2 the same)
(1) FBD:
The 200-kg crate in the figure is
suspended using the ropes AB θ
and AC. Each rope can withstand
a maximum force of 10 kN before
it breaks. If AB always remains
horizontal, determine the smallest
angle θ to which the crate can be
suspended before one of the FD = W = 200(9.81)
ropes breaks.
(2) Eq’ns of Equilibrium:
Σ 𝐹𝑥 = 0 𝐹𝐵 − 𝐹𝐶𝑥 = 0 Σ 𝐹𝑦 = 0 𝐹𝐶𝑦 − 𝐹𝐷 = 0
𝐹𝐵 − 𝐹𝐶 𝑐𝑜𝑠 θ = 0 𝐹𝐶 𝑠𝑖𝑛 θ − 1962 = 0
𝐹𝐵 = 𝐹𝐶 𝑐𝑜𝑠 θ (10,000) 𝑠𝑖𝑛 θ − 1962 = 0
1962
▸ From the eq’n, FB will always be less than FC, 𝑠𝑖𝑛 θ =
since FB is just equal to a fraction/portion of FC. 10000
Note that cos(θ) is always less than 1.
θ = 11.30
▸ Thus, rope of FC breaks first. Before the rope
breaks, FC is maximum which is equal to 10 kN.
▸ FC = 10 kN = 10,000 N
Sample Problem 3
Determine the required length of cord AC in the
figure so that the 8-kg lamp can be suspended
in the position shown. The undeformed length of
spring AB is lo,AB = 0.4 m, and the spring has a
stiffness of kAB = 300 N/m. 𝑙𝐴𝐶 𝑐𝑜𝑠 30 𝑙𝐴𝐵
Solution:
(2) ▸ From figure, by geometry:
(1) FBD: 2 − 𝑙𝐴𝐵
𝑙𝐴𝐶 𝑐𝑜𝑠 30 + 𝑙𝐴𝐵 = 2 𝑚 𝑙𝐴𝐶 = (eq’n 2)
cos(30)
(3) Eq’ns of Equilibrium:
Σ 𝐹𝑦 = 0 𝑇𝐴𝐶𝑦 − 𝑊 = 0
𝑇𝐴𝐶 𝑠𝑖𝑛 30 − 78.48 = 0
78.48
𝑇𝐴𝐶 = = 156.96 𝑁
sin(30)
Σ 𝐹𝑥 = 0 𝑇𝐴𝐵 − 𝑇𝐴𝐶𝑥 = 0
▸ 𝑊 = 8 9.81 = 78.48 𝑁
𝑇𝐴𝐵 − 𝑇𝐴𝐶 𝑐𝑜𝑠 30 = 0
(force on spring)
▸ 𝑇𝐴𝐵 = 𝑘𝑠 = 𝑘(𝑙𝐴𝐵 − 𝑙𝑜, 𝐴𝐵) 𝑇𝐴𝐵 = 156.96 𝑐𝑜𝑠 30 = 135.93
135.93
𝑇𝐴𝐵 = 300(𝑙𝐴𝐵 − 0.4) 𝑙𝐴𝐵 = + 0.4 = 0.853 𝑚 (from eq’n 1)
300
𝑇𝐴𝐵 2 − 0.853
𝑙𝐴𝐵 = + 0.4 (eq’n 1) 𝑙𝐴𝐶 = = 𝟏. 𝟑𝟐 𝒎 (from eq’n 2)
300 cos(30)
 Sample Problem 4
A 90-lb load is suspended from the hook shown in the figure.
If the load is supported by two cables and a spring having a
stiffness k = 500 lb/ft, determine the force in the cables and
the stretch of the spring for equilibrium. Cable AD lies in the
x–y plane and cable AC lies in the x–z plane.

Solution:
(1) FBD:
(2) Eq’ns of Equilibrium:
(Always start with the eq’n with least number of unknown)
Σ 𝐹𝑧 = 0 𝐹𝐶𝑧 − 90 = 0
3
𝐹𝐶 = 90 𝐹𝐶 = 150 𝑙𝑏
5
Σ 𝐹𝑥 = 0 𝐹𝐷𝑥 − 𝐹𝐶𝑥 = 0
4
𝐹𝐷 𝑠𝑖𝑛 30 − 𝐹𝐶 =0
5
4
𝐹𝐷 𝑠𝑖𝑛 30 = (150) 𝐹𝐷 = 240 𝑙𝑏
5
Σ 𝐹𝑦 = 0 𝐹𝐵 − 𝐹𝐷𝑦 = 0
(Using the spring eq’n)
𝐹𝐵 − 𝐹𝐷 𝑐𝑜𝑠(30) = 0
▸ 𝐹𝐵 = 𝑘𝑠
𝐹𝐵 = (240) 𝑐𝑜𝑠(30) 𝐹𝐵 = 207.8 𝑙𝑏
207.8 𝑙𝑏 = (500 𝑙𝑏/𝑓𝑡)(𝑠)
𝑠 = 0.416 𝑓𝑡
 (Note that a whole
Sample Problem 5 circle has 3600)
The 10-kg lamp in the figure is suspended from the
three equal-length cords. Determine its smallest vertical
distance s from the ceiling if the force developed in any
cord is not allowed to exceed 50 N.
Solution:
(1) FBD: (2) Practical Analysis of the Problem:
▸Since the cords are of equal
length and they are evenly
spaced on a plane parallel to
x-y plane, the three cords
experience the same force.
▸Using the maximum tension
force, T = 50 N
(3) Eq’ns of Equilibrium:
Σ 𝐹𝑧 = 0 3(𝑇𝑧) − 𝑊 = 0 (where, γ = angle
3[50 𝑐𝑜𝑠 γ ] − 98.1 = 0 between any of the
(Actually, the FBD makes the chord and the z-axis)
−1
98.1
problem looks more complicated) γ = cos = 49.160
3 50
▸ 𝑊 = 10 9.81 = 98.1 𝑁 (From the figure)
600 𝑚𝑚
tan γ = tan(49.160) = 𝑠 = 519 𝑚𝑚
𝑠
Sample Problem 6
Determine the force in each cable used to support the 40-lb crate shown
in the figure. (3) Eq’ns of Equilibrium:
Solution: (a)Σ 𝐹𝑦 = 0 𝐹𝐶𝑦 − 𝐹𝐵𝑦 = 0
(Their length on the
(1) FBD: concerned axis)
4 4
𝐹𝐶 − 𝐹𝐵 =0
89 89
𝐹 𝐶 = 𝐹𝐵 (Total length)

(b) Σ 𝐹𝑧 = 0 𝐹𝐶𝑧 + 𝐹𝐵𝑧 − 𝑊 = 0

8 8
𝐹𝐶 + 𝐹𝐵 − 40 = 0
89 89
8
2𝐹𝐵 − 40 = 0 𝐹𝐵 = 𝟐𝟑. 𝟔 𝒍𝒃
89
(2) Compute length of (Because FB and FC are equal) 𝐹 = 𝟐𝟑. 𝟔 𝒍𝒃
𝐶
chords:
(c) Σ 𝐹𝑥 = 0 𝐹𝐷 − 𝐹𝐵𝑥 − 𝐹𝐶𝑥 = 0
𝑙𝐴𝐵 = 𝑙𝐴𝐶 = 32 + 82 + 42
𝑙𝐴𝐵 = 𝑙𝐴𝐶 = 89 𝑓𝑡 3 3
𝐹𝐷 = 𝐹𝐵 + 𝐹𝐶 = 𝟏𝟓. 𝟎 𝒍𝒃
(𝑙𝐴𝐷 cannot be computed 89 89
and is not needed.)
 Sample Problem 7
Determine the tension in each cord used to
support the 100-kg crate shown in the figure.
(2) Eq’ns of Equilibrium:
(a) Σ 𝐹𝑧 = 0 𝐹𝐶𝑧 + 𝐹𝐷𝑧 − 𝑊 = 0
2
𝐹𝐷 + 𝐹𝐶 𝑐𝑜𝑠 60 − 981 = 0
3
2
𝐹𝐷 + 𝐹𝐶 𝑐𝑜𝑠 60 = 981 (eq’n 1)
3
Solution: 𝐹𝐷𝑦 − 𝐹𝐶𝑦 = 0
(b) Σ 𝐹𝑦 = 0
(1) FBD: 2
𝐹𝐷 − 𝐹𝐶 𝑐𝑜𝑠(180 − 135) = 0
3
2
𝐹𝐷 − 𝐹𝐶 𝑐𝑜𝑠(45) = 0 (eq’n 2)
3
(With 2 eq’ns – 2 unknowns)

𝐹𝐷 = 𝟖𝟔𝟐. 𝟎 𝑵 𝐹𝐶 = 𝟖𝟏𝟐. 𝟕 𝑵
(c) Σ 𝐹𝑥 = 0 𝐹𝐵 − 𝐹𝐷𝑥 − 𝐹𝐶𝑥 = 0
▸ 𝑊 = 100 9.81 = 981 𝑁
1
▸ length of chord AD 𝐹𝐵 − 𝐹𝐷 − 𝐹𝐶 𝑐𝑜𝑠 180 − 120 = 0
3
𝑙𝐴𝐷 = 12 + 22 + 22 = 3 1
𝐹𝐵 = 862 + 812.7 𝑐𝑜𝑠 60 = 𝟔𝟗𝟑. 𝟕 𝑵
3
 “The most effective way of learning the
principles of engineering mechanics is to
solve problems.”
ASSIGNMENT!
Answer the following problems from the Chapter 3 review
problems of our textbook in page 114-115: Problems 3-70, 3-71,
3-74, 3-78, and 3-79.

Write the solutions neatly and in an organized manner on short

bond papers (Do not encode). Scan them and save them to one
file as PDF.

Due date: October 18, 2021 (for ENS 161 – Ff)

October 19, 2021 (for ENS 161 – Uu1)