ENGINEERING SCIENCE 11
STATICS OF RIGID BODIES
MODULE 4
EQUILIBRIUM OF PARTICLES
Outline
4.1
Conditions for Static Equilibrium
4.2
Particle and Forces
4.3
Free-Body Diagrams
4.4 Common Types of Support in Equilibrium of Particles
ENGINEERING SCIENCE 11
STATICS OF RIGID BODIES
MODULE 4
EQUILIBRIUM
OF PARTICLES
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EQUILIBRIUM
OF PARTICLES
Introduction
We define previously that engineering statics deals with the application of force conditions
necessary to maintain equilibrium in engineering structures. The fundamental knowledge,
therefore, of the identification of a rigid body or group of bodies is an important tool in the
analysis of mechanical system in equilibrium. Likewise, the identification of the forces acting
on a rigid body, be it a tensile or compressive force, should be well established. Hence,
creating the whole picture of analyzing a rigid body through the application of equilibrium
equations from Newton’s First Law of Motion is essential. In this module, we will treat bodies
in consideration to be particles which means that we can assume forces to act at only one
point of application.
Objectives
At the end of this chapter, the student should be able to:




4.1
state the conditions for particle equilibrium
draw free-body diagram of particles acted upon by forces
describe the common types of particle support
solve problems involving particle equilibrium involving
 forces on particle supports
 multiple failure criteria
Conditions for Static Equilibrium
Static Equilibrium – the particle considered should be at rest, means a = 0
Newton’s First Law of Motion – when the resultant of all the forces acting on the
particle is zero, the particle is in equilibrium.
𝛴𝛴𝛴𝛴 = 0
4.2
Particle and Forces
𝛴𝛴𝐹𝐹𝑥𝑥 = 0
𝛴𝛴𝐹𝐹𝑦𝑦 = 0
As stated above, we have a static equilibrium if our particle has zero acceleration.
Here, a particle may represent:
 an individual particle of a real body or structure
 a portion of the real body or structure
 the entire body or structure
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EQUILIBRIUM
OF PARTICLES
When applying ∑ 𝑭𝑭 = 𝟎𝟎 to the particle, all forces that are applied to the particle
must be included
Forces have a number of sources, as follows:



due to interaction of the particle with its environment
• e.g. weight due to gravity, force of wind blowing against a structure
due to structural members that are attached to (or contain) the particle
• e.g. a cable attached to it
due to supports
• e.g. if a particle (or the body the particle represents) is glued to a surface,
the glue will usually apply forces to the particle
4.3 Free-Body Diagrams
A free body diagram (FBD) is a sketch showing the particle “free” from its
surroundings with all the forces acting on it. This is an essential representation or tool
to help ensure that all forces, including their line of action, acting on a body are
accounted.
Procedure for Drawing the Free-Body Diagram:
Step 1.
Step 2.
Step 3.
Draw the outlined shape of the particle, isolated from its surroundings.
Indicate all forces acting on the particle.
 External loads / Active forces- tend to set the particle in motion
 Reactive forces- caused by supports that prevent motion
 Weight
Identify each force
 Known forces- label with proper magnitude and direction
 Unknown forces- represent by letter
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EQUILIBRIUM
OF PARTICLES
Example 4.3a. Free-body diagram of the bicycle
An example below is a man pulling his bicycle (A). To isolate and focus on the bicycle,
the forces acting on it would be the known force caused by the effect of pulling from
the man and the unknown force due to the weight of the bicycle and the upward
reactive force on the ground (B). If the bicycle were treated as a particle, it can be
represented by a dot (C).
(A)
(B)
(C)
Example 4.3b. Free-body diagram of the skier and the skis
The example below is taken from the book of Plesha et al. (2010). It is a picture of a
skier being towed upward at constant velocity (A). The skier and the skis are isolated
from the surroundings. (B). In drawing the FBD, all forces acting on the skier and skis
are considered. Those forces are the weight of the skiers, reaction of the ground on
the skis, and the force of the rope on the skiers (C). And lastly, we can idealize the
skier as a particle with forces acting to maintain its equilibrium (D).
It is important to note that we do not need to include forces that are internal to the
FBD (e.g. forces that keeps skier’s boots and the skis). Since, the isolation of the body
from the surrounding do not include the separation of the boots from skis.
(A)
(B)
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(C)
EQUILIBRIUM
OF PARTICLES
(D)
4.4 Common Types of Support in Equilibrium of Particles
A. Springs – linear elastic springs
Springs are usually found as damping materials in automobiles, foam beds,
computer keyboards, clocks, pens, etc. They come in various sizes and shapes.
The forces produced from the spring is due to their deformation. Hence, the force
experienced by the spring is directly proportional to the spring constant of the
material and the distance displaced by the spring from its unloaded position. The
spring constant is dependent on the type of material used in the mechanical
system. As a convention, if the spring is compressed, the spring force becomes
negative and positive if stretched.
𝑭𝑭 = 𝒌𝒌𝒌𝒌
Where:
𝑘𝑘 = spring constant
𝑠𝑠 = measured from unloaded position (𝑙𝑙 − 𝑙𝑙0 )
𝑙𝑙 = final length
𝑙𝑙0 = un-deformed length
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EQUILIBRIUM
OF PARTICLES
B. Cables and Pulleys



assumed to have negligible weight, inextensible and perfectly flexible
can support only a tension or “pulling” force
• tension force in a continuous cable which passes over a frictionless
pulley has a constant magnitude
for any angle 𝜽𝜽, the cable is subjected to a constant tension 𝑻𝑻 throughout its
length
Consider the figures below. In figure A, if we consider that the pulley has friction,
then 𝑇𝑇1 ≠ 𝑇𝑇2 in general. This chapter in our syllabus assumed that our cables pass
over a frictionless pulley and considered weightless. Hence, we can assume that
the magnitude of the tension is constant for the entire cable (Figures B and C).
(A)
(B)
(C)
*Source: Plesha et al. (2010).
C. Bars


assumed to have negligible weight and inextensible, in both tension and
compression
can support both tension and compression
In the figure below, the structure is connected with a bar AC (Figure A). We can
assume that the bar experiences a compressive force due to the effect of the
downward force P, so FAC points toward point A (Figure B).
(A)
*Source: Plesha et al. (2010).
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ENGINEERING SCIENCE 11
STATICS OF RIGID BODIES
MODULE 4
EQUILIBRIUM
OF PARTICLES
SAMPLE PROBLEMS
Problem 1
If the mass of cylinder C is 40 kg, determine the mass of cylinder A and tension in cable DE
in order to hold the assembly in the position shown [Source: Hibbeler, 2013].
Required:
𝑚𝑚𝐴𝐴
Solution: FBD of ring E:
𝑇𝑇𝐵𝐵𝐵𝐵
Since the system is in equilibrium, the weight of
block A causes tension in cables ED and EB. The
cable EBC is continuous, hence we can assume that
the magnitude of the tensile force experience in EB
and BC is the same (𝑇𝑇𝐵𝐵𝐵𝐵 = 𝑇𝑇𝐸𝐸𝐸𝐸 ) with the assumptions
we initially made on cables. From this, we can
compute 𝑇𝑇𝐵𝐵𝐵𝐵 to be equal to the weight of the
cylinder C.
= 𝑊𝑊𝐶𝐶 = 40𝑘𝑘𝑘𝑘(9.806 𝑚𝑚⁄𝑠𝑠 2 ) = 392.24 𝑁𝑁
Applying equilibrium equations along x and y axes, we have
⇸ 𝛴𝛴𝐹𝐹𝑥𝑥 = −𝑇𝑇𝐷𝐷𝐷𝐷 + 𝑇𝑇𝐵𝐵𝐵𝐵 cos 30 = 0
⤉ 𝛴𝛴𝐹𝐹𝑦𝑦 = 𝑇𝑇𝐵𝐵𝐵𝐵 sin 30 − 𝑊𝑊𝐴𝐴 = 0
Solving for 𝑇𝑇𝐷𝐷𝐷𝐷 and 𝑊𝑊𝐴𝐴
𝑻𝑻𝑫𝑫𝑫𝑫 = 𝟑𝟑𝟑𝟑𝟑𝟑. 𝟔𝟔𝟔𝟔𝟔𝟔𝟔𝟔 𝑵𝑵
Hence,
𝑊𝑊𝐴𝐴 = 196.12 𝑁𝑁 = 𝑚𝑚(9.806 𝑚𝑚⁄𝑠𝑠 2 )
𝒎𝒎𝑨𝑨 = 𝟐𝟐𝟐𝟐 𝒌𝒌𝒌𝒌
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STATICS OF RIGID BODIES
MODULE 4
EQUILIBRIUM
OF PARTICLES
Problem 2
For the system shown:
a) Determine the stretch in
springs AC and AB for
equilibrium of the 2-kg block.
The springs are shown in the
equilibrium position.
b) If the unstretched length in
spring AB is 3 m and the
block is held in the
equilibrium position shown,
determine the mass of the
block at D.
To emphasize, the two
problems are separate but
refers to the same figure.
[Source: Hibbeler, 2013]
Required:
Solution:
𝑠𝑠𝐴𝐴𝐴𝐴 , 𝑠𝑠𝐴𝐴𝐴𝐴
In the first requirement of the problem, we need to find the stretch in springs
AC and AB. Hence, our strategy is to find the forces in AC and AB initially then
equate later to our spring equation to find the stretch in the springs.
 FBD of joint A:
The springs supports a 2-kg block at joint A,
hence by applying equilibrium equations, we can
obtain 𝐹𝐹𝐴𝐴𝐴𝐴 and 𝐹𝐹𝐴𝐴𝐴𝐴 .
⇸ 𝛴𝛴𝐹𝐹𝑥𝑥 = −𝐹𝐹𝐴𝐴𝐴𝐴 �
⤉ 𝛴𝛴𝐹𝐹𝑦𝑦 = 𝐹𝐹𝐴𝐴𝐴𝐴 �
3
3
4
� + 𝐹𝐹𝐴𝐴𝐴𝐴 � � = 0
3√2
3
5
� + 𝐹𝐹𝐴𝐴𝐴𝐴 � � − 2(9.806) = 0
3√2
5
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EQUILIBRIUM
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Obviously, these springs experienced tensile forces. We now use the spring
equation (𝐹𝐹 = 𝑘𝑘𝑘𝑘) and the corresponding spring constants to find the stretches
in the springs.
𝐹𝐹𝐴𝐴𝐴𝐴 = 15.8489 𝑁𝑁 = 𝑘𝑘𝐴𝐴𝐴𝐴 𝑠𝑠𝐴𝐴𝐴𝐴
Required:
Solution:
𝑚𝑚𝐷𝐷
𝐹𝐹𝐴𝐴𝐴𝐴 = 14.0009 𝑁𝑁 = 𝑘𝑘𝐴𝐴𝐴𝐴 𝑠𝑠𝐴𝐴𝐴𝐴
𝒔𝒔𝑨𝑨𝑨𝑨 = 𝟎𝟎. 𝟕𝟕𝟕𝟕𝟕𝟕𝟕𝟕 𝒎𝒎
𝒔𝒔𝑨𝑨𝑨𝑨 = 𝟎𝟎. 𝟒𝟒𝟒𝟒𝟒𝟒𝟒𝟒 𝒎𝒎
To find 𝑚𝑚𝐷𝐷 in the next problem, consider again the FBD of joint A below to
find the mass of block D. In this scenario, the unstretched length of AB is 3 m.
By looking at our FBD, we have a special triangle (3, 4, 5) at the orientation of
spring AB, hence we can compute the final stretch length and the
corresponding force of spring AB using the spring equation.
𝐹𝐹𝐴𝐴𝐴𝐴 = 𝑘𝑘𝐴𝐴𝐴𝐴 �𝑙𝑙𝐴𝐴𝐴𝐴 − 𝑙𝑙0,𝐴𝐴𝐴𝐴 � = 30(5 − 3) = 60 𝑁𝑁
Applying equilibrium equations,
⇸ 𝛴𝛴𝐹𝐹𝑥𝑥 = −𝐹𝐹𝐴𝐴𝐴𝐴 �
⤉ 𝛴𝛴𝐹𝐹𝑦𝑦 = 𝐹𝐹𝐴𝐴𝐴𝐴 �
Solving for 𝑚𝑚𝐷𝐷 ,
𝑚𝑚𝐷𝐷 =
3
3
4
� + 60 � � = 0
3√2
3
5
� + 60 � � − 𝑊𝑊𝐷𝐷 = 0
3√2
𝑊𝑊𝐷𝐷
9.806 𝑚𝑚⁄𝑠𝑠 2
5
= 𝟖𝟖. 𝟓𝟓𝟓𝟓𝟓𝟓𝟓𝟓 𝒌𝒌𝒌𝒌
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𝐹𝐹𝐴𝐴𝐴𝐴 = 48√2 𝑁𝑁
𝑊𝑊𝐷𝐷 = 84 𝑁𝑁
ENGINEERING SCIENCE 11
STATICS OF RIGID BODIES
MODULE 4
EQUILIBRIUM
OF PARTICLES
Problem 3
The spring assembly shown is used to support a 10-lb
block and block A. If the tension in cable  is 5 lb,
determine the following:
(a) the tension (in lb) of cables  and
(b) the change in length (in inches) of the spring
(c) the weight (in lb) of block A.
Solution:
2
60°
3
45°
A
The spring stiffness is 3000 lb/ft.
Required:
10 lb
1
𝑇𝑇1 , 𝑇𝑇3 , 𝑠𝑠, 𝑊𝑊𝐴𝐴
In this problem, a combination of spring and cable supports are shown. We
have two continuous cable system (T1 and T2). T2 tensile force is 5 lb while T1
and T3 are unknowns. To find these unknown forces, we can consider the FBDs
of the 10-lb block and block A below. Likewise, we can use these FBDs to
calculate the stretch in the spring at the 10-lb block and weight of block A.
FBD of the 10-lb block
FBD of block A
Because we have two equilibrium conditions, we can start with the FBD that
only has two unknowns. In this example, we can start with either of the two
FBDs because both only have unknowns.
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 Consider FBD of the 10-lb block:
Here, we assume that the spring experiences a tensile force. By applying
equilibrium equations, we can obtain T3 and Fs simultaneously. Likewise, using
the spring equation and the given spring constant, the change in the length of
the spring can be computed.
⇸ 𝛴𝛴𝐹𝐹𝑥𝑥 = −5 𝑐𝑐𝑐𝑐𝑐𝑐 45 + 𝑇𝑇3 = 0
⤉ 𝛴𝛴𝐹𝐹𝑦𝑦 = −5 𝑠𝑠𝑠𝑠𝑠𝑠 45 − 𝐹𝐹𝑠𝑠 − 10 = 0
𝑻𝑻𝟑𝟑 = 𝟑𝟑. 𝟓𝟓𝟓𝟓𝟓𝟓𝟓𝟓 𝒍𝒍𝒍𝒍
𝐹𝐹𝑠𝑠 = −13.5355 𝑙𝑙𝑙𝑙 = 13.5355, C
= 𝑘𝑘𝑘𝑘 = �3000
𝒔𝒔 = 𝟎𝟎. 𝟎𝟎𝟎𝟎𝟎𝟎𝟎𝟎𝟎𝟎 𝒊𝒊𝒊𝒊
𝑙𝑙𝑙𝑙
𝑓𝑓𝑓𝑓
��
1 𝑓𝑓𝑓𝑓
12 𝑖𝑖𝑖𝑖
� 𝑠𝑠
Because we have assumed the spring to be in tension but the value we is
negative, it means that the spring is in compression
 Consider FBD of block A:
As observed in T1, the cable passes over a pulley 3 times. Hence, we need to
represent this cable as 3T1. Therefore, we can now obtain T1 and WA using our
equilibrium equations along x and y axes.
⇸ 𝛴𝛴𝐹𝐹𝑥𝑥 = −3𝑇𝑇1 cos 60 + 5 𝑐𝑐𝑐𝑐𝑐𝑐 45 = 0
⤉ 𝛴𝛴𝐹𝐹𝑦𝑦 = 3𝑇𝑇1 sin 60 + 5 + 5 𝑠𝑠𝑠𝑠𝑠𝑠 45 − 𝑊𝑊𝐴𝐴 = 0
𝑻𝑻𝟏𝟏 = 𝟐𝟐. 𝟑𝟑𝟑𝟑𝟑𝟑𝟑𝟑 𝒍𝒍𝒍𝒍
𝑾𝑾𝑨𝑨 = 𝟏𝟏𝟏𝟏. 𝟔𝟔𝟔𝟔𝟔𝟔𝟔𝟔 𝒍𝒍𝒍𝒍
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EQUILIBRIUM
OF PARTICLES
PROBLEM 4
A pulley-and-cable assembly supports block W and requires a downward force at P for
equilibrium. If the weight of the block is 1000 N, determine the force experienced by the
bar and the springs and the necessary force P to keep the system in equilibrium.
Required:
Solution:
𝐹𝐹𝐶𝐶𝐶𝐶 , 𝐹𝐹𝑠𝑠,𝐴𝐴𝐴𝐴 , 𝐹𝐹𝑠𝑠,𝐹𝐹𝐹𝐹 , 𝑃𝑃
The tension of the cable that passes over pulley E is uniform throughout its
length. For the system to be in equilibrium, it means that the tension in that
cable is equal to the weight of the block.
For us to be able to solve for P, we start with a joint/ particle that has only two
unknowns and continue to analyze joints/ particles with only two unknowns.
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 Consider FBD of pulley C:
Here we assume that the bar experiences tension.
⇸ 𝛴𝛴𝐹𝐹𝑥𝑥 = −2𝑇𝑇𝐵𝐵𝐵𝐵 cos 25 + 1000 cos 42 = 0
⤉ 𝛴𝛴𝐹𝐹𝑦𝑦 = 2𝑇𝑇𝐵𝐵𝐵𝐵 sin 25 − 𝐹𝐹𝐶𝐶𝐶𝐶 − 1000 sin 42 = 0
𝑇𝑇𝐵𝐵𝐵𝐵 = 409.9848 𝑁𝑁
𝐹𝐹𝐶𝐶𝐶𝐶 = −322.5965 𝑁𝑁
𝑭𝑭𝑪𝑪𝑪𝑪 = 𝟑𝟑𝟑𝟑𝟑𝟑. 𝟓𝟓𝟓𝟓𝟓𝟓𝟓𝟓 𝑵𝑵, 𝑪𝑪
Since the sign for 𝐹𝐹𝐶𝐶𝐶𝐶 is negative, it means that the bar experiences a
compressive force.
 Consider FBD of pulley B:
It is assumed here that spring AB is in tension.
⇸ 𝛴𝛴𝐹𝐹𝑥𝑥 = 2𝑇𝑇𝐵𝐵𝐵𝐵 cos 25 − 𝐹𝐹𝑠𝑠,𝐴𝐴𝐴𝐴 cos 15 + 𝑇𝑇𝐵𝐵𝐵𝐵 sin 36 = 0
⤉ 𝛴𝛴𝐹𝐹𝑦𝑦 = −2𝑇𝑇𝐵𝐵𝐵𝐵 sin 25 + 𝐹𝐹𝑠𝑠,𝐴𝐴𝐴𝐴 sin 15 + 𝑇𝑇𝐵𝐵𝐵𝐵 cos 36 = 0
𝐹𝐹𝑠𝑠,𝐴𝐴𝐴𝐴 = 862.1694 𝑁𝑁
𝑇𝑇𝐵𝐵𝐵𝐵 = 152.5163 𝑁𝑁
𝑭𝑭𝒔𝒔,𝑨𝑨𝑨𝑨 = 𝟖𝟖𝟖𝟖𝟖𝟖. 𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏 𝑵𝑵, 𝑻𝑻
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 Consider FBD of pulley F:
It is assumed here that spring AB is in tension.
⇸ 𝛴𝛴𝐹𝐹𝑥𝑥 = −𝑇𝑇𝐵𝐵𝐵𝐵 sin 36 + 𝐹𝐹𝑠𝑠,𝐹𝐹𝐹𝐹 cos 54 = 0
⤉ 𝛴𝛴𝐹𝐹𝑦𝑦 = −𝑇𝑇𝐵𝐵𝐵𝐵 cos 36 − 𝐹𝐹𝑠𝑠,𝐹𝐹𝐹𝐹 sin 54 + 𝑇𝑇𝐺𝐺𝐺𝐺 = 0
𝐹𝐹𝑠𝑠,𝐹𝐹𝐹𝐹 = 152.5163 𝑁𝑁
𝑇𝑇𝐺𝐺𝐺𝐺 = 246.7765 𝑁𝑁
𝑭𝑭𝒔𝒔,𝑭𝑭𝑭𝑭 = 𝟏𝟏𝟏𝟏𝟏𝟏. 𝟓𝟓𝟓𝟓𝟓𝟓𝟓𝟓 𝑵𝑵, 𝑻𝑻
 Consider FBD of pulley H:
⇸ 𝛴𝛴𝐹𝐹𝑥𝑥 = −𝐹𝐹𝑠𝑠,𝐹𝐹𝐹𝐹 cos 54 + 3𝑇𝑇𝐻𝐻𝐻𝐻 cos 78.701 = 0
⤉ 𝛴𝛴𝐹𝐹𝑦𝑦 = 𝐹𝐹𝑠𝑠,𝐹𝐹𝐹𝐹 sin 54 + 3𝑇𝑇𝐻𝐻𝐻𝐻 sin 78.701 − 𝑃𝑃 = 0
𝑇𝑇𝐻𝐻𝐻𝐻 = 152.5158 𝑁𝑁
𝑷𝑷 = 𝟓𝟓𝟓𝟓𝟓𝟓. 𝟎𝟎𝟎𝟎𝟎𝟎𝟎𝟎 𝑵𝑵
4-15
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EQUILIBRIUM
OF PARTICLES
ENGINEERING SCIENCE 11
STATICS OF RIGID BODIES
MODULE 4
EQUILIBRIUM
OF PARTICLES
FOR ENRICHMENT
For problems with multiple failure criteria…


It is important to understand that it is unlikely that all members will
simultaneously be at their failure loads.
Generally, the preferable way to solve it is to determine the forces supported by
each member in terms of P, where P is yet to be determined.
• The feature of this approach is that you do not need to guess which of
several members will fail first.
Problem 5
The system shown consists of cables AB and AC, horizontal cable CE, and vertical bar CD.
If the cables and bar have the failure strengths shown, determine the largest load P that can
be supported by the system. [Source: Plesha et al., 2010]
Member strength
Required:
Solution:
AB
300 lb
AC
300 lb
CE
200 lb
CD
200 lb, compression
𝑃𝑃𝑚𝑚𝑚𝑚𝑚𝑚
Our strategy here is to find the forces in each cable in terms of force P and
then apply the failure criteria to determine the largest value of P.
The cables presented here are not connected over a pulley, hence, they
experience a unique tensile force from each other. As we can see, we decided
to divide our FBDs as joints A and C since the chosen body involve one or
more of the desired unknown quantities.
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MODULE 4
FBD of the joint A
EQUILIBRIUM
OF PARTICLES
FBD of the joint C
 Consider FBD of joint A:
Applying equilibrium equations along x and y axes, we can simultaneously solve for 𝐹𝐹𝐴𝐴𝐴𝐴 and
𝐹𝐹𝐴𝐴𝐴𝐴 in terms of P. To compute for P, we can equate the given failure strength of cables AB
and AC to their values in terms of P.
⇸ 𝛴𝛴𝐹𝐹𝑥𝑥 = −𝐹𝐹𝐴𝐴𝐴𝐴 cos 45 + 𝐹𝐹𝐴𝐴𝐴𝐴 cos 30 = 0
⤉ 𝛴𝛴𝐹𝐹𝑦𝑦 = 𝐹𝐹𝐴𝐴𝐴𝐴 sin 45 + 𝐹𝐹𝐴𝐴𝐴𝐴 sin 30 − 𝑃𝑃 = 0
𝐹𝐹𝐴𝐴𝐴𝐴 = 0.8965754722 𝑃𝑃 = 300 𝑙𝑙𝑙𝑙
𝑃𝑃 = 334.6065 𝑙𝑙𝑙𝑙
𝐹𝐹𝐴𝐴𝐴𝐴 = 0.7320508076 𝑃𝑃 = 300 𝑙𝑙𝑙𝑙
𝑃𝑃 = 409.8076 𝑙𝑙𝑙𝑙
 Consider FBD of joint C:
Using the same strategy in joint A, here we can also solve the values of P in terms of the
failure strength of cables CE and AC using our equilibrium equations.
⇸ 𝛴𝛴𝐹𝐹𝑥𝑥 = −𝐹𝐹𝐴𝐴𝐴𝐴 cos 30 + 𝐹𝐹𝐶𝐶𝐶𝐶 = 0
⤉ 𝛴𝛴𝐹𝐹𝑦𝑦 = −𝐹𝐹𝐴𝐴𝐴𝐴 sin 30 + 𝐹𝐹𝐶𝐶𝐶𝐶 = 0
𝐹𝐹𝐶𝐶𝐶𝐶 = 0.6339745962 𝑃𝑃 = 200 𝑙𝑙𝑙𝑙
𝑃𝑃 = 315.4701 𝑙𝑙𝑙𝑙
𝐹𝐹𝐶𝐶𝐶𝐶 = 0.3660254038 𝑃𝑃 = 200 𝑙𝑙𝑙𝑙
𝑃𝑃 = 546.4102 𝑙𝑙𝑙𝑙
Comparing the different values of P from our various failure criteria. We found that the
smallest P is 315.4701 lb. This means that when P is 315.4701 lb, cable CE experiences a
200-lb force. Greater than that value of P, the force in cable CE would be greater than the
force it can bear. Hence, to maintain structural integrity within the system, we must choose
this P. Hence, our maximum P is
𝑷𝑷𝒎𝒎𝒎𝒎𝒎𝒎 = 𝟑𝟑𝟑𝟑𝟑𝟑. 𝟒𝟒𝟒𝟒𝟒𝟒𝟒𝟒 𝒍𝒍𝒍𝒍
4-17
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ENGINEERING SCIENCE 11
STATICS OF RIGID BODIES
MODULE 4
EQUILIBRIUM
OF PARTICLES
PROBLEM 6
A pulley-and-cable assembly supports block W and requires a downward force at P for
equilibrium. Determine the largest tension at P if the tension in the cables were not to
exceed 500 N, the springs were not to stretch by more than 250 mm nor compress by 200
mm, and the tension in the bar were not to exceed 900 N nor the compression greater than
800 N. Also, calculate the deformation in the springs and force in the bar when P is at
maximum. (Use k = 4 kN/m.)
Required:
Solution:
𝑃𝑃𝑚𝑚𝑚𝑚𝑚𝑚
Our strategy here is to find the forces in each cable in terms of force P. We
start with the joint/ particle where force P acts.
4-18
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MODULE 4
EQUILIBRIUM
OF PARTICLES
 Consider FBD of pulley H:
By applying the equilibrium conditions, we can solve for 𝐹𝐹𝑠𝑠,𝐹𝐹𝐹𝐹 and 𝑇𝑇𝐻𝐻𝐻𝐻 in terms of P.
⇸ 𝛴𝛴𝐹𝐹𝑥𝑥 = −𝐹𝐹𝑠𝑠,𝐹𝐹𝐹𝐹 cos 54 + 3𝑇𝑇𝐻𝐻𝐻𝐻 cos 78.701 = 0
⤉ 𝛴𝛴𝐹𝐹𝑦𝑦 = 𝐹𝐹𝑠𝑠,𝐹𝐹𝐹𝐹 sin 54 + 3𝑇𝑇𝐻𝐻𝐻𝐻 sin 78.701 − 𝑃𝑃 = 0
𝐹𝐹𝑠𝑠,𝐹𝐹𝐹𝐹 = 0.2666053799𝑃𝑃 = (0.250𝑚𝑚)(4000𝑁𝑁/𝑚𝑚)
𝑇𝑇𝐻𝐻𝐻𝐻 = 0.2666045474𝑃𝑃 = 500 𝑁𝑁
𝑃𝑃 = 3750.8620 𝑁𝑁
𝑃𝑃 = 1875.4369 𝑁𝑁
Because the computed value of 𝐹𝐹𝑠𝑠,𝐹𝐹𝐹𝐹 is positive, it means that spring FH is in
tension. To solve for the maximum tensile force the spring can support, the largest
possible length of stretch 0.250𝑚𝑚 is multiplied to the spring constant.
 Consider FBD of pulley F:
Since we already have the value of 𝐹𝐹𝑠𝑠,𝐹𝐹𝐹𝐹 in terms of P, we can solve for 𝑇𝑇𝐵𝐵𝐵𝐵 and 𝑇𝑇𝐺𝐺𝐺𝐺
in terms of P by applying the equilibrium equations.
⇸ 𝛴𝛴𝐹𝐹𝑥𝑥 = −𝑇𝑇𝐵𝐵𝐵𝐵 sin 36 + 𝐹𝐹𝑠𝑠,𝐹𝐹𝐹𝐹 cos 54 = 0
⤉ 𝛴𝛴𝐹𝐹𝑦𝑦 = −𝑇𝑇𝐵𝐵𝐵𝐵 cos 36 − 𝐹𝐹𝑠𝑠,𝐹𝐹𝐹𝐹 sin 54 + 𝑇𝑇𝐺𝐺𝐺𝐺 = 0
𝑇𝑇𝐵𝐵𝐵𝐵 = 0.2666053799𝑃𝑃 = 500𝑁𝑁
𝑇𝑇𝐺𝐺𝐺𝐺 = 0.43137655663𝑃𝑃 = 500 𝑁𝑁
𝑃𝑃 = 1159.080115 𝑁𝑁
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ENGINEERING SCIENCE 11
STATICS OF RIGID BODIES
MODULE 4
EQUILIBRIUM
OF PARTICLES
 Consider FBD of pulley B:
⇸ 𝛴𝛴𝐹𝐹𝑥𝑥 = 2𝑇𝑇𝐵𝐵𝐵𝐵 cos 25 − 𝐹𝐹𝑠𝑠,𝐴𝐴𝐴𝐴 cos 15 + 𝑇𝑇𝐵𝐵𝐵𝐵 sin 36 = 0
⤉ 𝛴𝛴𝐹𝐹𝑦𝑦 = −2𝑇𝑇𝐵𝐵𝐵𝐵 sin 25 + 𝐹𝐹𝑠𝑠,𝐴𝐴𝐴𝐴 sin 15 + 𝑇𝑇𝐵𝐵𝐵𝐵 cos 36 = 0
𝐹𝐹𝑠𝑠,𝐴𝐴𝐴𝐴 = 1.507111055𝑃𝑃 = (0.250𝑚𝑚)(4000𝑁𝑁/𝑚𝑚)
𝑇𝑇𝐵𝐵𝐵𝐵 = 0.7166719732𝑃𝑃 = 500 𝑁𝑁
𝑃𝑃 = 663.5211 𝑁𝑁
𝑃𝑃 = 697.6693𝑁𝑁
 Consider FBD of pulley C:
Here we assume that the bar experiences tension.
⇸ 𝛴𝛴𝐹𝐹𝑥𝑥 = −2𝑇𝑇𝐵𝐵𝐵𝐵 cos 25 + 1000 cos 42 = 0
⤉ 𝛴𝛴𝐹𝐹𝑦𝑦 = 2𝑇𝑇𝐵𝐵𝐵𝐵 sin 25 − 𝐹𝐹𝐶𝐶𝐶𝐶 − 1000 sin 42 = 0
𝑇𝑇𝐶𝐶𝐶𝐶 = 1.748045247𝑃𝑃 = 500 𝑁𝑁
𝐹𝐹𝐶𝐶𝐶𝐶 = −0.5639132491𝑃𝑃 = 800 𝑁𝑁
𝑃𝑃 = 286.0338 𝑁𝑁
𝑃𝑃 = 1418.6580 𝑁𝑁
Since the sign for 𝐹𝐹𝐶𝐶𝐶𝐶 is negative, it means that the bar experiences a compressive
force and the largest compressive force it can support is 800 N.
Therefore the maximum P that the system can support is 286.0338 N. Greater than
this value, cable CE would already have failed.
4-20
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ENGINEERING SCIENCE 11
STATICS OF RIGID BODIES
MODULE 4
EQUILIBRIUM
OF PARTICLES
KNOWLEDGE CHECK!
1. Which of the following statements is/are TRUE?
A) Pulleys can support both tension and compression.
B) Bars, pulleys, and cables can support tensile forces.
C) Cables can support compressive forces.
D) Bars support only tension force.
2. Cables are assumed to ________.
A) have constant weight
B) be extensible
C) be perfectly flexible
D) all of the choices
3. A particle with unknown weight (W) is being acted upon by forces from a compressed
spring (Fspring) along the x-axis and cable (Fcable) acting 45° N of E. Which of the following
is the correct free-body diagram?
A)
C)
B)
D)
4. Which of the following statement(s) is/are true about bars?
I. They support both compressive and tensile forces.
II. The change in their length is assumed to be negligible.
III. Their weight is proportional to the weight they carry.
A) I only
B) I and II only
C) II and III only
D) All of the statements are true
5. For the figure on the right, what is the spring constant if F = 8N?
A) 40 N/m
B) -40 N/m
C) 1.6 N∙m
D) -1.6 N∙m
4-21
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MODULE 4
EQUILIBRIUM
OF PARTICLES
PRACTICE PROBLEMS
1. A pulley-and-cable assembly as shown on the
annexed figure supports block W and for
equilibrium, it requires two blocks – one
weighing 3 slugs and the other 5 slugs.
Calculate the following:
a. tension (in lb) of the cable that passes
over pulleys A and B
b. mass (in slug) of block W
c. deformation (in ft) of the spring CE
[Use g = 32.2 ft/s2, 1 slug = 32.2 lb]
Guide Questions:
a. How many FBDs are required to analyze the given problem? How to compute the
tension in cable AB?
b. Upon obtaining TAB, which FBD is next to consider? Can we assume that weight of
block W is the same as the tension in cable CF?
Answer: TABC= 48.3 lb, mW, min = 4.3764 slug; sCE = -1.4164 ft
2. For the system shown, the
weight of block P is 80%
of the weight of load W. If
the applied force F is 100
N, determine the
following: k = 4000 N/m)
a. weight (in N) of
load W
b. deformation (in
mm) of spring J
c. force (in N) in bars
AB and HI
Answer: W= 105.7222 N, sJ = 10.7062 mm, FAB= 102.7917 N, C, FHI= 39.8402 N, T
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OF PARTICLES
3. The system shown below is in equilibrium. Spring 3 is known to have a change in length
of 10 cm. The magnitude of the compressive force in Spring 2 is 15% more than the
magnitude of force P, while the magnitude of the compressive force in Bar 1 is 100 N
less than that in Spring 2. If the spring constants of Springs 2 and 3 are 625 and 1200
N/m, respectively, determine the following:
a. magnitude of forces P (in N) and F (in N)
b. tension (in N) in cable ABC
c. mass (in kg) of Block A
d. compressive force in Bar 1
e. change in length (in cm) of Spring 2
Draw the complete FBD’s of Block A, Hook C and Pulley E.
Answer: 𝑃𝑃 = 131.2831 𝑁𝑁, 𝐹𝐹 = 87.1569 𝑁𝑁, 𝑇𝑇𝐴𝐴𝐴𝐴𝐴𝐴 = 110.7024 𝑁𝑁, 𝑚𝑚𝐴𝐴 = 15.6338 𝑘𝑘𝑘𝑘,
𝐹𝐹1 = 50.9756 𝑁𝑁, 𝑠𝑠2 = 24.1561 𝑐𝑐𝑐𝑐
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STATICS OF RIGID BODIES
MODULE 4
EQUILIBRIUM
OF PARTICLES
CHALLENGE YOURSELVES!
1. Determine the maximum weight of the bucket that
the wire system can support so that no single wire
develops a tension exceeding 200 lb. [Source:
Hibbeler, 2013]
Guide Questions:
a. The system is connected with rings at B and E to
support the load at E. What are the FBDs of the
wire system?
b. Since our wire system can support only up to
200 lb of tensile force in each wire, can we have a scenario that force in AB can handle
200 lb only so that we can compute for the tensile forces in wires BC and BE? Similarly,
can we have the same analogy for the other wires (e.g. wires BC and BE)? Which case
produce a tensile force not exceeding the 200 lb limit in each wire?
Answer: Wmax = 115.4701 N
2. What is the maximum weight W
that the system could support if
the maximum force that a cable
could handle is 500 kN and the
mass of the block at J is 300 kg?
Also, compute for the initial
length of the springs if their
spring constant is 80 kN/mm? BC
is collinear with CG.
Answer:
Wmax= 442.9084 kN,
lJ,0= 95.3337 mm,
lE,0= 164.9973 mm
4-24
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MODULE 4
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OF PARTICLES
REFERENCES
BEER, F.P., JOHNSTON, R.E., MAZUREK, D.F., CORNWELL, P.J., and SELF, B.P. 2016.
Vector mechanics for engineers. Statics and dynamics. 11th ed. McGraw-Hill Education.
HIBBELER, R.C. 2013. Engineering Mechanics: Statics 13th ed. Pearson Prentice Hall.
MERIAM, J.L. and KRAIGE L.G. 2002. Engineering Mechanics Volume 1: Statics. 5th ed. John
Wile & Sons, Inc.
PLESHA, M.E., GRAY, G.L., and COSTANZO, F. 2010. Engineering Mechanics: Statics.
McGraw Hill Co.
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