ch.03 equilibrium of a particle

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03. Equilibrium of a Particle

HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien

Engineering Mechanics Statics 3.01 Equilibrium of a Particle

Chapter Objectives

To introduce the concept of the free-body diagram for a particle

To show how to solve particle equilibrium problems using the equations of equilibrium



1. Condition for Equilibrium of a Particle

- Newtons First Law of Physics: If the resultant force on a particle is zero, the particle will remain at rest or will continue

at constant speed in a straight line

= 0 Equilibrium

- Equilibrium: a key concept in statics is that of equilibrium. If an

object is at rest, we will assume that it is in equilibrium and that

the sum of the forces acting on the object equal zero

= = 0

Equations for 2D Equilibrium: = 0, = 0

Equations for 3D Equilibrium : = 0, = 0, = 0



2. The Free-Body Diagram

- A free-body diagram (FBD): drawing that shows the particle

with all the forces that act on it

- Two types of connections often encountered in particle

equilibrium problems

Springs: can be used to apply forces of tension/compression

Cables and Pulleys

+ Ideal pulleys simply change the direction of a force

+ The tension on each side of an ideal pulley is the same

+ The tension is the same everywhere in a given rope or cable if ideal pulleys are used




- A Procedure for Drawing a Free-body Diagram

Imagine the particle to be isolated or cut free from its surroundings

Show all the forces that act on the particle

+ Active forces: they want to move the particle

+ Reactive forces: they tend to resist the motion

Identify each force and show all known magnitudes and directions. Show all unknown magnitudes and/or directions

as variables




- Example 3.1 The sphere has a mass of 6 and is supported as shown. Draw a free-body diagram of the sphere,

the cord , and the knot at Solution

- Knot

- Cord

- Sphere



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3. Coplanar Force Systems

- If a particle is subjected to a system of coplanar forces that lie

in the plane, then each force can be resolved into its and components

= 0

+ = 0

= 0 = 0




- Procedure for Analysis

+ Free-Body Diagram

Establish the , axes in any suitable orientation

Label all the known/unknown force magnitudes and directions

The sense of an unknown magnitude force can be assumed

+ Equations of Equilibrium

Apply the equations of equilibrium, = 0 and = 0

If more than two unknowns exist and the problem involves a spring, apply = to relate the spring force to the deformation of the spring

Since the magnitude of a force is always (+), then if the solution for a force yields a () result, this indicates its sense is the reverse of that shown on the free-body diagram




- Example 3.2 Determine the tension in cables and necessary to support the 60 cylinder

Solution

Free-body diagram




Equation of equilibrium

+ = 0: 450

4

5 = 0

+ = 0: 450 +

3

5 60 9.81 = 0

The tension in cables

: = 420

: = 475.66




- Example 3.3 The 200 crate is suspended using the ropes and . Each rope can withstand a maximum force of 10 before it breaks. If always remains horizontal, determine the smallest angle to which the crate can be suspended before one of the ropes breaks

Solution


+ = 0: + = 0

+ = 0: 200 9.81 = 0

The smallest angle

= 11962

10 103= 11.310 11.30

= 10 10311.310 = 9.81 103




- Example 3.4 Determine the required length of cord so that the 8 lamp can be suspended in the position shown. The undeformed length of spring is = 0.4

Solution


300 = 0

300 78.5 = 0

= 157.0

= 135.9

The stretch of spring

= = 0.453

The stretched length = + = 0.4 + 0.453 = 0.853

Distance from to 2 = 300 + 0.853 = 1.32



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Fundamental Problems

- F3.1: The crate has a weight of 550. Determine the force in each supporting cable




- F3.2: The beam has a weight of 700. Determine the shortest cable that can be used to lift it if the maximum force the cable can sustain is 1500




- F3.3: If the 5 block is suspended from the pulley and the sag of the cord is = 0.15, determine the force in cord . Neglect the size of the pulley




- F3.4: The block has a mass of 5 and rests on the smooth plane. Determine the unstretched length of the spring




- F3.5: If the mass of cylinder is 40, determine the mass of cylinder in order to hold the assembly in the position shown




- F3.6: Determine the tension in cables , , and , necessary to support the 10 and 15 traffic lights at and , respectively. Also, find the angle



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4. Three-Dimensional Force Systems

- The necessary and sufficient condition for particle equilibrium

= 0

+ + = 0

= 0 = 0

= 0




- Procedure for Analysis

+ Free-Body Diagram

Establish the , , axes in any suitable orientation

Label all the known/unknown force magnitudes and directions

The sense of an unknown magnitude force can be assumed

+ Equations of Equilibrium

Apply the equations of equilibrium, = 0, = 0, = 0

If the 3D geometry appears difficult, express each force on the free-body diagram as a Cartesian vector, substitute these

vectors into = 0, set the , , components equal to zero

Since the magnitude of a force is always (+), then if the solution for a force yields a () result, this indicates its sense is the reverse of that shown on the free-body diagram




- Example 3.5 A 90 load is suspended from the hook. If the load is supported by two cables and a

spring having a stiffness = 500/, determine the force in the cables and

the stretch of the spring for equilibrium.

Cable lies in the plane and

cable lies in the plane

Solution

Free-body diagram


= 0: 300

4

5 = 0

= 0: 300 + = 0

= 0: 3

5 90 = 0





= 0: 300

4

5 = 0

= 0: 300 + = 0

= 0: 3

5 90 = 0

Solving the above equations

= 150

= 240

= 207.8

The stretch of the spring

=

==207.8

500= 0.416




- Example 3.6 The 10 lamp is suspended from the three equal-length cords. Determine its smallest vertical distance

from the ceiling if the force developed in any cord is not allowed to exceed 50

Solution

Free-body diagram


= 0: 3 50

109.81= 0

= 198.1

150= 49.160

From the shaded triangle

=600

49.160= 519




- Example 3.7 Determine the force in each cable used to

support the 40 crate

Solution

Free-body diagram


= 3 4 + 8

(3)2+(4)2+82

= 0.318 0.424 + 0.848

= 3 + 4 + 8

(3)2+42 + 82

= 0.318 + 0.424 + 0.848

= , = 40



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Equilibrium required

+ + + = 0

0.318 0.318 +

+ 0.424 +0.424

+ 0.848 +0.848 40 = 0

Equating the respective , , components to zero yields

= 0: 0.318 0.318 + = 0

= 0: 0.424 +0.424 = 0

= 0: 0.848 +0.848 40 = 0

Solving the above equations give

= = 23.6

= 15.0




- Example 3.7 Determine the tension in each cord used to support

the 100 crate

Solution

Free-body diagram


=

= 1200 + 135

0 + 600

= 0.5 0.707 + 0.5

= + 2 + 2

(1)2+22 + 22

= 0.333 + 0.667 + 0.667

= 981




Equilibrium required

+ + + = 0

0.5 0.333

+ 0.707 +0.667

+ 0.5 +0.667 981 = 0

Equating the respective , , components to zero yields

= 0: 0.5 0.333 = 0

= 0: 0.707 + 0.667 = 0

= 0: 0.5 + 0.667 981 = 0

Solving the above equations give

= 694, = 813, = 862




- F3.7: Determine the magnitude of forces 1, 2 , 3 so that the particle is held in equilibrium




- F3.8: Determine the tension developed in cables , ,







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- F3.11: The 150 crate is supported by cables , , and . Determine the tension in these wires



ch.03 equilibrium of a particle

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