Mechanic Machine

JJ205 ENGINEERING MECHANICS

CHAPTER 3

EQUILIBRIUM OF PARTICLE

PREPARED BY:

AMRAN BIN AWANG @ MUDA

DEPARTMENT OF MECHANICAL ENGINEERING

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.

Chapter Outline

Condition for the Equilibrium of a Particle The Free-Body Diagram

Coplanar Systems

CONDITION FOR THE EQUILIBRIUM OF A PARTICLE

Particle at equilibrium if- At rest

- Moving at constant a constant velocity

Newton’s first law of motionΣF = 0

where ΣF is the vector sum of all theforces acting on the particle

Newton’s second law of motion

ΣF = ma

When the force fulfill Newton's first law of motion,

ma = 0

a = 0

therefore, the particle is moving in constant velocity or at rest

The Free-Body Diagram

Best representation of all the unknownforces (ΣF) which acts on a body

A sketch showing the particle “free” from the surroundings with all the forces acting on it

Consider two common connections in this subject Spring

Cables and Pulleys

Spring

Linear elastic spring: change in length is directly proportional to the force acting on it

- spring constant or stiffness k: defines the elasticity of the spring

- Magnitude of force when spring is elongated or compressed F = ks

where s is determined from the difference in spring’s deformed length l and its undeformed length lo

s = l - lo

- If s is positive, F “pull”

onto the spring

- If s is negative, F “push”

onto the spring

EXAMPLE

Given lo = 0.4m and k = 500N/m

To stretch it until l = 0.6m, A force, F = ks

=(500N/m)(0.6m – 0.4m) = 100N is needed

To compress it until l = 0.2m,

A force, F = ks

=(500N/m)(0.2m – 0.4m)

= -100N is needed

- Cables (or cords) are assumed to havenegligible weight and they cannot stretch

-A cable only support tension or pulling force

- Tension always acts in thedirection of the cable

- Tension force in a continuouscable must have a constantmagnitude for equilibrium

Cables and Pulley

For any angle θ, the cable is subjected to a constant tension T throughout its length

1. Draw outlined shape • Isolate particle from its surroundings

2. Show all the forces• Indicate all the forces• active forces: set the particle in motion• Reactive forces: result of constraints and

supports that tend to prevent motion

Procedure for Drawing a FBD

3. Identify each forces

- Known forces should be labeled with proper magnitude and direction

- Letters are used to represent magnitude and directions of unknown forces

A spool is having a weight W which is suspended from the crane bottom

Consider FBD at A since these forces act on the ring

Cables AD exert a resultant force of W on the ring

Condition of equilibrium is used to obtained TB and TC

Each term must be expressed in the same units Eg: s = vt + ½ at2 where s is

position in meters (m), t is time in seconds (s), v is velocity in m/s and a is acceleration in m/s2

Regardless of how the equation is evaluated, it maintains its dimensional homogeneity

All the terms of an equation can be replaced by a consistent set of units, that can be used as a partial check for algebraic manipulations of an equation

The bucket is held in

equilibrium by the cable

Force in the cable =

weight of the bucket

Isolate the bucket for

FBD

Two forces acting on

the bucket, weight W

and force T of the cable

Resultant of forces = 0

W = T

Example 3.1

The sphere has a mass of 6kg and issupported. Draw a free-body diagram ofthesphere, the cordCE and the knot at C.

FBD at Sphere Two forces acting,weight and theforce on cord CE. Weight of 6kg(9.81m/s2) = 58.9N

SolutionCord CE Two forces acting, forceof the sphere and forceof the knot Newton’s Third Law: FCEis equal but opposite FCE and FEC pull the cordin tension For equilibrium, FCE =FEC

FBD at Knot

Three forces acting, force by cord CBA, cord CE

and spring CD

Important to know that

the weight of the sphere

does not act directly on

the knot but subjected to

by the cord CE

Coplanar Systems

A particle is subjected to coplanar forces in

the x-y plane

Resolve into i and j components for

equilibrium

ΣFx = 0

ΣFy = 0

Scalar equations of equilibrium

require that the algebraic sum

of the x and y components to

equal o zero

Coplanar Systems

Scalar Notation

Sense of direction = an algebraic sign that corresponds to the arrowhead direction of the component along each axis

For unknown magnitude, assume arrowhead sense of the force

Since magnitude of the force is always positive, if the scalar is negative, the force is acting in the opposite direction

Consider the free-body diagram of the particle subjected to two forces

Assume unknown force F acts to the right for equilibriumΣFx = 0 ; + F + 10N = 0F = -10N

Force F acts towards the left for equilibrium

The chain exerts three forces on the ring at A.

The ring will not move, or will move with constant velocity, provided the summation of the forces along the y axis is zero

With any force known, the magnitude of other two forces are found by equations of equilibrium

Example 3.2Determine the tension in cables AB and AD for equilibrium of the 250kg engine.

FBD

Solution

+→ ΣFx = 0;

+↑ ΣFy = 0;

Solving,

TB = 4.90kN

TD = 4.25kN

*Note: Neglect the weights of the cables since they

are small compared to the weight of the engine

Example 3.3If the sack at A has a weight of 20N (≈ 2kg), determine the weight of the sack at B and the force in each cord needed to hold the system in the equilibrium position shown.

Example 3.4Determine the required length of the cord AC so that the 8kg lamp is suspended. The undeformed length of the spring AB is l’AB = 0.4m, and the spring has a stiffness of kAB = 300N/m.

Chapter Summary

Two Dimensional

Two scalars equations ΣFx = 0 and ΣFy =

0 is applied to x, y coordinate system

If the solution is negative, the sense of

the force is opposite to that shown on

the FBD.

If problem involves a linear spring,

stretch or compression can be related to

force by

F = ks