4. Equilibrium of a Particle;

THE FREE-BODY DIAGRAM & COPLANAR FORCE

SYSTEMS

Engineering Mechanics

Statics

Mechanical Systems Engineering -2016

Lecture Objectives:

Students will be able to :

a) Draw a free body diagram (FBD), and,

b) Apply equations of equilibrium to solve a 2-D problem.

The crane is lifting a load. To decide if the straps

holding the load to the crane hook will fail, you

need to know the force in the straps. How could

you find the forces?

Engineering Applications

Straps

T

W

For a given force exerted on the boat’s

towing pendant, what are the forces in the

bridle cables? What size of cable must

you use?

Remember the following terms :

- Coplanar force system or 2D system – The forces lie on a sheet of paper (plane).

- A particle – It has a mass, but a size that can be neglected.

- Equilibrium – A condition of rest or constant velocity (V=constant or V=0).

This is an example of a 2-D or coplanar

force system. If the whole assembly is in

equilibrium, then particle A is also in

equilibrium.

To find the tensions in the cables for a

given weight of the cylinder, we need to

learn how to draw a free body diagram

and apply the equations of equilibrium.

Condition for the Equilibrium of a Particle

A particle is said to be in Equilibrium when it remains at rest if

originally at rest, or has a constant velocity if originally in motion.

0

amF

This means that the acceleration a = 0.

The Free-Body Diagram (FBD)

Free Body Diagrams are very important things to know how to draw and use.

- What is a free-body diagram?

It is a drawing that shows all external forces acting on a particle.

- Why is it useful?

It is key to being able to write the equations of equilibrium,

which are used to solve for the unknowns (usually forces or

angles).

How ?

Active forces: trying to move the particle.

Reactive forces: trying to resist the motion.

Note : Cylinder mass = 40 kg

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

A

3. Identify each force and show all known magnitudes and directions.

Show all unknown magnitudes and / or directions as variables .

FC = 392.4 N (What is this?)

FB

FD

30˚

2. Show all the forces that act on the particle.

FBD at A

A

y

x

Equations of Equilibrium (E-of-E) in a 2D System

Or, written in a scalar form,

Fx = 0 and Fy = 0

These are two scalar equations of equilibrium (E-of-E).

They can be used to solve for up to two unknowns.

Since particle A is in equilibrium, the

net force at A is zero:

or:

FBD at A

A

In general, for a particle in equilibrium,

(a vector equation)

FBD at A

A

FB

FD A

FC = 392.4 N

y

x 30˚

0

DCB FFF

0

F

jijFiF yx

00

Example:

Write the scalar E-of-E:

+ Fx = FB cos 30º – FD = 0

+ Fy = FB sin 30º – 392.4 N = 0

FBD at A

A

FB

FD A

FC = 392.4 N

y

x 30˚

Solving the second equation gives: FB = 785 N

From the first equation, we get: FD = 680 N ←

Notes about Springs & Cables

With a frictionless pulley,

T1 = T2. Spring Force = spring constant * deformation,

or

F = k s

T1

T2

We normally assume that cables (or cords), springs, rings, hooks … etc,

have a negligible (0) weight, unless otherwise stated.

1.When a particle is in equilibrium, the sum of forces acting on it equals _____.

A) A constant B) A positive number C) Zero D) A negative number

F2

20 lb

F1

C

50°

2. Using this FBD of Point C, the sum of forces in the x-direction ( Fx ) is ____ .

A) F2 sin 50° – 20 = 0

B) F2 cos 50° – 20 = 0

C) F2 sin 50° – F1 = 0

D) F2 cos 50° + 20 = 0

x

Questions:

3. Draw the FBD of the particle A, y= 6 m, AC= 8.49 m, m= 200 kg

Solution:

m=200 kg, AB=1.5m, y=0.75m.

FBA=? , FBC=?

First, let’s draw the FBD.

B θ= 30

FBA

FBC

W =200(9.81) N

x

y

W=m.g=200(9.81) N

Notice : kg is an SI unit

Classroom Exercise:

Classroom Exercise:

Solution:

1- Draw a FBD for Point C

FCD

FBC

30 lb

y

x 30

15

2- The E-of-E are:

+ Fx = FBC cos 15º – FCD cos 30º = 0

+ Fy = FCD sin 30º – FBC sin 15º – 30 = 0

FBC = 100.4 lb

FCD = 112.0 lb

Example: Cylinder E weighs 30 lb and the geometry

is as shown. Find: Forces in the cables (FBC , FCD ,

FBA ) and weight of cylinder F (WF).

3- Draw a FBD for Point B FBC =100.4 lb

FBA

WF

y

x

15 45

4- The E-of-E are:

Fx = FBA cos 45 – 100.4 cos 15 = 0

Fy = FBA sin 45 + 100.4 sin 15 – WF = 0

FBA = 137 lb

WF = 123 lb

3D Force System

In 3-D, when a particle is in equilibrium:

This vector equation will be satisfied only when:

,0,0,0

0

zyx

zyx

FFF

kFjFiF

Example

Given: A 600 N load is supported by three

cords with the geometry as shown.

Find: The tension in cords AB, AC and AD.

Solution:

FBD at A

FC FD

A

600 N

z

y 30˚

FB

x

1 m

2 m

2 m

FB = FB (sin 30 i + cos 30 j) N

= {0.5 FB i + 0.866 FB j} N

FC = – FC i N

FD = FD (rAD /rAD)

= FD { (1 i – 2 j + 2 k) / (12 + 22 + 22)½ } N

= { 0.333 FD i – 0.667 FD j + 0.667 FD k } N

Now equate the respective i , j , k components to

zero.

Fx = 0.5 FB – FC + 0.333 FD = 0

Fy = 0.866 FB – 0.667 FD = 0

Fz = 0.667 FD – 600 = 0

FD = 900 N

FB = 693 N

FC = 646 N

Answers:

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