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Department of Mechanical Engineering Chapter 3 Equilibrium of concurrent forces

Chapter 3 Equilibrium of concurrent forces

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Page 1: Chapter 3 Equilibrium of concurrent forces

Department of Mechanical Engineering

Chapter 3Equilibrium of concurrent forces

Page 2: Chapter 3 Equilibrium of concurrent forces

Department of Mechanical Engineering

ObjectiveTo know how to apply the principle of

force equilibrium in static analysis

To practice how to construct free-body diagrams

Page 3: Chapter 3 Equilibrium of concurrent forces

Department of Mechanical Engineering

Some assumptions (just for this chapter)

The loaded object can be reduced into a dimensionless particle

The shape of the object does not affect the response

The mass of the object is concentrated to this particle

The forces are concurrent The lines of action pass the particle point

Page 4: Chapter 3 Equilibrium of concurrent forces

Department of Mechanical Engineering

Force equilibrium (mechanical eql.)

(Mechanical) equilibrium requires that the concurrent forces that act on the body satisfy

The particle in a equilibrium system must satisfy

Since both must be satisfied, the material point then must have zero acceleration, a = 0

0==∑FR

aFR .m==∑

Page 5: Chapter 3 Equilibrium of concurrent forces

Department of Mechanical Engineering

Free body diagrams

Inside the FBD– A portion of the body of interest or– A full body of interest or– A group of bodies of interest– Body forces

On the boundary of the FBD– “Replacement” forces/distributed forces

FBD

Page 6: Chapter 3 Equilibrium of concurrent forces

Department of Mechanical Engineering

How to make the FBD Select and draw bodies or parties of interest to be shown in the

FBD Identify and draw the body and surface forces applied on the

bodies Discard contacting bodies that are not wanted and replace them

with the forces they exert to the body of interest Cut the body parts that are not wanted and replace them with

internal forces they exert to the body of interest Decide and locate a coordinate system

Page 7: Chapter 3 Equilibrium of concurrent forces

Department of Mechanical Engineering

Discarded bodies and the replacement forces

Contacting bodies

String/cable/axially loaded members

Body parts

Forces normal and/or tangential to the FBD surfaces

Forces with lines of actions that coincide with the member axis

A distributed force called stress

Page 8: Chapter 3 Equilibrium of concurrent forces

Department of Mechanical Engineering

Example: basic problem

Page 9: Chapter 3 Equilibrium of concurrent forces

Department of Mechanical Engineering

The free body diagram

P

F W

N

N = normal force perpendicular to the removed surfaceW = weight of the box downwardP = the applied forceF = friction force opposite to the direction of the motion/applied forces

3

8

Page 10: Chapter 3 Equilibrium of concurrent forces

Department of Mechanical Engineering

Example: contact problem

Frictions on the same contact pointbut different FBDs are in the opposite direction.

Page 11: Chapter 3 Equilibrium of concurrent forces

Department of Mechanical Engineering

Example: cable tension

TCA

TCD

TCB

T = Cable tension outward from the FBD

Cable can only sustain tension

Page 12: Chapter 3 Equilibrium of concurrent forces

Department of Mechanical Engineering

Example: Pulley problem

TAB TBC

For pulley:TAB = TBC = T

Weight

Page 13: Chapter 3 Equilibrium of concurrent forces

Department of Mechanical Engineering

Page 14: Chapter 3 Equilibrium of concurrent forces

Department of Mechanical Engineering

Example 1: Static analysis

P

FW

N

3

8

Page 15: Chapter 3 Equilibrium of concurrent forces

Department of Mechanical Engineering

Static analysis Some simplification:

– Note that in this case, F = 0 (frictionless/ smooth surface)

– The problem is 2D

P

W

N

3

8

Page 16: Chapter 3 Equilibrium of concurrent forces

Department of Mechanical Engineering

Static analysis Select and position

a coordinate system

P

F W

N

3

8

x

yor

Page 17: Chapter 3 Equilibrium of concurrent forces

Department of Mechanical Engineering

Static analysis Find a force system that is parallel to

the selected coordinate system and equal to the original system ~ find vectorial components of the forces in the selected system

P

W

N

3

8

x

y

θ

Px

Py

Ny

NX

Wθθ

θθ

cossin

sincos

NNNNPPPP

yx

yx

==

==

x

y

Page 18: Chapter 3 Equilibrium of concurrent forces

Department of Mechanical Engineering

Static analysis Sum the forces

(according to their directions) and equate each of them to zero

x

yPx

Py

Ny

NX

W

0

0

=−+=

=−=

∑∑

WNPF

NPF

yyy

xxx

Page 19: Chapter 3 Equilibrium of concurrent forces

Department of Mechanical Engineering

Static analysis For this problem, W = 25 lbs P = ? and N = ? 2 unknowns 2 equations

x

yPx

Py

Ny

NX

W0

0

=−+=

=−=

∑∑

WNPF

NPF

yyy

xxx

θθ

θθ

cossin

sincos

NNNNPPPP

yx

yx

==

==

Page 20: Chapter 3 Equilibrium of concurrent forces

Department of Mechanical Engineering

Example: cable tension

TCATCD

TCB

T = Cable tension outward from the FBD

W

TCD

Note the direction of TCD in the two FBDs

Page 21: Chapter 3 Equilibrium of concurrent forces

Department of Mechanical Engineering

Example: cable tensionThe equilibrium equations3 unknowns3 equations

TCATCD

TCB

T = Cable tension outward from the FBD

W

TCD

Note the direction of TCD in the two FBDs

30sin60sin30cos60cos

CACDCB

CACB

CD

TTTTT

WT

+==

=

Page 22: Chapter 3 Equilibrium of concurrent forces

Department of Mechanical Engineering

Example: Maximum capacity

Note: Cables A and B do not experience the same tension

Page 23: Chapter 3 Equilibrium of concurrent forces

Department of Mechanical Engineering

Example: multiple free body diagrams

Note: 4 unknowns = 3 tensions + 1 angle4 equations = 2 x 2 equationsRemember: tension direction is always leaving the FBD

Page 24: Chapter 3 Equilibrium of concurrent forces

Department of Mechanical Engineering

Example: double pulley

Page 25: Chapter 3 Equilibrium of concurrent forces

Department of Mechanical Engineering

Example:

Page 26: Chapter 3 Equilibrium of concurrent forces

Department of Mechanical Engineering

Find F2 and F3 so that R = 0

Example

Page 27: Chapter 3 Equilibrium of concurrent forces

Department of Mechanical Engineering

Find cable tension and normal force exerted by the rod BC to the collar

Note: the rod is smooth, no friction

Example

Page 28: Chapter 3 Equilibrium of concurrent forces

Department of Mechanical Engineering

How much (vertical) force the man must apply if TCD = 2000 lb?

Example

Page 29: Chapter 3 Equilibrium of concurrent forces

Department of Mechanical Engineering

If the light is 1000 kg, what are the cable tensions?

Example

Page 30: Chapter 3 Equilibrium of concurrent forces

Department of Mechanical Engineering

Rx=Σ Fx = 0

In two-dimensions , only two of these equations are needed

To solve a problem involving a particle in equilibrium, draw a free-body diagram showing all the forces acting on the particle. The conditions which must be satisfied for particle equilibrium are

Σ Fx = 0 Σ Fy = 0

The particle is in equilibrium when the resultant of all forces acting on it is zero.

Ry = Σ Fy = 0 Rz = Σ Fz = 0

Summary