10-1 CHAPTER 10 Force and Force-Related Parameters © 2011 Cengage Learning Engineering. All Rights...

10-1

CHAPTER 10Force and Force-Related Parameters

© 2011 Cengage Learning Engineering. All Rights Reserved.

10-2

Outline

In this chapter we will

• explain what we mean by force

• discuss Newton’s laws in mechanics

• explain tendencies of unbalanced mechanical forces

• quantify the tendencies of a force acting at a distance (moment), over a distance (work), and over a period of time (impulse)

• discuss mechanical properties of materials


10-3

What Is Force?

Force represents the interaction between 2 objects

Objects: car and person

Interaction: person pushing the car


10-4

What Is Force?

Force exerted by your hand on a lawn mower.

Force exerted by bumper hitch on the trailer.


10-5

Definition of Force

All forces are defined by their magnitudes, their directions, and the point of applications

Examples to demonstrate the effect of magnitude, direction, and the point of application of a force on the same object


10-6

Tendencies of a Force

Examples to demonstrate the tendencies of a force


10-7

Force is a derived parameter

Units of force

Units of Force

ma onaccelerati mass force

N 4.448lb 1

s

ftsluglblb

s

ft 1slug 1lb 1

s

mkgN

s

m 1kg 1 N 1

f

2f2f

22


10-8

Types of Forces

Spring forces and Hooke’s Law

kxF x


10-9

Example 10.1 – Spring Constant

Given: a spring as shown; dead weight is attached to the end of the spring and the corresponding deflection is measured.

Find: the spring constant

Results of the Experiments


10-10

Example 10.1 – Spring Constant

Solution:

Spring constant k is determined by calculating the slope of a force-deflection line

N/mm 54.0

mm936

N9.46.19

deflectionin change

forcein change slope

k


10-11

Types of Forces

• Friction forces dry friction

viscous (fluid) friction

NF max

weight

normalforce

frictionforce

appliedforce


10-12

Example 10.2 – Friction Force

newtons 12206.0max NF

Given: the coefficient of static friction between a book and a desk surface is 0.6. The book weighs 20 N. A horizontal force of 10 N is applied to the book.

Find: would the book move? If not, find the friction force and magnitude of the horizontal force F to set the book in motion.

Solution:

since Fmax > 10 N, the book would not move. It would take a 12 N horizontal force to set the book in motion.

weight

normalforce

frictionforce

appliedforce


10-13

Newton’s Laws in Mechanics

• First law an object at rest remains at rest if no

unbalanced forces acting on it an object in motion with a constant velocity,

and if there are no unbalanced forces acting on it, the object will continue to move with the same velocity


10-14


Second law

amF


10-15


• Third law For every action there is a reaction The forces of action and reaction have the same

magnitude and act along the same line, but they have opposite directions


10-16

• Universal law of gravitation attraction

Two masses attract each other with a force that is equal in magnitude and opposite in direction

G = 6.673 x10-11 m3/kg.s2

weight of an object is the force that is exerted on the mass of the object by earth’s gravity


221

r

mGmF


10-17


Weight of an object having a mass m on the earth is

mgW

R

GMg

R

mGMW

earth

earth

earth

earth

22

let ,

Mearth = 5.97 x 1024 kgRearth = 6378 x 103 mg = 9.8 m/s2


10-18

Example 10.3 – Newton’s Laws

Given: an exploration vehicle with a mass of 250 kg on Earth

Find: the mass of the vehicle on Moon (gMoon = 1.6 m/s2) and on planet Mars (gMars = 3.7 m/s2); weight of vehicle on Moon and Mars

Solution:

The mass of the vehicle is the same on Earth, Moon and Mars. The weight of the vehicle is determined from W = mg,

N925sm .73kg 250 :MarsOn

N400sm .61kg 250 :MoonOn

N2450sm 8.9kg 250 :EarthOn

2

2

2

W

W

W


10-19


Given: The space shuttle orbits the Earth at altitudes ranging from 250 km to 965 km; mass of an astronaut is 70 kg on Earth

Find: The g value and the weight of the astronaut in orbit

Solution:

Space shuttle orbiting at an altitude of 250 km above Earth’s surface.

N 635sm 07.9kg 70

sm 07.9m10250106378

kg 1097.5skg

m10673.6

2

2233

242

311

2Earth

W

R

GMg


10-20


Solution (continued):

Space shuttle orbiting at an altitude of 965 km above the Earth’s surface.

N 517sm .387kg 70

sm 38.7m10965106378

kg 1097.5skg

m10673.6

2

2233

242

311

2Earth

W

R

GMg

The near weightless conditions that you see on TV are created by the orbital speed of the shuttle. For example, when the shuttle is orbiting at an altitude of 935 km at a speed of 7744 m/s, it creates a normal acceleration of 8.2 m/s2. It is the difference between g and normal acceleration [(9.8 – 8.2) m/s2] that creates the condition of weightlessness.


hinge

10-21

Moment, Torque – Force Acting at a Distance

Moment is the measure of the tendency of a force acting about an axis or a point

Moment is a vector

Example – opening and closing a doorTo open or close a door, we apply a pulling or pushing force on the doorknob. This force will make the door rotate about its hinge. In mechanics, this tendency of force is measured in terms of a moment of a force about an axis or a point.


10-22

Moment, Torque – Force Acting at a Distance

0M

FdM

FdM

C

2B

1A

Calculating moment of a force

A d1

F

d2

B

C

Line of action


10-23

Example 10.5 – Moment, Torque

Given: the locations and forces applied as shown

Find: the resulting moment of the forces about point O

Solution:

0.07 cos 35o

0.1 cos 35o

mN 55.13

m 35cos1.0N 100m 35cos07.0N 50m 05.0N 50

OM


10-24

Example 10.6 – Moment, Torque

Given: two forces are applied as shown

Find: moment about points A, B, C, and D

Solution:

Note the two forces have equal magnitude and act in opposite directions.

This pair of forces is called a couple.

mN 10m 25.0N 100m 35.0N 100

mN 10m 15.0N 100m 25.0N 100

mN 100N 100m 1.0N 100

mN 10m 1.0N 1000N 100

D

C

B

A

M

M

M

M


10-25

Internal Force

When an object is subjected to an external force, internal forces are created inside the material to hold the material and the components together.

An example of internal force. When you try to pull apart the bar, inside the bar material, internal forces develop that keep the bar together as one piece


10-26

Reaction Force

Reaction forces are developed at the supporting boundaries to keep the object held in position as planned


10-27

Reaction Force

Examples to demonstrate various support conditions and how they influence the behavior of an object.


10-28

Work – Force Acting Over a Distance

Mechanical work is defined as the component of the force – that moves the object – times the distance the object moves

distance, d

force

work = F x d

units: N.m, J, lb.ft….


10-29

Work Done on a Car

Work done on the car by the force F is equal to W1-2 = (F cos θ)(d)

d


10-30

time

workpower

units: J/s (watts), N.m/s lb.ft/s

Mechanical Power

Power is the time rate of doing work. It represents how fast you want to do the work.


10-31

Example 10.7 – Work

Given: a box with weight of 100 N on the ground as shown

Find: work required to lift the box 1.5 m above the ground

Solution:

mN 150m 5.1N 100 W


10-32

Example 10.7 – Power

Given: a box with weight of 100 N on the ground as shown. We want to lift the box in 3 seconds

Find: power required

Solution:

watts50s 3

J 150

time

work power


10-33

Pressure – Force Acting Over an Area

area

force pressure

Pressure provides a measure of intensity of a force acting over an area.

Units: Pa (N/m2), kPa, MPa, GPa psi, ksi

contact area

weight


10-34

Pressure

An experiment demonstrating the concept of pressure using bricks (a) and (b)


10-35

Common Units of Pressure

psf 1000ksf 1 ft 1

lb 1psf 1

psi 1000ksi 1 in 1

lb 1psi 1

Pa 1000kPa 1 m

N 1Pa 1

2

2

2


10-36

Pressure – Pascal’s Law

Pascal’s law states that for a fluid at rest, pressure at a point is the same in all directions


10-37


For a fluid at rest, the pressure increases with the depth of fluid.

ghP

P = fluid pressure at a point (Pa) = density of fluid (kg/m3)g = acceleration due to gravity (m/s2)h = height of fluid column (m)


10-38


2

3

3

sm9.8gravity todueon accelerati

mkg fluid ofdensity

)(mobject theof volume

(N) forcebuoyancy

g

V

ρVgFB

Buoyancy is the force that a fluid exerts on a submerged object.


10-39

Example 10.8 – Pressure

ft 4338.0

in 144

ft 1ft

s

ft 2.32

ft

slug94.1

in

lb2

2

232

h

hP

ghP

Given: a water of height h in a water tower

Find: develop a table showing the water pressure in a pipeline located at the base of the water tower relating to h; use the table to determine h if a pressure of 50 psi is desired

Solution:We will first develop the equation for pressure P as a function of height h


10-40

Example 10.8 – Pressure


Substituting values of h from 10 ft to 240 ft yields the pressure in the following table:

For a pressure of 50 psi, the water level in the tower should be approximately 120 ft


10-41

Pressure in Hydraulic Systems

In an enclosed fluid system, the pressure is nearly constant throughout the system. The relationship between the forces F1 and F2 could be established.

11

22

2

2

1

12121

2

22

1

11

,PP since

FA

AF

A

F

A

FPP

A

FP

A

FP


10-42

Pressure in Hydraulic System

reservoir

To raise the load – push the arm of the hand pump down – fluid will be pushed into the load cylinder, which in turn creates a pressure that is transmitted to the load piston, and consequently the load is raised.

To lower the load – open the release valve. The fluid will return to the reservoir and the load is lowered. Open to

lower load

Push to raise load


10-43

Pressure in Hydraulic System

To raise the load – move the control lever UP – fluid will be pushed into the load cylinder, which in turn creates a pressure that is transmitted to the load piston, and consequently the load is raised.

To lower the load – push the control lever down. The fluid will return to the reservoir and the load is lowered.

Gear or rotary pump

Up to lo

addown to

unload


10-44

Example 10.12 – Hydraulic System

Given: the hydraulic system shown, g = 9.81 m/s2

Find: the load that can be lifted by the hydraulic system.

Solution:

kg 900 m/s 81.9kg N 8829

N 8829N 981m 05.0

m 15.0

N 981m/s 81.9kg 100

22

22

2

2

11

22

211

mmF

FA

AF

gmF


10-45

Example 10.12 – Hydraulic System


Alternate approach

kg 900kg 100cm 5

cm 152

2

121

22

2

121

22

211

22

mR

Rm

gmR

RgmF

A

AF

The second approach is preferred. We can see clearly the relationship between m1 and m2.


10-46

Stress

area

force stress

Stress provides a measure of the intensity of a force acting over an area.

units: Pa (N/m2), kPa, MPa, GPa psi, ksi

area

area toparallelcomponent force stressshear

area

area tonormalcomponent force stress normal

Normal stress is often called pressure


Strain

• Strain is a description of the deformation of a body in terms of the relative displacement of the particles in the body.

That is, how much the body stretches.

• Deformation, or strain, is typically caused by the application of an external force.

Engineering Fundamentals, By Saeed Moaveni, Fourth Edition, Copyrighted 2011 10-47

10-48

• Modulus of elasticity

a measure of how easily material stretches

• Modulus of Rigidity (also called shear modulus)

a measure of how easily material twists

• Tensile strength

• Compressive strength

Mechanical Properties of Materials


10-49

Mechanical Properties of Materials

Modulus of elasticity, E – measures how easily a material will stretch when pulled (subject to a tensile force) or how well a material will shorten when pushed (subject to compressive force).

When subjected to the same force, which piece of material will stretch more?

Esteel > Ealuminum > Erubber

L

L L


10-50

Tensile Test of Metal Specimen

Tensile test set up

Original specimen Final

specimen


10-51

Stress-Strain Diagram

Stress-strain diagram for a mild steel sample

σe = elastic stress (σY)u = upper-yield stress(σY)l = lower-yield stressσY = ultimate stressσf = fracture stress


10-52

Modulus of Elasticity

Ax

FLE

L

xE

A

FE

yields, Efor solve

law sHooke'

x


10-53

Other Mechanical Properties of Materials

• Modulus of rigidity or shear modulus measures how easily a material can be

twisted or sheared is used to select materials for shafts or

rods subjected to twisting torques value is determined using a torsional test

machine


10-54

Other Mechanical Properties of Materials

• Tensile strength or ultimate strength maximum tensile force per unit of original

cross-sectional area of specimen

• Compressive strength maximum compressive force per unit of

cross-sectional area of specimen


10-55

Modulus of Elasticity and Shear Modulus of Selected Materials


10-56

Strength of Selected Materials


10-57

Strength of Selected Materials


10-58

We will consider the strength of the material as the design factor in this example.

Aluminum alloy or structural steel material with yield strength of 50 MPa and 200 MPa, respectively, could carry the load safely.

Example 10.13 – Material Properties

Given: a structural member and a load of 4000 N distributed uniformly over the cross-sectional area of the member

Find: select a material to carry the load safely

Solution:

MPa 16m 005.0m 0.05

N 4000


10-59

Summary

• You should have a good understanding of force and its common units.

• You should know the types of forces.

• You should understand the tendency of an unbalanced force, to rotate or to translate.

• You should know that the application of forces can lengthen, shorten, bend, and twist objects.

• You should be familiar with Newton’s laws in mechanics.


10-60

Summary

• You should understand pressure and should know Pascal’s law for static fluids relationship between fluid pressure and depth

of fluid how hydraulic systems work fluid properties: viscosity, bulk modulus of

compressibility

• You should know the mechanical properties of materials: modulus of elasticity, shear modulus, tensile and compressive strength


10-61

Summary

• You should know how a material behaves under applied forces

• You should understand the effect of an applied force on an object

stress: intensity of a force acting within the material Internal forces: forces created inside the material to

hold the material and the components together moment: force acting at a distance work: force acting over a distance linear impulse: force acting over a period of time


10-1 CHAPTER 10 Force and Force-Related Parameters © 2011 Cengage Learning Engineering. All Rights...

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Transcript of 10-1 CHAPTER 10 Force and Force-Related Parameters © 2011 Cengage Learning Engineering. All Rights...