1. CHAPTER 10 Force and Force-Related Parameters
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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
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3. What Is Force? Force represents the interaction between 2 objects
Objects: car and person
Interaction: person pushing the car
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4. What Is Force? Force exerted by your hand on a lawn mower.
Force exerted by bumper hitch on the trailer.
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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
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6. Examples to demonstrate the tendencies of a force
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7. Units of Force Force is a derived parameter Units of force
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8. Types of Forces Spring forces and Hooke’s Law
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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
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10. Example 10.1 – Spring Constant
Solution:
Spring constant k is determined by calculating the slope of a force-deflection line
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11. Types of Forces Friction forces dry friction viscous (fluid) friction
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12. Example 10.2 – Friction Force
Given: the coefficient of static friction between a book and a desk surface is 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.
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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
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14. Newton’s Laws in Mechanics
Second law
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15. Newton’s Laws in Mechanics
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
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16. Newton’s Laws in Mechanics
Universal law of gravitation attraction
Two masses attract each other with a force that is equal in magnitude and opposite in direction
G = 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
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17. Newton’s Laws in Mechanics
Weight of an object having a mass m on the earth is
Mearth = 5.97 x 1024 kg
Rearth = 6378 x 103 m
g = 9.8 m/s2
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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,
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19. Example 10.4 – Newton’s Laws
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.
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20. Example 10.4 – Newton’s Laws
Solution (continued):
Space shuttle orbiting at an altitude of 965 km above the Earth’s surface.
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.
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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
hinge
Example – opening and closing a door
To 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.
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22. Moment, Torque – Force Acting at a Distance
Calculating moment of a force A d1 F d2 B C
Line of action
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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
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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.
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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
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26. Reaction Force
Reaction forces are developed at the supporting boundaries to keep the object held in position as planned
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27. Reaction Force
Examples to demonstrate various support conditions and how they influence the behavior of an object.
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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
work = F x d
units: N.m, J, lb.ft….
distance, d
force
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29. Work Done on a Car d
Work done on the car by the force F is equal to W1-2 = (F cos θ)(d)
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30. Mechanical Power
Power is the time rate of doing work. It represents how fast you want to do the work.
units: J/s (watts), N.m/s
lb.ft/s
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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:
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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:
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33. Pressure – Force Acting Over an Area
Pressure provides a measure of intensity of a force acting over an area.
contact area
weight
Units: Pa (N/m2), kPa, MPa, GPa
psi, ksi
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34. Pressure
An experiment demonstrating the concept of pressure using bricks (a) and (b)
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35. Common Units of Pressure
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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
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37. Pressure – Pascal’s Law
For a fluid at rest, the pressure increases with the depth of fluid.
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)
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38. Pressure – Pascal’s Law
Buoyancy is the force that a fluid exerts on a submerged object.
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39. Example 10.8 – Pressure 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
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40. Example 10.8 – Pressure Solution (continued):
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
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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.
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42. Pressure in Hydraulic System
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.
Push to raise load
reservoir
Open to lower load
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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.
down to unload
Up to load
Gear or rotary pump
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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:
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45. Example 10.12 – Hydraulic System
Solution (continued):
Alternate approach
The second approach is preferred. We can see clearly the relationship between m1 and m2.
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46. Stress
Stress provides a measure of the intensity of a force acting over an area.
units: Pa (N/m2), kPa, MPa, GPa
psi, ksi
Normal stress is often called pressure
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47. 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

48. Mechanical Properties of Materials
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
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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
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50. Tensile Test of Metal Specimen
Tensile test set up
Original specimen
Final specimen
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51. Stress-Strain Diagram
Stress-strain diagram for a mild steel sample
σe = elastic stress
(σY)u = upper-yield stress
(σY)l = lower-yield stress
σY = ultimate stress
σf = fracture stress
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52. Modulus of Elasticity x
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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
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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
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55. Modulus of Elasticity and Shear Modulus of Selected Materials
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56. Strength of Selected Materials
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57. Strength of Selected Materials
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58. 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:
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.
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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.
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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
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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
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