2014 SEPT - MECHANICS REG REVIEW.pdf

Rowel Allan Rocaberte, REE, ECE

R

Beam

- A slender bar that carries transverse loading, that is, the applied forces are perpendicular to the bar

- Horizontal structural element that is capable of withstanding load primarily by resisting bending

Cantilever Beam

- Anchored at only one end

Overhanging Beam

- Asymmetrical placing of supports

In a beam, the internal force system consists of a shear force and a bending

moment acting on the cross section of the bar.

The internal forces give rise to two kinds of stresses on a transverse section of a beam: (1) axial stress that is caused by the bending moment and (2) shear stress due to the shear force.

R. Rocaberte || MECH 313 © 1st Sem 2013-2014

No need to discuss derivation. Dati naman hindi na.

R

Beams are classified according to their supports.

Fixed Beam

Simply Supported Beam


. .

P

P


R

Beams are classified according to their supports.

Cantilever Beam

Overhanging Beam


. .

ω

R M

ω P

R1 R2


R

Loading Principles:

Uniformly Distributed Load Effective Resultant &

Moment Arm


ω

L

R = ωL

L/2


R

Loading Principles:

Uniformly Varying Load Effective Resultant &

Moment Arm


ω

L

R = ½ ωL

2/3 L


R

Loading Principles:

Combination of Distributed Effective Resultant &

and Uniformly Varying Loads Moment Arm


ω2

L

R2 = ½ (ω2-ω1)L

2/3 L

ω1

ω2 - ω1

L/2

R1 = ω1L


R

Shear and Moment Diagrams

- Analytical tools used in conjunction with structural analysis to help perform structural design by determining the value of shear force and bending moment at a given point of an element

- Convenient visual references to the internal forces in a beam, in particular, they identify the maximum values of τ and M.


Read also the slides on Beams by M. S. Sivakumar from Indian Institute of Technology Madras as

supplement


R

Shear and Moment Diagrams

Sign conventions


External Loads

Shear Force

Bending Moment

Positive Negative

ω P M

ω P

M

V

V

V

V

M M M M


55. Which of the following support or connection has two unknowns?

A. Roller Support C. Short Link

B. Rocker Support D. Hinge Support

56. A simple beam, 10 m long carries a concentrated load of 500 kN at the midspan. What is the maximum moment of the beam in kN-m?

A. 250

B. 500

C. 1000

D. 1250

57. Determine the maximum moment of the beam loaded as shown in kN-m.

A. 125

B. 175

C. 875

D. 1250

5m 10m

200 kN

R1

R2

5m

100 kN

58. It occurs when a member carries a load perpendicular to its long axis while being supported in a stable manner.

A. Direct shear C. Torsional shear

B. Bending stress D. Torsional stress

Bending Stress a.k.a Moment Stress

RADIUS OF GYRATION

59. What is the radius of gyration of a 5-inch square with respect to its centroidal axis?

A. 1.44 in C. 10.39 in

B. 0.866 in D. 0.5 in square

ak =

12

4

2

square

aI 12k = =A a

5k = = 1.44 in

12

SECTION MODULUS

60. What is the section modulus of a 5-inch square with respect to its centroidal axis?

A. 20.83 cm3 C. 25.7 cm3

B. 49.5 cm3 D. 57.2 cm3

4

squaresquare

3

square

33

square

Ise mo =

y

aI 12z = =ay

2

az =

6

5z = = 20.83 in

6

61. What is the formula of the radius of gyration, k, of a circular cross-section with diameter D?

A. B.

C. D.

1k D

18

1k D

12

1k D

4

1k D

12

62. What is the formula of the section modulus, z, of a rectangle?

A. B.

C. D.

3bhZ

6

2bhZ

36

3bhZ

36

2bhZ

6

63. What is the moment of inertia of a cylinder of radius 5 m and mass of 5 kg?

A. 62.5 kg-m2 B. 80 kg-m2

C. 72.5 kg-m2 D. 120 kg-m2

- oldest branch of Physics.

- study of the bodies and systems and the forces acting on them.

1. Which of the following refers to the branch of mechanics dealing with the motion of bodies?

A. Kinematics

B. Kinetics

C. Dynamics

D. Statics

Force

• The scientific definition for force is simply a push or a pull. For example, when you do homework you exert a force on your pen or pencil because you push and pull it across the paper.

Net Force

• The sum of the forces is called the net force. In this case the net force is an unbalanced force. An unbalanced force is a force that changes an object’s motion or causes it to accelerate. The arrows show different forces and their direction, the wider the arrow the stronger the force.

Force and Force Systems

• Collinear Forces are forces that act on the same line of action

• Parallel Forces are forces that are of the same angle to one another. Couple is a pair of parallel forces of the same magnitude but opposite in direction.

• Frictional Force is a force that always acts in opposite direction to the applied force.

Force and Force Systems

• Coplanar Forces are forces lying on the same plane.

• Concurrent Forces are forces that meet in one common point.

• Non-concurrent Forces are forces that do not meet in one common point.

A quantity that has magnitude (and unit) only.

E.g. speed, distance, volume, current & etc.

A quantity that has magnitude, (unit) and direction. E.g. force gravity, displacement, acceleration, momentum, etc.

SCALARS VECTORS

Mass A load has a mass of 5kg.

Weight / Force A force of 15 N acts on a body in an upward direction.

Distance The train has traveled a distance of 80 km.

Displacement An airplane flies a distance of 100km in an easternly direction.

Speed Velocity A car moves 60km/hr, 35⁰ east of north.

Time The car has reached its destination after 1 hr.

Acceleration

Addition: A + B = B + A Subtraction: A – B Multiplication: Dot Product: A · B Cross Product: A x B

The sum of two or more vectors is represented by a single vector called resultant.

This resultant vector may be found by using:

– Graphical Method

– Pythagorean Theorem

– Component method

Rowel Allan Rocaberte, REE, ECE [email protected]

+63926 – 740 – 1530

MECH 312 RROCABERTE © 2013

CABLES

Used in numerous engineering applications (suspension bridges, power transmission lines, cable supporting heavy trolleys or telephone lines)

Incapable of developing internal forces other than TENSION

TWO TYPES:

1. Parabolic

2. Catenary

Front slide photo taken from http://catenary-project.wikispaces.com/Golden+Gate+Bridge


PARABOLIC CABLES: Symmetric

- Loading is distributed uniformly along the horizontal

L = span; distance between supports (m, ft)

d = sag ; maximum vertical displacement (m, ft)

S = total length of the cable (m, ft) Photo taken from http://www.solutioninn.com/engineering/civil-engineering/statics/the-cable-is-subjected-to-the-parabolic-loading-w-w0-18722-2xa

L

d S

5

6

3

42

7

256

5

32

3

8

L

d

L

d

L

dLS

ω



ω = load per horizontal length (N/m, lb/ft)

T = tension at the supports (N, lb)

H = tension at the lowest point (N, lb)

S/2

T

H

Photo taken from http://www.solutioninn.com/engineering/civil-engineering/statics/the-cable-is-subjected-to-the-parabolic-loading-w-w0-18722-2xa

L/2

2

2

2

1

LωHT

ω

2

Lω

FBD:

T

H

2

Lω



Photo taken from http://www.solutioninn.com/engineering/civil-engineering/statics/the-cable-is-subjected-to-the-parabolic-loading-w-w0-18722-2xa

d

LωH

8

2

042

)(0

dHLLω

cwM AT

H

L/2

2

Lω

d


S/2

PARABOLIC CABLES: Unsymmetric

- x and y axes are placed on the lowest point

- analyzed like two separate symmetric cables

Photo taken from http://www.mathalino.com/forum/strength-materials/parabolic-cable

O

L

dA

dB

2

LA2

LB


2. A pipeline crossing a river is suspended from a steel cable stretched between two posts 100 m apart. The weight of the pipe is 14 kg/m while the cable weighs 1 kg/m assumed to be uniformly distributed horizontally. If the allowed sag is 2 m, determine the tension of the cable at the post.

A. 9047.28 kg B. 9404.95 kg

C. 9545.88 kg D. 9245.37 kg

L = 100m; w = 15kg/m; d = 2m H = wL2/8d H = (15)(100)2 / 8(2) = 9375 kg T2 = H2 + [(1/2)wL]2

T = sqrt [93752 + (1/2 x 15 x 100)2 ] = 9404.95 kg

3. A certain cable is suspended between two supports at the same elevation and 500 ft apart. The load is 500 lbs per horizontal foot including the weight of the cable. The sag of the cable is 30 ft. Calculate the total length of the cable.

A. 503.21 ft B. 504.76 ft

C. 505.12 ft D. 506.03 ft

The Catenary Curve MECH 312 RROCABERTE © 2013

Gateway Arch in St. Louis, Missouri [1]

Spider’s web [2]

Hanging chain [3]

Lace [5]

Freely-hanging transmission lines [4]

a

xay cosh

Dance Step Exercise:

Hawi – Sine (wave) – Wax ng Hair)

Galileo claimed that the curve of a chain hanging under gravity should be a parabola. In 1669, his claim was proved to be wrong.

Galileo

• called the father of modern science precisely because he initiated the comparison between theory and experiment

“Big Bang is the Day without Yesterday” - Msgr. George Lemaitre

• Lemaitre told Einstein, “Your mathematics is correct, but your physics is abominable!”

- Loading is distributed uniformly along the length of the cable

L = span (m, ft)

y = sag(m, ft)

S = total length of the cable (m, ft)

CATENARY CABLES: Symmetric

Photo taken from http://lecture.civilengineeringx.com/structural-analysis/structural-steel/suspension-bridge-analysis/

L

y

S

T T

ω


A B

SAO = length of the cable from A to O (m, ft)

x = horizontal span from A to O (m, ft)

H = tension at the lowest point (N, lb)

ω = weight per unit length of the cable (N/m, lb/ft)



y SAO

T

ω


A

x

H O

H

xω

ω

HSAO sinh

1cosh

H

xω

ω

Hy

T = tension at the supports (N, lb)

Recall equation of a catenary



T


x

H

22

AOSωHT

AOSω

FBD:

T

H

AOSω

yωHT

a

xay cosh H

xωHT cosh

y

Note:

For symmetric catenary cables,

If , you may solve catenary cable problem as

a parabolic cable.

(A small sag-to-span ratio means that the cable is tight, and the uniform distribution of weight along the cable is not very different from the same load intensity distributed uniformly along the horizontal.)

CATENARY CABLES: Symmetric MECH 312 RROCABERTE © 2013

Ly10

1

AOSS

xL

2

2

Similar to parabolic (unsymmetric):

- x and y axes are placed on the lowest point

- analyzed like two separate symmetric cables

CATENARY CABLES: Unsymmetric


L

TB

TA


A

B

x

y

O SAO

SOB

4. A cable which has a mass of 0.6 kg/m and is 240 m long is to be suspended with a sag of 24 m. Determine the tension at midlength, maximum tension and maximum span.

A. 233.55 m

B. 240 m

C. 265.216 m

D. 190.98 m


22

AOSωHT

yωHT

H = 1696.39 T = 1837.75

H

xωHT cosh

x = 116.78 L = 2x = 233.55 m

MECH 312 UNIT 1


+63926 – 740 – 1530

Speed

The average speed of an object is equal to the total distance traveled in a given unit of time. It is a scalar quantity because it has no direction .

t

dvs

Speed and velocity are often used interchangeably in ordinary conversations. In physics, there is a clear distinction between these two concepts.

Velocity We define velocity as the time rate of change of

position.

The velocity of an object moving along a straight path is equal to the slope of the d-against-t graph. When the graph is a straight line, the velocity is constant.

Velocity The average velocity of an object is the total

displacement divided by the elapsed time.

t

dvave

Velocity

Instantaneous velocity is the velocity of the object at a particular instant. Its magnitude is equal to the slope of the line tangent to the point corresponding to the given time t.

iedt

dsv ˆ

vectorgentunitei tanˆ

Acceleration We define acceleration as the time rate of

change of velocity.

average instantaneous

acceleration acceleration

t

vaave

dt

vda

Rectilinear Translation Rectilinear Translation

A type of motion in which a body moves in a straight line or is moving in the direction parallel to its displacement.

Uniform Motion

A motion with constant speed or velocity [a = 0]

Uniformly Accelerated Motion

A motion with constant change in velocity or of uniform acceleration [a is (+) is accelerating or speeding up; a is (-) if decelerating or slowing down]

D1.

Kinematic Differential Equations of Motion

s = displacement

v = velocity

a = acceleration

t = time

dt

dsv )1(

dt

dva )2(

dvvdsa )3(a

dv

v

ds

andfrom

),2()1(

Rectilinear Translation For uniformly accelerated motion along a

straight horizontal path:

atvv

asvv

attvs

o

o

o

2

2

1

22

2

s = displacement vo = initial velocity v = final velocity a = acceleration t = elapsed time Uniform velocity: a = 0, v = vo

Starting from rest: vo = 0 Stop at a point: v = 0

5. From the speed of 100 kph, a car decelerates at the rate of 15 m/min/sec along a straight road. Which of the following gives the distance travelled at the end of 40 sec.

A. 3800 m

B. 911.112 m

C. 91.111 m

D. 455.56 m

V0 = 100kph x 1000

/3600 = 27.7778 m/s

a = -15/60 mps2 = -0.25

mps2

s = 27.7778(40) + 0.5(-

0.25)(40)2 = 911.112 m

Practice Problem: An airplane lands on a carrier deck at 200 mi/h and is brought to a stop uniformly, by an arresting device, in 50 ft. Find the time required to stop.

• A. 0.34 sec C. 0.46 sec

• B. 0.21 sec D. 0.86 sec

There’s no other way but up!

0.34 sec

Practice Problem: A sports car starting from rest can attain a speed of 60 mi/hr in 8 seconds. A runner can do a 100-yard dash in 9.8 seconds. Assume that the runner is moving with uniform speed and that the car starts at the instant he passes it. How far will both travel until the car overtakes the runner?

• A. 180.3 ft C. 200.3 ft

• B. 190.3 ft D. 170.3 ft


170.3 ft

Practice Problem: What average net thrust must a 17-ton airplane have in order to reach an altitude of 5,000 ft and a speed of 600 mi/hr at an airline distance of 10 mi from its starting point?

• A. 11,000 lb C. 14,000 lb

• B. 13,000 lb D. 12,000 lb


11,000 lb

6. An airplane acquires a take-off velocity of 150 mph on a 2-mile runway. If the plane started from rest and the acceleration remains constant, what is the time required to reach take-off speed?

A. 40 s B. 45 s

C. 58 s D. 96 s

MECH 312 UNIT 1


+63926 – 740 – 1530

Free Falling Body - straight line free fall

- motion in vertical direction

ga

For free fall: where g = acceleration due to gravity = 9.81 m/s2 = 32.2 ft/s2

Free Falling Body Same formulas for rectilinear motion

gtvv

gyvv

gttvy

o

o

o

2

2

1

22

2

Free Falling Body

from A to B: y (+) v (+) g (-) from B to C: y (+) v (-) g (-) from C to D: y (-) v (-) g (-)

B (highest point)

A C (reference point)

D

Free Falling Body

y (+) above the reference point

y (-) below the reference point

v (+) upward motion

v (-) downward motion

g (-) always negative

B (highest point)

A

D

C (reference point)

7. A stone is thrown down from the top of the cliff 150 m high with an initial speed of 30 m/sec. How long will it take to reach the bottom?

A. 3 sec B. 2.5 sec

C. 4.56 sec D. 3.26 sec

8. A ball is thrown vertically upward from the ground and a student gazing out of a window sees it moving upward pass him at 5 m/s. The window is 10 m above the ground. How high does the ball go above the ground?

A. 1.276 m

B. 11. 276 m

C. 5.276 m

D. 15.276 m


10

222

hH

ghvv o

Practice Problem: A stone is dropped down a well and 5 seconds later the sound of the splash is heard. If the velocity of sound is 1120 ft/s, what is the depth of the well? Ans. 353.31 ft


MECH 312 UNIT 1


+63926 – 740 – 1530

Projectile Motions

• A projectile is an object or body thrown with an initial velocity and whose motion is influenced by the pull of gravity

Projectile Motions

• Projectile motion is the motion of a body (the projectile) with a constant acceleration.

• It is principally determined by two types of motion: vertical and horizontal motions.

• Trajectory is the path of a projectile. Projectiles follow a parabolic path.

222

2

cos2

cos

cos2

1

θvxavv

θvtavv

tθvattvx

oxoxx

oxoxx

oox

Flight of Projectile

θ

xmax

vo

voy

vox

y = 0 y = 0

vy = 0

ymax

gyθvyavv

gtθvtavv

gttθvtatvy

oyoyy

oyoyy

oyoy

2sin2

sin

2

1sin

2

1

222

22

θvv

θvv

ga

a

motionprojectileFor

ooy

oox

y

x

sin

cos

0

,

Range of a Projectile

g

θvR

g

θθv

g

θvθv

tθvtvxR

g

θvt

gttθv

yRangeforgttvy

o

ooo

oox

o

o

o

2sin

cossin2sin2cos

cos

sin2

2

1sin0

0,;2

1

2

2

max

2

2

Height of a Projectile

g

θvH

gHθv

vHeighttheat

gHvv

o

o

fy

oyfy

2

sin

2sin0

0,max

2

22

22

22

If final position is not the same as the original level, we have values for both x and y

θv

gxθxy

θv

gx

θ

θxy

θv

xg

θv

xθvy

θv

xtlet

gttθvy

tθvx

o

o

oo

o

o

o

o

22

2

22

2

2

2

cos2tan

cos2

1

cos

sin

cos2

1

cossin

cos

2

1sin

cos

Guide Question

9. A shot is fired at an angle of 45˚ with the horizontal and a velocity of 300 fps. Calculate the range of the projectile.

A. 3500 yd

B. 2800 yd

C. 1471 yd

D. 932 yd

Guide Question

10. A projectile leaves at a velocity of 50 m/s at an angle of 30˚ with the horizontal. Find the maximum height that it could reach in meter.

A. 41.26

B. 28.46

C. 31.86

D. 51.26

11. A projectile is fired from a cliff 300 m high with an initial velocity of 400 m/s. If the firing angle is 30° from the horizontal, compute the horizontal range of the projectile.

A. 15.74 km B. 14.54 km

C. 12.31 km D. 20.43 km

y=xtanθ-gx2/2(vocosθ)2

-300=(tan30)x-9.81/2(400cos30)2 x2

x=14.63km

at = r α

at = r ω2

12. The normal acceleration of a particle on the rim of a pulley 10 ft in diameter is constant at 1200 fps2. Which of the following gives the speed of the pulley in rpm.

a. 77.4597 b. 15.4919

c. 147.9371 d. 73.9686

an = v2/r, 1200 = v2/5,

v = 77.4597 fps

v = rω, 77.4597 = 5ω, ω

= 15.4919

rad/sec*60/2Π =

147.9371 rpm

13. A discus thrower moves the discus in a circle of radius 80.0 cm. At a certain instant, the thrower is spinning at an angular speed of 10.0 rad/s and the angular speed is increasing at 50.0 rad/s2. At this instant, find the tangential and centripetal components of the acceleration of the discus and the magnitude of the acceleration. a. 89.1 m/s2 b. 89.2 m/s2 c. 89.3 m/s2 d. 89.4 m/s2

2 2

tan

2 2 2

2 2 2

tan

(0.800 )(50.0 / ) 40.0 /

(10.0 / ) (0.800 ) 80.0 /

89.4 /

rad

rad

a r m rad s m s

a r rad s m m s

a a a m s

MECH 312 UNIT 2


+63926 – 740 – 1530

Newton’s Laws of Motion

- The basis for extending the kinetics of a particle to a body composed of a system of particles.

“particle” – denotes an object of point size

“body” – denotes a system of particles which form an object of appreciable size

Any rigid body, regardless of its size, may be considered to be a particle if all of its parts move in identical parallel paths.


Newton’s Laws of Motion

First Law (Law of Inertia)

A body acted on by no net force moves with constant velocity (which may be zero) and zero acceleration.

Second Law (Law of Acceleration)

If a net external force acts on a body, the body accelerates. The direction of acceleration is the same as the direction of the net force.

Third Law (Law of Interaction)

If body A exerts a force on body B, then body B exerts a force on body A. These two forces are equal in magnitude but are opposite in direction; a.k.a. “Law of Action and Reaction”


aFFne t m

Newton’s Laws of Motion for a Particle

Inertia is the resistance a body offers to a change in its motion.

The reference frame or set of axes in which Newton’s laws are valid is one having a fixed origin and fixed directions of the axes. It is called an inertial, Newtonian, Galilean frame of reference.

Absolute motion - motion with respect to an inertial frame.


Every particle in the universe attracts every other particle with a force which is directly

proportional to the product of the masses of two particles and inversely proportional to the square of the distance between the centers of

the masses.

Newton’s Law of Universal Gravitation

Fg = gravitational force

G = 6.673 x 10-11 m3/kg-s2 (CONST 39)*

m1 & m2 = masses of the particles (kg)

r = center-to-center distance (m)

1 2

2g

m mF G

r

Gravitational Force

*Do not confuse G with acceleration g due to gravity.

Effective Force on a Particle

The effective force on a particle is defined as the resultant force on a particle, e.g. R or ma.

Inertia force the reaction caused by a resultant force

force numerically equal to ma but directed oppositely to the acceleration

Dynamic equilibrium: when inertia force is considered to act on a particle together with the resultant force, the particle is in a state of equilibrium.


D’Alembert’s Principle

The resultant of the external forces applied to a body (rigid or nonrigid) composed of a system of particles is equal to the vector summation of the of the effective forces acting on all particles.

The impressed forces acting on any body are in dynamic equilibrium with the inertia forces of the particles of the body.


P2

P1

W

332211

21

amamamR

PPWR

Newton’s Laws of Motion for a Particle

Newton’s second Law for any particle may now be expressed as

If the effective forces miai are reversed and considered to act on each respective particle of the system that the system will be in a state of balance known as dynamic equilibrium.


n

i

ii

n

i

i m11

aF

Guide Question

14. Riders in a bus are pushed forward during a sudden stop. Which law of motion provides an explanation?

A. Law of Inertia

B. Law of Interaction

C. Law of Universal Gravitation

D. Hooke’s Law

Guide Question

15. The force required to maintain an object at a constant speed in free space is equal to ____.

A. The weight of the object

B. The mass of the object

C. Zero

D. The force required to stop it

Guide Question

16. A rock is dropped out of the window of a moving car. At the same time a ball is dropped from the rest of the same height. Neglecting air resistance, which will reach the ground first?

A. Rock will hit the ground first.

B. Ball will hit the ground first.

C. Both will hit at the same time.

D. Neither will hit the ground.

MECH 312 UNIT 3


+63926 – 740 – 1530

Translation: Analysis as a Particle

The fundamental equation of kinetics for a particle


aaR

FR

g

Wm

0

:

yy

xx

aF

aF

g

W

g

W

motionrrectilineaFor

tt

nn

aF

aF

g

W

r

v

g

W

g

W

motionrcurvilineaFor

2

:

17. Two blocks A and B are released from rest on a 30ᴼ incline when they are 60 ft apart. The coefficient of friction under the upper block A is 0.20 and that under the lower block B is 0.40. Compute the acceleration of block A as it moves down the incline. A. 11.52 m/s2 B. 12.52 m/s2m C. 14.52 m/s2 D. 10.52 m/s2


μA = 0.2

μB = 0.4 30o

Practice Problem: Two blocks A and B are released from rest on a 30ᴼ incline when they are 60 ft apart. The coefficient of friction under the upper block A is 0.20 and that under the lower block B is 0.40. Compute the acceleration of block B as it moves down the incline. Ans. 4.946 m/s2


μA = 0.2

μB = 0.4 30o

Practice Problem: Two blocks A and B are released from rest on a 30ᴼ incline when they are 60 ft apart. The coefficient of friction under the upper block A is 0.20 and that under the lower block B is 0.40. Compute the elapsed time until the blocks touch. Ans. 4.56 sec


μA = 0.2

μB = 0.4 30o

Centripetal Force

A force that makes a body follow a curved path: it is always directed orthogonal to the velocity of the body.

A force which keeps a body moving with a uniform speed along a circular path and is directed along the radius towards the center.

from Latin centrum meaning "center" and petere, meaning "to seek"

Centrifugal Force

Represents the effects of inertia that arise in connection with rotation and which are experienced as an outward force away from the center of rotation.

from Latin centrum, meaning "center", and fugere, meaning "to flee"

Centrifugal / Centripetal Force

FC = centrifugal/centripetal force

m = mass

v = velocity

r = radius

2

C

mVF

r

Guide Question

18. A 50,000 N car traveling with a speed of 150 km/hr rounds a curve whose radius is 150 m. Find the centrifugal force.

A. 65 kN

B. 38 kN

C. 70 kN

D. 59 kN

Banking of Highway Curves (Case I)

Car is rotating (circular) radius

of curvature r with velocity v.


N

W

Nμr

v

g

W

2

gr

vμ

2

W v2 g r

Fc=

f=μN

N

W

NW

Fc f

r

Banking of Highway Curves (Case II)

When v = vrated = 0 (car is not moving), upward frictional force is introduced to prevent skidding.


θ N

W

N W θ

N

Nμ

N

fθ tan

θμ tan

f=μN

Banking of Highway Curves (Case III)

When v = vrated, no tendency to slip up or down the road (ideal banking of curve)


θ N

W

W v2 g r

Fc=

N W

Fc

θ

W

r

v

g

W

W

Fθ c

2

tan

gr

vθ

2

tan θ = ideal angle of banking

v = rated speed of the curve

Banking of Highway Curves (Case IV)

When v > vrated, downward frictional force f is introduced to prevent skidding.


θ R

W W v2 g r

Fc=

R W

Fc

W

r

v

g

W

W

Fθ c

2

tan Φ

N

θ φ

N

f

θ+φ

gr

vθ

2

tan Φ

Φtan

..

μ

fcμ

Banking of Highway Curves (Case V)

When v < vrated, upward frictional force f is introduced to prevent skidding.


θ

R

W W v2 g r

Fc= R W

Fc

W

r

v

g

W

W

Fθ c

2

tan Φ

N θ

φ

N

f θ-φ

gr

vθ

2

tan Φ

Φtan

..

μ

fcμ

19. A highway curve has a super elevation of 7 degrees. What is the radius of curvature of the curve such that there will be no lateral pressure between the tires and the roadway at a speed of 40 mph? A. 670 ft B. 770 ft C. 870 ft D. 970 ft


Example 5 20. The rated speed of a highway of 200 ft radius is 30 mph. If the coefficient of kinetic friction between the tires and the road is 0.60, what is the maximum speed at which a car can round the curve without skidding? A. 57.35 mph B. 102.34 mph C. 78.33 mph D. 23.83 mph


21. Find the angle of banking for a highway curve of 300 ft radius designed to accommodate cars travelling at 100 mph, if the coefficient of friction between the tires and the road is 0.6. What is the rated speed of the curve? A. 51.08 mph B. 45.334 mph C. 67.21 mph D. 55.9 mph


Test Yourself 1. vrated is used in what particular formula? 2. Formula for ideal banking of curves 3. Formula used when there is no frictional force

between the tires and the ground (no lateral pressure, no side thrust, etc.)

4. When a car is moving at its maximum, what is the formula?

5. When you consider the coefficient of friction between the tires and the road, what angle can we get? θ or Ф?

6. I want to know the coefficient of friction between my car’s tires and the ground. I parked my car on an inclined surface, θ from the horizontal. How do I determine for μ?


Answers to Test Yourself 1.tan θ = v2/gr 2.tan θ = v2/gr 3.tan θ = v2/gr 4.tan (θ + Ф) = v2/gr 5.Ф if car is moving not equal to the rated speed θ if car is not moving at all 6. μ = tan θ


Test Yourself Practice Problem: Determine the angle of super elevation for a highway curve of 600 ft radius so that there will be no side thrust for a speed of 45 mph. Ans. 12. 71°


UNIT 4


+63926 – 740 – 1530

MECH 321: DYNAMICS OF RIGID BODIES (Summer 2013)

Angular Motion. Fixed Axis Rotation.

Fixed-axis rotation is defined as that motion of a rigid body in which the particles move in circular paths with their centers on a fixed straight line that is called the axis of rotation.


r

r

ωo

ωf

α θ = angular displacement, rad

ω0 = initial angular velocity, rad/s

ωf = final angular velocity, rad/s

α = angular acceleration, rad/s2

Uniform velocity:

α = 0, ωf = ω0

Starting from rest:

ω0 = 0

Stop at a point:

ωf = 0

θ

2

2

dt

θd

dt

ωdα

dt

θdω

Angular Motion. Fixed Axis Rotation.

Kinematic Differential Equations of Rotation


r ωo

ωf

α

θ

αra

dt

ωdr

dt

dv

ωrv

dt

θdr

st

ds

θrs

a vo

vf

s

2

22

ωra

r

ωr

r

va

αra

dt

dva

N

N

T

T

tαωω

αθωω

tαtωθ

of

of

o

2

2

1

22

2

Rotation with constant angular acceleration

For uniformly accelerated motion along a circular path:

atvv

asvv

attvs

of

of

o

2

2

1

22

2

αra

ωrv

θrs

Rectilinear Motion Rotation (Related by)


22. A flywheel of radius 14 inches is rotating at the rate of 1000 rpm. How fast does a point on the rim travel in ft/sec? A. 122.17 B. 1466.04 C. 100 D. 39.05


v=rω v=14in(1000rev/min)(1ft/12in)(2pi/1rev)(1min/60sec) v=122.17ft/s

23. When the angular velocity of a 4-ft diameter pulley is 3 rad/s, the total acceleration of a point on its rim is 30 fps2. Determine the angular acceleration of the pulley at this instant?

A. 9 rad/s2

B. 12 rad/s2

C. 15 rad/s2

D. 14 rad/s2


2

222

2

/12 sradα

αra

aaa

ωra

T

nt

N

24.. A flywheel rotating at 200 rev/min slows down at a constant rate of 2 rad/s2. How many revolutions does it make in the process?

A. 10.47 B. 17.5

C. 109.7 D. 62.83


Work-Energy Theorem

Rowel Allan Rocaberte, REE, ECE

[email protected]

+63926 – 740 – 1530

Work – energy required to move an object by a distance

Potential Energy – energy possessed by an object by virtue of its motion (energy at rest), PE = mgh

Kinetic Energy – energy possessed by an object by virtue of its motion (energy in motion), KE = ½ mv2

Power – rate at which energy/work is transferred or consumed

Recall:

Fvdt

dsF

time

WorkP

Work – product of the component of force in a direction of displacement

Recall:

θ θ

P

θPsWork cos

+W if direction of W is in the direction of displacement -W if direction of W is opposite the direction of displacement

• Relates the force, displacement and velocity in a given system

• Removes the analysis of internal forces like tension

Work-Energy Method

Principle: The work done in a translating body is equal to the

sum of the change in kinetic energy of the body.

ifnet KEKEKEW

Derivation of Formula

ds

dvva

dvvdsa

ag

WF

KEsF

vvmsF o

22

2

1

22

00

2o

v

v

s

vvg

WsF

dvvg

WdsF

ds

dvv

g

WF

resultant work

change in kinetic energy

Practice Problem: Suppose a 30.0-kg package on the roller belt conveyor system is moving at 0.500 m/s. What is its kinetic energy?

Ans. 3.5 J

KE = 1/2mv 2 KE=0.5(30.0 kg)(0.500 m/s) 2

KE=3.75 kg⋅m2/s 2 =3.75 J.

Practice Problem: Suppose that you push on the 30.0-kg package with a constant force of 120 N through a distance of 0.800 m, and that the opposing friction force averages 5.00 N. (a) Calculate the net work done on the package. (b) Solve the same problem as in part (a), this time by finding the work done by the applied force. Find the work done by the friction.

Ans. (a) 92.0 J; (b) 96.0 J; (c) -4.0 J

(b) The applied force does work: Wapp = Fappdcos(0º)=Fappd = (120 N)(0.800 m) = 96.0 J

(c) So the amounts of work done by gravity, by the normal force, by the applied force, and by friction are, respectively Wgr = 0 Wn = 0 Wapp = 96.0 J Wfr= −4.00 J The total work done: W total= Wgr + Wn + Wapp + Wfr = 92.0

Energy Stored in a Spring

F = axial force

x = elongation

k = spring constant

2

2

1

2

1kxEFxE springspring

25. A large coil spring with a spring constant k = 120 N/m is elongated, within its elastic range by 1 m. Compute the stored energy of the spring in N-m.

A. 60

B. 40

C. 20

D. 120

26. The combined mass of car and passengers travelling at 72 km/hr is 1500kg. Find the kinetic energy of this combined mass.

A. 300kJ

B. 330kJ

C. 305kJ

D. 310kJ

K=mv2 = (1500kg)(72km/hr)2 (1000m/km)2 2k (2)(1kg-m/N-sec2 )(3000 sec/hr)2 = 300,000 J or 300 kJ

27. A Foucault pendulum swings to 3.0 in above the ground at the highest points and is practically touching the ground at the lowest point. What is the maximum velocity of the pendulum?

A. 4 ft/s C. 5 ft/s

B. 2 ft/s D. 10 ft/s

lost gainedPE =KE

v = 2gh

UNIT 7


+63926 – 740 – 1530

MECH 321: DYNAMICS OF RIGID BODIES (1st Sem 2013-2014)

Front slide photo taken from http://origami-blog.origami-kids.com/paper-airplanes-glossary-of-terms.htm

Definition

Momentum

- Measure of the motion of a body

- Quantity of motion that an object has

- Product of mass and velocity

Impulse

- Produced when force is applied over time periods

- Product of force and the time


mvP

FtI

Impulse-Momentum Theorem

Impulse equals the change in momentum


12 vvmFt

dt

dv

g

WF

dt

dva

ag

WF

o

v

v

t

vvg

WtF

dvg

WdtF

dvg

WdtF

o

0

Law of Conservation of Momentum

Special Case (when P and KE are conserved) :

u2 – u1 relative velocity of separation after collision

v1 – v2 relative velocity of approach before collision


collisionaftercollisionbefore PP

1221 uuvv D6.

Coefficient of Restitution

The coefficient of restitution (e), or bounciness

of an object is a fractional value representing the

ratio of relative velocities after and before an impact.

An object with an e of 1 collides elastically, while an object with an e < 1 collides inelastically.


Coefficient of Restitution

Collision Remark

e = 1 Perfectly elastic P and KE is conserved

e < 1 Inelastic collision P is conserved; KE is not conserved

e = 0 Perfectly plastic P and KE is not conserved (e.g. colliding particles stick together)

21

12

vv

uue

21

12

vv

uue

1

2

tan

tan

e

o

r

h

he

Sign Convention:

Before Collision

After collision

Law of Conservation of Momentum

Loss in Kinetic Energy


100%

initial

initialfinal

loss

initialfinalloss

KE

KEKEKE

KEKEKE

m1 m2

m1 m2

v1 v2

u2 u1

+v -v

22112211 umumvmvm

28. If the coefficient of restitution is zero, the impact is __________.

A. partially plastic

B. perfectly plastic

C. perfectly elastic

D. partially elastic

Guide Question

29. A ball is dropped from a height of 20 m upon a stationary slab. If the coefficient of restitution is 0.40, how high will the ball rebound?

A. 3.2 m

B. 4.6 m

C. 5.2 m

D. 8.0 m

30. The 10 kg and 20 kg bodies are approaching each other with the velocities shown. If e = 0.60, what will be the velocity of each body directly after impact? (a) u1 = -13.33 m/s, u2 = 10 m/s (b) u1 = 6.0 m/s, u2 = 7.5 m/s (c) u1 = 10.25 m/s, u2 = 10 m/s (d) u1 = -13.33 m/s, u2 = 16.7 m/s


21

12

vv

uue

signstheofnote

umumvmvm 22112211

31. A 200-lb block in contact with the ground is acted upon by a horizontal force equal to 100 lb. The coefficient of kinetic friction is 0.20. In what time will the velocity of the block be increased from 4 fps to 10 fps?

A. 0.621 s

B. 5.1 sec

C. 1.23 sec

D. 8 ms


sec621.0

12

t

vvmtF

32. A gun is shot in a 0.50 kN block which is hanging from a rope of 1.8 m long. The weight of the bullet is equal to 5 N with a muzzle velocity of 320 m/s. How high will the block swing after it was hit by the bullet?

A. 0.51 m

B. 0.53 m

C. 0.32 m

D. 0.12 m

m1v1 + m2v2 = m1u1 + m2u2

m1v1 + m2v2 = mTu

500(0) + (5/9.81)(320) = (500+5)/9.81 x u u = 3.168 m/s PE = KE mgh = mu2/2 h = u2/2g h = 3.1682/(2x9.81) = 0.512 m

UNIT 8


+63926 – 740 – 1530

MECH 321: DYNAMICS OF RIGID BODIES (Summer 2013)

Frequency of SHM

Simple

pendulum

Conical

pendulum

L

h

Frequency of SHM

Helical

spring

in SHM

m

k = spring constant

m = mass

a= acceleration

x = displacement

General

formula

33. Calculate the period of a simple pendulum connected to a string of length 2 meters.

A. 2.1 sec C. 1.9 sec

B. 1.7 sec D. 2.8 sec

34. What is the frequency of a 0.6-kN/m helical spring in simple harmonic motion if a body of 3 kg is attached to its end.

A. 1.75 Hz

B. 2.0 Hz

C. 2.25 Hz

D. 1.9 Hz

Documents

2014 SEPT - MECHANICS REG REVIEW.pdf