Chap 3 Mechanics

8/10/2019 Chap 3 Mechanics

1/45

110Loading Next Page

GENERAL ENGINEERING & APPLIED SCIENCES

CHAPTER 3Engineering MechanicsGEASGEAS

ENGINEERINGMECHANICS

EngineeringMechanicsis a branch of the physical sciences that is concerned

with the state of rest or motion of bodies subjected to the action of forces.

ST TICS

is concerned with the equilibrium of a body that is either at rest or inmotion with constant velocity.

DYN MICSdeals with the accelerated motion of the body.

KINEM TICSis a branch of dynamics, which treats only the geometric aspect of

the motion, whileKINETICSis the analysis of the forces causing the motion.

STATICS

I. THE STATIC FORCE SYSTEMConcurrent Forces are forces whose lines of action all pass through a

common point.Coplanar Forces are forces lying in one plane.

Equilibrium is defined as the condition in which the resultant of allforces acting on a body is zero.

Torque (or moment) about an axis, due to a force, is a measure of theeffectiveness of the force in producing rotation aboutthat axis.

Weight of an object is the force with which gravity pullsdownward upon it.

Center of gravity of an object is the point at which the entire weight ofthe object is concentrated.

ENGINEERINGMECHANICS

STATICS DYNAMICS

KINETICS KINEMATICS


2/45

GENERAL ENGINEERING AND APPLIED SCIENCES

111

CHAPTER3ngineering Mechanics

Loading Next Page

II. RESULTANT OF TWO OR MORE COPLANAR, CONCURRENT FORCES Parallelogram Method: (for two coplanar, concurrent forces)

Cosine Law:

2 2 2

1 2 1 2R F F 2FF cos= +

Sine Law:

2F R

sin sin=

For more than 2 forces: (Use the Component Method)

( ) ( )

22

x y

y1

x

R F F

Ftan

F

= +

=

III. FORCES IN EQUILIBRIUM TYPES OF EQUILI RIUM

Static equilibrium is the condition of a body at rest and remains at rest

under the action of concurrent forces.

Translational equilibriumis the condition of a body in motion with

constant velocity.

THE TWO CONDITIONS FOR EQUILI RIUM

First or the Force ConditionThe vector sum of all forces acting on the body must be zero .

x yF 0 F 0= =

Second or the Torque Condition

The sum of all the torques acting on the body must be zero

0 =

2F

1F

R

1F

2F

3F


3/45

112


CHAPTER 3-MECHANICSGEASGEAS

Loading Next Page

CABLES

I. PARABOLIC CABLES

Taking summation of moment at A:

( )L

H d2 4

=

=

Thus,

Where:T= tension at the supportH= tension at the lowest pointd= sag = weight per unit lengthL= span or distance between supports

2

Tension at the lowest point:

LH

8d

=

2

2

Tension at the support:

LT H

2

= +

2 4

3

Approximate lenght of cable :

8d 32dS L

L 5L= +

d

L

L / 2 L / 2

(N/m)

T T

H

W

L / 2

L / 2

(N/m)

T T

L / 4 L / 4

W

H


4/45


113


Loading Next Page

II. CATENARY(For Symmetrical & Unsymmetrical Supports)

Where:

T= tension at the supportH= tension at the lowest point = weight per unit lengthy= height of the support

=c minimum clearance from the ground

1 2S & S are half lengths of the cable

L= span or distance between supports

d

L

x x

2T 1T

H

S S

2y1y

c

1

1 1

Tension at the support T :

T y=

2

2 2


T y=

( )

1

22

1 1


T H S= +

( ) ( )

2

2 2

2 2


T H S= +

1 11

Dis tance Between supports :S y

x c lnc

+ =

2 2

2

Distance Between supports :S y

x c lnc

+ =

( ) ( ) ( )2 2 2

1 1

Relationship among S,y & c :

S y c=

( ) ( ) ( )2 2 2

2 2

Relationship among S,y & c :

S y c=


5/45

114



Loading Next Page

FRICTION

FRICTION ON BLOCK

F N

Ftan

N

=

= =

MAXIMUM ANGLE OF INCLINE WITHOUT CAUSING THE BODY TO SLIDE DOWN:

1tan = =

Where, in&:

F frictional force

N normal force

P the applied force

R total surface reaction

coefficient of friction

angle of friction

angle of the incline

=

=

=

=

=

=

=

BELT FRICTION

1 1

2 2

T Te ln

T T

= =

Where:

= the coefficient of friction

= angle of contact in radians

1T = tension in the tight side

2T = tension in the slack side

N

W

P

fF N=

N

W

1T

2T


6/45


115


Loading Next Page

oV fV

s

DYNAMICS

I. RECTILINEAR MOTION - Motion in a Straight Line)

Uniform Motion - (constant speed / zero acceleration)

s vt=

Where:s= distancev= uniform speed or velocityt= time

Uniformly Accelerated Motion - (velocity increases uniformly)Equations of Motion:

2o1

s v t at2

=

f ov v at =

2 2f ov v 2as =

Where:

s= distance traveled or displacement

ov = original velocity ; ( ov 0, if from rest= )

fv = final velocity ; f(v 0,if to stop)=

a= acceleration ( 2 2m/ s or ft /s )

t= time, (seconds)

v v

o fv v v= =

s

ov fv

s

Use ( )+ if : f o(v v )>

Use ( ) if : f o(v v )<


7/45

116



Loading Next Page

II. FREE-FALLING BODY- Motion Under gravity)

Important Equations:

2o1

h v t gt2

=

f ov v gt =

2 2f ov v 2gh =

Where:

h= height

ov = original velocity

f

v = final velocity

g= acceleration due to gravity

2 2 2

m ft cm9.81 32.2 981

s s s= = =

t= time

Note:

Use ( )+ for g going down Use ( ) for g going up

Maximum Height Attained by a Body if projected straight upward:

( )2

omax

vh

2g=

Time taken to reach the highest point if projected vertically upward:

ovtg

=

Time taken to reach the ground if dropped from a height h:

2ht

g= ; ( )ov 0=


8/45


117


Loading Next Page

Time of flight in going to the maximum height and falling back to the point

where it was projected:

o2vTg

=

III. PROJECTILE MOTION

For problems involving projectile motion, resolve the initial velocity ( ov )

into two perpendicular components as follows:

Horizontal velocity : ox ov v cos=

Vertical velocity : oy ov v sin=

Then, apply formulas for kinematics in one dimension.

AT ANY TIME, t: Horizontal motion: ( xa 0= )

( )ox od v t v cos t= =

Vertical motion: ( y 2m

a g 9.81s

= = )

( )2 2oy o1 1

h v gt v sin t gt2 2

= =

xv

yv

oyv

oxv

ov v

R

Point ofimpact

Projectile @ maxh

maxh

d


9/45

118



Loading Next Page

TIME TAKEN TO REACH THE HIGEST POINT, t:

ov sintg

=

TIME OF FLIGHT, T:

The time of flight is the time taken by the projectile to return to theoriginal horizontal level.

o2v sinT 2tg

= =

RANGE, R:Range is the distance covered in the horizontal direction in the totaltime of its flight.

( )2o

o

v sin 2R v cos T

g

= =

AT MAXIMUM HEIGHT, maxh : ( yv 0= )

At the highest point, the vertical velocity of the projectile is zero (asin the case of a body thrown vertically upwards), its velocity is onlyin the horizontal direction.

( )2

o

max

v sin

h 2g

=

VELOCITY AT ANY POINT V:

The velocity of a projectile at any point of its path is given by theresultant of its vertical and horizontal velocities at that point.

( ) ( )22

x yv v v= +


10/45


119


Loading Next Page

GENERAL EQUATION OF PROJECTILE

( )

2

2

o

gdh dtan

2 v cos =

Note:

Use (+), if the point of impact is above the point of release.Use (-), if the point of impact is below the point of release.

CASE I- Point of Impact is Above the Point of Release

CASE 2 -Point of Impact is Below the Point of Release

y

y


11/45

120



Loading Next Page

IV. ROTARY MOTION

Important Notation:

= angular displacement, radians = angular velocity, rad/s = angular accelerationt= time

Uniform Angular Motion:

t =

Uniformly Accelerated Angular Motion:

2o

1t t

2 =

f o t =

( ) ( )22f o 2 =

Relationship Between Angular and Tangential Quantities:

s r

v r

a r

=

=

=

Provided, , , and are in radians.

v

rr

s

r

r


12/45


121


Loading Next Page

REFP

a

f

V. NEWTONS LAWS OF MOTION

First Law: The Law of Inertia

Newtons first law of motion states that if the body is at rest, it will remainat rest, if it is in motion, it will remain in motion with constant speed in astraight line unless there is a net force acting upon it.

netF 0=

Second Law: The Law of Acceleration

If a net force F acting on the body of mass m is not zero, the bodyaccelerates in the direction of the force. The acceleration a isproportional to the force and inversely proportional to the mass of the

body.F

a or F mam

= =

Third Law: The Law of Action and Reaction

To every action there is always an equal and opposite reaction.

DAlemberts Principle: (Jean le Rond dAlemberts)

The resultant of the external forces applied to a body (rigid or non-rigid)and reversed effective force (REF) is equal to zero.

Equations:

WREF P f 0 or REF a ma

g = = =

Where:

m mass W weight a acceleration= = =

Note:

Reversed Effective Force (REF) is always opposite in the direction ofacceleration.


13/45

122



Loading Next Page

Before Impact

1m 2m

2v1v

After Impact'

2v'

1v

1m 2m

VI. IMPULSE AND MOMENTUM

ImpulseImpulse is the product of force and the time it acts.

I Ft=

MomentumMomentum is the product of the mass and the velocity of the body.

P mv=

Impulse - Momentum Theorem

Impulse is equal to the change in momentum.

2 1Ft mv mv=

The Law of Conservation of MomentumWhen two bodies of masses m1and m2collide, the total momentumbefore impact is equal to the total momentum after impact.

Equation:

before impact after impact

' '

1 1 2 2 1 1 2 2

P P

m v m v m v m v

=

+ = +

v

m

F

t


14/45


123


Loading Next Page

Types of Collision

Collision refers to the mutual action of the molecules, atoms, andetc., whenever they encounter one another.1. Elastic colli sion is a collision which conserves kinetic energy2. Inelastic collisi on is a collision which does not conserve

energy.3. Perfectly inelastic collision is the collision which the object

sticks together afterward. In such collisions the KE loss ismaximum.

Coefficient of Restitution, e:

' '

2 1

1 2

v vRelative velocity of recession

e Re lative velocity of approach v v

= =

Where:

e 0= for perfectly inelastic collisione 1= for perfectly elastic collision

Special Cases:

If a ball is dropped from a height ho upon a floor and rebounds to aheight of hr, the coefficient of restitution between the ball andthe floor is:

r

o

he

h=

Where:

rh = height of rebound

oh = original height

If a ball is thrown at an angle 1with the normal to a smooth surfaceand rebounds at an angle 2:

1

2

tane

tan

=

oh

rh

1 2

Normalline


15/45

124



Loading Next Page

VII. UNIFORM CIRCULAR MOTION

Centripetal Force

22

C

mvF mr

r= =

Centripetal acceleration

22

C

va r

r= =

Note:

Centripetal force must be directed toward the center of the circularpath.

The Conical Pendulum

2 2

max

CF r vtan

W g gr

Tension in the cord:

WT

cos

Period, t :

ht 2

gfor t : L h

= = =

=

=

=

Where:

CF= centrifugal reactionT= tension of the cordr= radius of the circular pathh= heightL length of the cord=

r

h

T

CF

W

CF


16/45


125


Loading Next Page

VIII. BANKING OF CURVES

Ideal Angle of Banking

21 vtan

gr

=

For maximum velocity, v of the car withou t skidding

( )2v

tangr

+ =

Where:

= angle of banking = angle of frictionr= radius of the curvev velocity=

For Horizontal Rotating Flat form:

2 2r vtan

g gr

= = =

Also:

v r g=

Where:

= angle of frictionr= radiusv= velocityg= acceleration due to gravity

CF CF

r


17/45


18/45


127


Loading Next Page

TEST - 3

1. A collision in which the total Kinetic energy after collision is less thanbefore collision is called

A. off center collisionB. inelastic collisionC. straight line collisionD. elastic collision

2. Whenever a net force acts on a body, it produces acceleration in thedirection of the resultant force, an acceleration that is directlyproportional to the mass of the body. This theory is popularly known as

A. Newtons Second Law of MotionB. Newtons First Law of MotionC. Hookes Law of EquilibriumD. Faradays Law of Forces

3. To maximize the horizontal range of the projectile, which of the followingapplies?

A. Maximize velocityB. Maximize the angle elevation and velocityC. Maximize the angle of elevationD. The tangent function of the angle of trajectory must be equal to one

4. The moment of inertia of a plane figure, ____.

A. increases as distance of the axis moves farther from the centroidB. is maximum at the centroidal axisC. is zero at the centroidal axisD. decreases as the distance of the axis moves farther from the

centroid


19/45

128



Loading Next Page

5. A measure of a resistance of the body it offers to any change in itsangular velocity, determined by its mass and distribution of its massabout the axis rotation is known as ________.

A. moment of inertiaB. frictionC. torsionD. angular acceleration

6. Momentum is the product of mass and ______.

A. accelerationB. velocityC. forceD. time

7. Centrifugal force is __________ .

A. directly proportional to the radius of the curvatureB. directly proportional to the square of the tangential velocityC. inversely proportional to the square of the tangential velocityD. directly proportional to the square of the weight of the object

8. According to this law, The force between two charges varies directly asthe magnitude of each charge and inversely as the square of thedistance between them.

A. Law of Universal GravitationB. Coulombs LawC. Newtons LawD. Inverse Square Law

9. When the total kinetic energy of a system is the same as before and aftercollision of two bodies, it is called

A. Plastic collisionB. Inelastic collisionC. Elastic collisionD. Static collision


20/45


129


Loading Next Page

10. In a polar coordinate system, the length of the ray segment from a fixedorigin is known as ______.

A. amplitudeB. radius vectorC. hypotenuseD. minimum point

11. Momentum is a property related to the objects ______.

A. motion and massB. mass and accelerationC. motion and weight

D. weight and velocity

12. The study of motion without reference to the force that causes the motionis known as ______.

A. staticsB. dynamicsC. kineticsD. kinematics

13. Varignons theorem is used to determine ______.

A. location of centroidB. moment of inertiaC. mass moment of inertia

D. moment of area

14. The periodic oscillations either up or down or back and fourth motion inthe straight line is known as_______.

A. transverse harmonic motionB. resonanceC. rotational harmonic motionD. translational harmonic motion


21/45

130



Loading Next Page

15. A freely falling body is a body in rectilinear motion and withconstant_____.

A. velocityB. speedC. decelerationD. acceleration

16. When the total kinetic energy of the system is the same as before andafter the collision of two bodies, it is called

A. Static collisionB. Elastic collisionC. Inelastic collisionD. Plastic collision

17. What is the charge in the gravitational attraction between an orbiting

object and the earth if the distance between them is doubled?

A. no changeC. one halfB. doubleD. one fourth

18. With reference to the thermodynamic diagram of temperature entropy(TS), what is represented by the area under the diagram?

A. work doneB. enthalpyC. temperature differenceD. heat transferred

19. What is the standard acceleration due to gravitational force?

A. 32 ft/sec/secB. 980 ft/sec/secC. 32 m/sec/secD. 98 ft/sec/sec


22/45


131


Loading Next Page

20. Which of the following collisions is an elastic collision?

A. Two bodies move towards each other, collide and then move awayfrom each other. There is a rise in temperature

B. Two bodies collide and the sound of collision is heard by a blindman

C. Two steel balls collide such that their kinetic energy is conservedD. A man jumps on to a moving cart

21. A mass is revolving in a circle which is in the plane of paper. Thedirection of centripetal acceleration is along the radius:

A. away from the center radiusB. toward the centerC. at right angle to angular velocityD. none of the above

22. When a mass is rotating in a plane about a fixed point, its angularmomentum is directed along:

A. the radiusC. a line at an angle of 45o to the plane of the rotationB. the tangent to orbitD. the axis of rotation

23. A solid iron sphere A rolls down an inclined plane, while an identicalhollow sphere B slides down the plane in a frictionless manner. At the

bottom of the inclined plane, the total kinetic energy of sphere A is:

A. less than that of BC. more than that of BB. equal to that of BD. sometime more andsometimes less


23/45

132



Loading Next Page

24. Which of the following has the highest moment of inertia when each ofthem has the same mass and the same radius?

A. A hollow sphere about one of its diameters.B. A solid sphere about one of its diameters.C. A disc about its central axis perpendicular to the plane of the discD. All of the above have the same moment of inertia

25. When a planet moves around the sun,A. the angular momentum remains conservedB. the angular speed remains constantC. the linear velocity remains constantD. the linear momentum remains constant

26. What keeps an earth satellite moving on its orbit?

A. Gravitational attraction between satellite and earthB. Ejection gases from the exhaust of the satelliteC. burning of fuelD. Gravitational attraction of sun

27. The value of universal gravitational constant G depends upon:

A. nature of material of two bodiesB. heat content of two bodiesC. acceleration of two bodiesD. None of these

28. If the mass of an object could be doubled, then its inertia would be

A. halved

B. doubledC. unchangedD. quadrupled

29. It shows the forces acting on an isolated object.

A. force diagramB. schematic diagramC. free body diagramD. force polygon


24/45


133


Loading Next Page

30. A wagon is uniformly accelerating from rest. The net force acting on thewagon is

A. zeroB. increasingC. constantD. decreasing

31. If the mass of an object were doubled, its acceleration due to gravitywould be

A. doubled alsoB. unchangedC. halved

D. fivefold32. Which of the following is not a vector quantity?

A. forceB. energyC. weightD. velocity

33. Which of the following is not a scalar quantity?

A. timeB. workC. temperatureD. displacement

34. The resultant of two concurrent forces is minimum when the anglebetween them is

A. 0 degreeB. 90 degreesC. 45 degreesD. 180 degrees


25/45

134



Loading Next Page

35. As the angle between two concurrent forces decreases from 180

o

, theirresultant

A. decreasesB. increasesC. unchangedD. cannot be determined

36. The maximum number of components that a single force may beresolved into is

A. oneB. twoC. threeD. four

37. The momentum of an object is the product of its

A. mass and displacementB. mass and velocityC. force and displacementD. force and time

38. If the direction of an objects momentum is west, the direction of thevelocity of the object is

A. eastB. westC. northD. south

39. The direction of an objects momentum is always the same as thedirection of the objects

A. inertiaB. massC. weightD. velocity


26/45


135


Loading Next Page

40. The moment of inertia of a triangle with respect to the base how manytime its moment of inertia with respect to its centroidal axis

A. 1/2B. 3C. 1/4D. 5

41. When to objects collide, which of the following is always true?

A. the velocity of each object does not changeB. there is no change in the displacement of each objectC. there is no net change in the kinetic energy of each object

D. there is no net change in the total momentum of the objects

42. The study of motion with reference to the force that causes the motion is

A. ballisticsB. kinematicsC. kineticsD. dynamics

43. What is the moment of inertia of a circle of radius r?A. r

4/4

B. r4/12

C. r4/15

D. r4/16

44. The moment of inertia of a rectangle with respect to the base is howmany times its moment of inertia with respect to the centroidal axis?

A. 2B. 3C. 4D. 5


27/45

136



Loading Next Page

45. The moment of inertia of a triangle with respect to its base b is

A. b2h

2/12

B. bh3/6

C. bh3/12

D. bh3/3

46. The moment of inertia of a circle with respect to its tangent is how manytimes its moment of inertia with respect to its centroidal axis?

A. 2B. 3C. 4D. 5

47. Moment of inertia is also called

A. Moment of massB. Moment of centerC. second moment of areaD. moment of volume

48. One newton is equivalent to

A. kg-m/sB. kg m/m/sC. kg m/s/sD. m/s

2per kilogram

49. These are forces whose lines of action all pass through a common point.

A. collinear forcesB. coupleC. coplanar forcesD. concurrent forces

50. The radial distance from the axis to the point of application of the force iscalled

A. radius vectorB. lever armC. normalD. displacement


28/45


137


Loading Next Page

1. A mango falls from a branch 5 meters above the ground. With whatspeed in meters per second does it strike the ground? g=10m/s

2

Solution:

From:

( ) ( )

22

f ov v 2gh= +

Where:

ov 0 (free fall)=

Substitute:

( ) ( )f

f

v 2gh 2 10 5

v 10 m / s

= =

=

2. An automobile accelerates at a constant rate of 15mi/hr to 45 mi/hr in 15seconds while traveling in a straight line. What is the averageacceleration?

Solution:

From: f ov v

at

=

Where:

f

o

mi 5280 ft 1hr v 45 66 ft / s

hr mi 3600s

mi 5280 ft 1hr v 15 22 ft / s

hr mi 3600s

= =

= =

Then,

266 22a 2.93 ft / s15

= =

Solved Problems

In Mechanics


29/45

138



Loading Next Page

3. A 50-kilogram block of wood rests on the top of the smooth plane whoselength is 3m and whose altitude is 0.8 m. How long will it take for theblock to slide to the bottom of the plane when released?

Solution:

Given:

ov 0

m 50 kg

s 3 m

h 0.8 m

=

=

=

=

Solving for the acceleration of the box:

Take summation of forces along the incline equal to zero:

2

mgsin ma

a gsin

0.8a 9.81

3

a 2.62 m / s

=

=

=

=

Solving for time, t:

( )

2

o

2

1s v t at

2

13 0 2.62 t2

t 1.51 sec .

= +

= +

=

mgsin

N3 m 0.8 m

W

ov 0 ( from rest)=


30/45


139


Loading Next Page

4. A baseball is thrown from a horizontal plane following a parabolic pathwith an initial velocity of 100m/s at an angle of 30 above the horizontal.How far from the throwing point will the ball attain its original level?

Solution:

Given:

ov 100 m/ s

30

=

=

From:

( )

( ) ( )

2

o

2

v sin2R

g

100 sin 2 30R

9.81

R 883 m

=

=

=

5. A block weighing 500 KN rests on a ramp inclined at 25 degrees with thehorizontal. The force tending to move the block down the ramp is

Solution:

Given:

W 500 kN

25

=

=

Let:F= the force along the incline that tends

to move the block down the rampThen,

F W sin

F 500 sin 25

F 211 kN

=

=

=

W25

F W sin=


31/45

140



Loading Next Page

6. How far does an automobile move while its speed increases uniformlyfrom 15 kph to 45 kph in 20 sec?

Solution:

o

f

15km 1000m 1hr v 4.17 m/ s

hr km 3600s

45km 1000m 1hr v 12.5 m / s

hr km 3600s

= =

= =

From:

( )

f ov vS vt t2

4.17 12.5

S 202

S 166.7 m

+ = =

+

= =

7. A rotating wheel has a radius of 2 feet and 6 inches. A point on the rim ofthe wheel moves 30 feet in 2 seconds. Find the angular velocity of thewheel.

Solution:

Solving for the tangential velocity:

v s t= tangential velocity

30v 15 ft / s

2

= =

From:v r

15 2.5

6 rad/s

=

=

=

ov 15 kph=

S

fv 45 kph=


32/45


141


Loading Next Page

( )1 2m m+

cv

Before Impact

1m

2m

2v 1v

After Impact

8. A 16 gm mass is moving at 30 cm/sec., while a 4gm mass is moving inan opposite direction at 50 cm/sec. They collide head on and sticktogether. Their velocity after collision is

Solution:

Let:

cv = combined velocity after collision

From the Law of Conservation of Momentum:Momentum Before Impact = Momentum After Impact

( )

( ) ( ) ( )1 1 2 2 1 2 c

c

c

m v m v m m v

16 0.30 4 0.50 16 4 v

v 0.14 m / s

+ = +

+ = +

=

9. A ball is dropped from a building 100 m high. If the mass of the ball is 10gm, after what time will the ball strike the earth?

Solution:

( ) ( )

2

o

2

1h v t gt

2

1100 0 t 9.81 t

2

t 4.52 sec .

= +

= +

=


33/45

142



Loading Next Page

10. Determine the angle of super elevation for a 200 m highway curve sothat there will be no side thrust at a speed of 90 kph.

Solution:

From:2v

tan banking angle fromulagr

=

Substitute given and solve for :

[ ]( )

2

90km 1000m 1hr

hr km 3600stan

9.81m/ s 200m

17.67

=

=

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

Solution:

2mvCF

r=

Where:

km 1000m 1hr v 150

hr 1km 3600s

v 41.67 m/ s

=

=

Thus, solving for FC:

( ) ( )( ) ( )

250,000 41.67

CF9.81 150

CF 59 kN

=

=


34/45


143


Loading Next Page

12. A 40 gm rifle with a speed of 300 m/s is fired into a ballistic pendulum ofmass 5 kg suspended from a chord 1 m long. Compute the verticalheight through which the pendulum rises.

Solution:

Let:

bm = mass of the bullet ; Bm = mass of the block

cv = combined velocity of the block

and the bullet

c b Bm m m

combined mass of the block

and the bullet

= +

=

Using the Law of Conservation of Momentum:

( ) ( ) ( ) ( )b b B B c c

c

c

m v m v m v

0.04 300 5 0 0.4 5 v

v 2.38 m/ s

+ =

+ = +

=

From the law of conservation of energy:

( )

( ) ( )( )

bottom top

2

c c c

2 2

c

KE PE

1m v m gh

2

v 2.38h

2g 2 9.81

h 0.2887 m or 28.87 cm

=

=

= =

=


35/45

144



Loading Next Page

13. A projectile is thrown with a speed of 100 ft/sec in a direction 30

o

abovethe horizontal. Determine the highest point to which it rises.

Solution:

At maximum height, fyv 0= :

( ) ( )

( ) ( )

( )

2 2

fy oy max

2 2

oy o

max

2

max

max

v v 2gh

v v sinh

2g 2g

100sin30h

2(32.2)

h 38.81m

=

= =

=

=

14. A missile is fired with a speed of 100 fps in a direction 30 degrees abovethe horizontal. Determine the maximum height to which it rises?

Solution:

( )

( ) ( )( )

2 2

o

max

2 2

max

max

v sinh

2g

100 sin30h

2 32.2

h 38.8

39 ft.

=

=

=

15. A golf ball weighs 1.6 ounces. If its velocity immediately after being

driven is 225 fps, what was the impulse of the blow in slug-feet/sec?

Solution:

1lbW 1.6oz 0.1lb

16oz

= =

From:

( )

( )

f o

Impulse change in momentum

I m v v

0.1I 225 0

32

I 0.703 slugs ft / s

=

=

=

=


36/45


37/45

146



Loading Next Page

18. A car accelerates uniformly from standstill to 80 mi/hr in 5 seconds. Whatis its acceleration?

Solution:

o

f

f

v 0 (from rest)

5280 ft 1hr v 80mph

mi 3600s

v 117.33 ft / s

=

=

=

From:

f ov v at =

Substitute and solve for a:

2

117.33 0 a(5)

a 23.47 ft / s

=

=

19. A DC-9 jet with a takeoff mass of 120 tons has two engines producingaverage force of 80,000 N during takeoff. Determine the planesacceleration down the runway if the takeoff time is 10 seconds.

Solution:

Formula:

Fa

m=

Substitute:

( )

2

2 80,000a

120,000

a 1.33 m / s

=

=

20. What is the acceleration of a point on a rim of a flywheel 0.8 m indiameter turning at the rate of 1400 rad/min?

Solution:2v

ar

=

But:v r=

Substitute to equation :

( )( )

2 2

2 2r rad 1min

a r 0.4 1400 217.77 m / sr min 60s

= = = =

Note:

1 ton 1000kg=


38/45


39/45

148



Loading Next Page

23. A ball was thrown upward with an initial velocity of 50ft/s. How high doesit go?

Solution:

From:

( ) ( )2 2

f ov v 2gh =

Where:

fv 0 ; at max imum height=

Substitute:

( ) ( )2

0 50 2 32 h

h 39 ft

=

=

24. A projectile is fired upward with muzzle velocity of 200 m/sec at an angle

of 30 degrees with the level ground. With what velocity will it hit theground in m/sec?

Solution:

Solving for the time of flight to reach maxh :

ov sintg

200sin30t

9.81

t 10.19 sec

=

=

=

Considering free - fall from maxh :

fy oy oy

fy

fy

v v gt;(v 0)

v 9.81(10.19)

v 100

= =

=

=

Solving for fxv :

fx o

fx

fx

v v cos

v 200cos30

v 173.2

=

=

=

Thus,

( ) ( ) ( ) ( )2 2 2 2

fy fxv v v 100 173.2

v 200 m / s

= + = +

=

maxh

ov

fv 0=

fxv

fyv fv

30

t t

2t


40/45


149


Loading Next Page

25. A 40 kg block rests at the top of an inclined smooth plane whose lengthis 4 m and whose height is 0.5 m. How long will it take for the block toslide to the bottom of the plane when released?

Solution:

Given:

ov 0

m 40 kg

s 4 m

h 0.5 m

=

=

=

=

Solving for the acceleration of the box:

net

2

F ma

mgsin ma

a gsin

0.5a 9.81

4

a 1.23 m / s

=

=

=

=

=

Solving for time, t:

( )

2

o

2

1s v t at

2

14 0 1.23 t2

t 2.55 sec

= +

= +

=

mgsin

N4 m 0.5 m

W

ov 0 ( from rest)=


41/45

150



Loading Next Page

26. A block weighing 500 kN rests on a ramp inclined at 39 degrees with thehorizontal. What is the force that tends to move the block down theramp?

Solution:

Given:

W 500 kN

39

=

=

Let:

F= the force along the incline that tendsto move the block down the ramp

Then,F W sin

F 500 sin 39

F 314.6 kN

=

=

=

27. A golf ball after being struck by a golf club at ground level, departs at anangle of 45 deg with the horizontal fairway with a velocity of 100 fps.How far (in yards) will the ball lands from the point of departure?

Solution:

Formula:

( )

2

o

2

v sin 2R

g

100 sin 90R

32.2

1ydR 310.6 ft

3 ft

R 103.5 yd

=

=

=

=

W25

F W sin=

45 =

R


42/45


151


Loading Next Page

28. An absentminded driver traveling at 75 mph suddenly sees a checkpointahead blocking the roadway. The driver applies the brakes following a0.75 second delay. After the brakes are applied, the car decelerates at4.2 m/sec

2. Determine the stopping distance.

Solution:

1.609km 1000m 1hr v 75mph

mi km 3600sec

v 33.52 m / s

=

=

Solving for 1s : (distance travelled by the car at constant speed )

1

1

1

s vt

s 33.52(.75)

s 25.14 m

=

=

=

Solving for 2s : distance travelled with constant deceleration)

( ) ( )

( )( )

2 2

f o 2

2

2

2

v v 2as

33.52s 2 4.2

s 133.76

=

=

=

Thus, the total distance travelled is:

1 2S s s

25.14 133.76

158.9 m

= +

= +

=

1s 2s

v 75 mph= v 75 mph= v 0=


43/45


44/45


153


Loading Next Page

31. A wheel revolving at 300 rpm decelerates at 5 rad/sec^2. How longbefore the wheel stops?

Solution:

o

o

2 rad 1min300rpm

rev 60s

31.42 rad/ s

=

=

From:

f o

f

t

0 31.42t

5

t 6.28 sec

=

= =

=

32. An automobile tire is 30 inches in diameter. How fast in rpm does thewheel turn on the axle when the automobile maintains a speed of 50mph?

Solution:

Given:

1ftd 30 in 2.5 ft

12in

5280 ft 1hr v 50 mph 73.33 ft / s

mi 3600s

= =

= =

From:v r

v 73.33

2.5r

2

1rev 60s58.664 rad/ s

2 rad min

560.2 rpm

=

= =

= =


45/45

Chap 3 Mechanics

Documents

Transcript of Chap 3 Mechanics