Chap 3 Mechanics
Short Description
mech...
Description
CHAPTER 3 – Engineering Mechanics
GEAS
GENERAL ENGINEERING & APPLIED SCIENCES
ENGINEERING MECHANICS Engineering Mechanics is a branch of the physical sciences that is concerned with the state of rest or motion of bodies subjected to the action of forces. ENGINEERING MECHANICS STATICS
DYNAMICS
KINETICS
KINEMATICS
STATICS is concerned with the equilibrium of a body that is either at rest or in motion with constant velocity. DYNAMICS deals with the accelerated motion of the body. KINEMATICS is a branch of dynamics, which treats only the geometric aspect of the motion, while KINETICS is the analysis of the forces causing the motion. STATICS I. THE STATIC FORCE SYSTEM Concurrent 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 all forces acting on a body is zero.
Torque (or moment)
about an axis, due to a force, is a measure of the effectiveness of the force in producing rotation about that axis.
Weight
of an object is the force with which gravity pulls downward upon it.
Center of gravity
of an object is the point at which the entire weight of the object is concentrated. 110
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GENERAL ENGINEERING AND APPLIED SCIENCES CHAPTER 3
II.
Engineering Mechanics
RESULTANT OF TWO OR MORE COPLANAR, CONCURRENT FORCES n Parallelogram Method: (for two coplanar, concurrent forces) Cosine Law:
F2
R2 = F12 + F22 − 2FF 1 2 cos α
θ
F1
For more than 2 forces: (Use the Component Method)
F2
F2 R = sin θ sin α
o
R=
(∑F ) + (∑ F ) 2
x
2
y
F3
⎛ ∑ Fy ⎞ θ = tan−1 ⎜ ⎜ ∑ F ⎟⎟ x ⎠ ⎝ III. FORCES IN EQUILIBRIUM TYPES OF EQUILIBRIUM ➊
α
R
Sine Law:
F1
β
α θ
Static equilibrium is the condition of a body at rest and remains at
rest
under the action of concurrent forces. ➋
Translational equilibrium is the condition of a body in motion with constant velocity.
THE TWO CONDITIONS FOR EQUILIBRIUM ➊
First or the Force Condition
The vector sum of all forces acting on the body must be zero.. "
➋
∑ Fx = 0
∑ Fy = 0
Second or the Torque Condition
The sum of all the torques acting on the body must be zero
∑τ = 0 111
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CHAPTER 3 -MECHANICS
GEAS
GENERAL ENGINEERING & APPLIED SCIENCES
CABLES I.
PARABOLIC CABLES L L/2
Tension at the sup port :
L/2
2
⎛ ωL ⎞ 2 T= ⎜ ⎟ +H ⎝ 2 ⎠
W
T
H
Approximate lenght of cable : S=L+
2
T
d
4
8d 32d − L 5L3
ω (N / m)
Taking summation of moment at A: L/2
= H ( d) =
L/4
ω⎛L ⎞ 2 ⎜⎝ 4 ⎟⎠
T
L/2
L/4
T
W
Thus,
H
Tension at the lowest po int : H=
ω (N / m)
ωL2 8d
Where: T = tension at the support H = tension at the lowest point d = sag ω = weight per unit length L = span or distance between supports
112
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GENERAL ENGINEERING AND APPLIED SCIENCES CHAPTER 3
Engineering Mechanics
II. CATENARY (For Symmetrical & Unsymmetrical Supports)
L
Tension at the sup port T1 :
x
x
T1 = ωy1 T1
Tension at the sup port T2 : T2 = ωy 2
d
S
T2
S
H
y1
y2
c
Dis tan ce Between supports :
Dis tan ce Between sup ports :
⎛S +y ⎞ x1 = c ln ⎜ 1 1 ⎟ ⎝ c ⎠
⎛ S + y2 ⎞ x 2 = c ln ⎜ 2 ⎟ ⎝ c ⎠
Tension at the sup port T1 : T1 = H2 + ( ωS1 )
Tension at the sup port T2 :
2
Re lationship among S, y & c :
( S1 )
2
= ( y1 ) − ( c ) 2
( H)
T2 =
2
+ ( ωS2 )
2
Re lationship among S, y & c :
( S2 )
2
2
= ( y2 ) − (c ) 2
2
Where: T = tension at the support H = tension at the lowest point ω = weight per unit length y = height of the support c = minimum clearance from the ground S1 & S2 are half lengths of the cable L = span or distance between supports
113
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CHAPTER 3 -MECHANICS
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GENERAL ENGINEERING & APPLIED SCIENCES
n
FRICTION ON BLOCK
FRICTION W P
F = μN tan φ = μ =
F N
Ff = μN
φ
N
o
MAXIMUM ANGLE OF INCLINE WITHOUT CAUSING THE BODY TO SLIDE DOWN:
θ = φ = tan−1 μ N
θ
Where, in n & o :
W
F = frictional force N = normal force P = the applied force R = total surface reaction μ = coefficient of friction φ = angle of friction θ = angle of the incline
p
BELT FRICTION
T1 = eμφ T2
⎛T ⎞ ln ⎜ 1 ⎟ = μφ ⎝ T2 ⎠
Where: μ = the coefficient of friction φ = angle of contact in radians
φ
T1 T2
T1 = tension in the tight side T2 = tension in the slack side
114
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GENERAL ENGINEERING AND APPLIED SCIENCES CHAPTER 3
I.
Engineering Mechanics
DYNAMICS RECTILINEAR MOTION - (Motion in a Straight Line)
Vo
Vf
s Uniform Motion - (constant speed / zero acceleration)
" s = vt
v
Where: s = distance v = uniform speed or velocity t = time
v
s
v = vo = v f
Uniformly Accelerated Motion - (velocity increases uniformly) Equations of Motion: ➊ s = v ot ±
vo
1 2 at 2
vf
s
➋ v f − v o = ±at
Use ( + ) if : (v f > v o ) Use ( −) if : (v f < v o )
➌ v 2f − v o2 = ±2as
Where: s = distance traveled or displacement v o = original velocity ; ( v o = 0, if from rest )
v f = final velocity
; (v f = 0,if to stop) 2
a = acceleration ( m / s or ft / s2 ) t = time, (seconds) 115
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CHAPTER 3 -MECHANICS
GEAS
GENERAL ENGINEERING & APPLIED SCIENCES
II.
FREE-FALLING BODY - (Motion Under gravity)
Important Equations: ➊ h = vot ±
1 2 gt 2
➋ v f − v o = ±gt ➌ v 2f − v o2 = ±2gh Where: h = height v o = original velocity
v f = final velocity g = acceleration due to gravity m ft cm = 9.81 2 = 32.2 2 = 981 2 s s s t = time
" Note:
Use ( + ) for “g” going down
➊
Maximum Height Attained by a Body if projected straight upward:
" hmax = ➋
( v o )2 2g
Time taken to reach the highest point if projected vertically upward: " t=
p
Use ( −) for “g” going up
vo g
Time taken to reach the ground if dropped from a height h: "t=
2h ; g
( vo = 0)
116
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GENERAL ENGINEERING AND APPLIED SCIENCES CHAPTER 3
➍
Engineering Mechanics
Time of flight in going to the maximum height and falling back to the point where it was projected: 2v " T= o g
III. PROJECTILE MOTION
Projectile @ hmax
v oy
vx
vo θ
v
hmax
v ox
vy
Point of impact
d
R
For problems involving projectile motion, resolve the initial velocity ( v o ) into two perpendicular components as follows: Horizontal velocity
: v ox = v o cos θ
Vertical velocity
: v oy = v o sin θ
Then, apply formulas for kinematics in one dimension. ➊
AT ANY TIME, t:
Horizontal motion: ( ax
=0)
" d = v ox t = ( v o cos θ ) t
Vertical motion: ( a y = g = 9.81 " h = v oy −
m s2
)
1 2 1 gt = ( v o sin θ ) t − gt 2 2 2
117
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CHAPTER 3 -MECHANICS
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➋
TIME TAKEN TO REACH THE HIGEST POINT, t:
" t= ➌
v o sin θ g
TIME OF FLIGHT, T: The time of flight is the time taken by the projectile to return to the original horizontal level.
" T = 2t =
➍
2v o sin θ g
RANGE, R: Range is the distance covered in the horizontal direction in the total time of its flight. " R = ( v o cos θ ) T =
➎
v o2 sin 2θ g
AT MAXIMUM HEIGHT, hmax : ( v y = 0 )
At the highest point, the vertical velocity of the projectile is zero (as in the case of a body thrown vertically upwards), its velocity is only in the horizontal direction. " hmax = ➏
( v o sin θ )2 2g
VELOCITY AT ANY POINT, V: The velocity of a projectile at any point of its path is given by the resultant of its vertical and horizontal velocities at that point.
" v=
( v x )2 + ( v y )
118
2
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GENERAL ENGINEERING AND APPLIED SCIENCES CHAPTER 3
Engineering Mechanics
GENERAL EQUATION OF PROJECTILE " ± h = d tan θ −
gd2 2 ( v o cos θ )
2
" 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
y
θ
CASE 2 - Point of Impact is Below the Point of Release
θ
y
119
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CHAPTER 3 -MECHANICS
GEAS
GENERAL ENGINEERING & APPLIED SCIENCES
IV.
ROTARY MOTION
s θ
r
r
9
;
Important Notation: θ = angular displacement, radians ω = angular velocity, rad/s α = angular acceleration t = time
Uniform Angular Motion:
" θ = ωt ;
Uniformly Accelerated Angular Motion:
" θ = ωo t ±
1 2 αt 2
" ωf − ωo = ±αt
"
;
( ωf )2 − ( ωo )2 = ±2αθ
Relationship Between Angular and Tangential Quantities:
" s = r×θ " v = r×ω " a = r×α
v ω
r θ r
Provided, θ, ω, and α are in radians.
120
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GENERAL ENGINEERING AND APPLIED SCIENCES CHAPTER 3
V.
Engineering Mechanics
NEWTON’S LAWS OF MOTION First Law: The Law of Inertia Newton’s first law of motion states that if the body is at rest, it will remain at rest, if it is in motion, it will remain in motion with constant speed in a straight line unless there is a net force acting upon it.
" Fnet = 0 Second Law: The Law of Acceleration If a net force “F” acting on the body of mass “m” is not zero, the body accelerates in the direction of the force. The acceleration “a” is proportional to the force and inversely proportional to the mass of the body. F a= or F = ma m Third Law: The Law of Action and Reaction To every action there is always an equal and opposite reaction. ;
D’Alembert’s Principle: (Jean le Rond d’Alembert’s)
The resultant of the external forces applied to a body (rigid or non-rigid) and reversed effective force (REF) is equal to zero. a P
REF f
Equations:
REF − P − f = 0 Where: m = mass
or
REF =
W = weight
W a = ma g a = acceleration
"Note: Reversed Effective Force (REF) is always opposite in the direction of acceleration. 121
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CHAPTER 3 -MECHANICS
GEAS
GENERAL ENGINEERING & APPLIED SCIENCES
VI.
IMPULSE AND MOMENTUM n
Impulse Impulse is the product of force and the time it acts. F
I = Ft o
Momentum t Momentum is the product of the mass and the velocity of the body.
v
P = mv
p
m
Impulse - Momentum Theorem Impulse is equal to the change in momentum.
Ft = mv 2 − mv1 q
The Law of Conservation of Momentum When two bodies of masses m1 and m2 collide, the total momentum before impact is equal to the total momentum after impact. v2
v1
Before Impact m1 v1'
m2 After Impact
v 2'
m2
m1
Equation:
Pbefore impact = Pafter impact m1v1 + m2 v 2 = m1v1' + m2 v '2 122
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GENERAL ENGINEERING AND APPLIED SCIENCES CHAPTER 3
Engineering Mechanics
r
Types of Collision Collision – refers to the mutual action of the molecules, atoms, and etc., whenever they encounter one another. 1. Elastic collision – is a collision which conserves kinetic energy 2. Inelastic collision – 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 is maximum.
s
Coefficient of Restitution, e:
e=
t
Re lative velocity of recession v '2 − v1' = Re lative velocity of approach v1 − v 2
Where: e = 0 → for perfectly inelastic collision e = 1 → for perfectly elastic collision Special Cases: If a ball is dropped from a height “ho” upon a floor and rebounds to a height of “hr”, the coefficient of restitution between the ball and the floor is:
e=
hr ho
ho
Where: hr = height of rebound
hr
ho = original height
e=
Normal line
If a ball is thrown at an angle θ1 with the normal to a smooth surface and rebounds at an angle θ2: tan θ1 tan θ2
θ1
123
θ2
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CHAPTER 3 -MECHANICS
GEAS
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VII. UNIFORM CIRCULAR MOTION n
Centripetal Force FC =
o
mv 2 = mrω2 r
Centripetal acceleration
r v2 aC = = rω2 r
" Note: Centripetal force must be directed toward the center of the circular path. p
The Conical Pendulum CF rω2 v 2 = = W g gr Tension in the cord : W T= cos θ Period, t : tan θ =
t = 2π
θ T FC
CF
h g
h
W
for tmax : L = h
Where: CF = centrifugal reaction T = tension of the cord r = radius of the circular path h = height L = length of the cord
124
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GENERAL ENGINEERING AND APPLIED SCIENCES CHAPTER 3
Engineering Mechanics
VIII. BANKING OF CURVES n
Ideal Angle of Banking
⎛ v2 ⎞ θ = tan−1 ⎜ ⎟ ⎝ gr ⎠
o
For maximum velocity, v of the car without skidding
tan ( θ + φ ) =
v2 gr
Where: θ = angle of banking φ = angle of friction r = radius of the curve v = velocity For Horizontal Rotating Flat form:
tan φ = μ =
rω2 v 2 = g gr r
Also:
FC
v = rμg
CF
Where: φ = angle of friction r = radius v = velocity g = acceleration due to gravity 125
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CHAPTER 3 -MECHANICS
GEAS
GENERAL ENGINEERING & APPLIED SCIENCES
IX.
ROTATION OF RIGID BODIES
The Kinetic Energy of Rotation : 1 " KEr = Iω2 2 where: KEr = Kinetic energy of rotation I = moment of inertia ω = angular velocity, (rad / s)
ω
Total Kinetic Energy : " KET = KEr + KEt KET =
1 2 1 Iω + mv 2 2 2
Where: KEt = Kinetic energy of translation X.
NEWTON’S LAW OF UNIVERSAL GRAVITATION
R
"F=G
M2
m1m2 R2 M1
Where: G = Gravitational constant G = 6.67 × 10 −11 N ⋅ m2 / kg2 m = mass R = center-to-center distance between two masses
126
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GENERAL ENGINEERING AND APPLIED SCIENCES CHAPTER 3
Engineering Mechanics
TEST - 3
1.
A collision in which the total Kinetic energy after collision is less than before collision is called A. B. C. D.
2.
Whenever a net force acts on a body, it produces acceleration in the direction of the resultant force, an acceleration that is directly proportional to the mass of the body. This theory is popularly known as A. B. C. D.
3.
off center collision inelastic collision straight line collision elastic collision
Newton’s Second Law of Motion Newton’s First Law of Motion Hooke’s Law of Equilibrium Faraday’s Law of Forces
To maximize the horizontal range of the projectile, which of the following applies? A. Maximize velocity B. Maximize the angle elevation and velocity C. Maximize the angle of elevation D. 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 centroid B. is maximum at the centroidal axis C. is zero at the centroidal axis D. decreases as the distance of the axis moves farther from the centroid
127
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CHAPTER 3 -MECHANICS
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5.
6.
A measure of a resistance of the body it offers to any change in it’s angular velocity, determined by its mass and distribution of its mass about the axis rotation is known as ________. A. moment of inertia B. friction C. torsion D. angular acceleration Momentum is the product of mass and ______. A. acceleration B. velocity C. force D. time
7.
Centrifugal force is __________ . A. B. C. D.
8.
According to this law, “ The force between two charges varies directly as the magnitude of each charge and inversely as the square of the distance between them”. A. B. C. D.
9.
directly proportional to the radius of the curvature directly proportional to the square of the tangential velocity inversely proportional to the square of the tangential velocity directly proportional to the square of the weight of the object
Law of Universal Gravitation Coulomb’s Law Newton’s Law Inverse Square Law
When the total kinetic energy of a system is the same as before and after collision of two bodies, it is called A. Plastic collision B. Inelastic collision C. Elastic collision D. Static collision
128
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GENERAL ENGINEERING AND APPLIED SCIENCES CHAPTER 3
Engineering Mechanics
10. In a polar coordinate system, the length of the ray segment from a fixed origin is known as ______. A. amplitude B. radius vector C. hypotenuse D. minimum point 11. Momentum is a property related to the object’s ______. A. B. C. D.
motion and mass mass and acceleration motion and weight weight and velocity
12. The study of motion without reference to the force that causes the motion is known as ______. A. statics B. dynamics C. kinetics D. kinematics 13. Varignon’s theorem is used to determine ______. A. location of centroid B. moment of inertia C. mass moment of inertia D. moment of area 14. The periodic oscillations either up or down or back and fourth motion in the straight line is known as_______. A. B. C. D.
transverse harmonic motion resonance rotational harmonic motion translational harmonic motion
129
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CHAPTER 3 -MECHANICS
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15. A freely falling body is a body in rectilinear motion and with constant_____. A. velocity B. speed C. deceleration D. acceleration 16. When the total kinetic energy of the system is the same as before and after the collision of two bodies, it is called A. Static collision B. Elastic collision C. Inelastic collision D. 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. C. B. D.
no change one half double one fourth
18. With reference to the thermodynamic diagram of temperature – entropy (TS), what is represented by the area under the diagram? A. B. C. D.
work done enthalpy temperature difference heat transferred
19. What is the standard acceleration due to gravitational force? A. 32 ft/sec/sec B. 980 ft/sec/sec C. 32 m/sec/sec D. 98 ft/sec/sec
130
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GENERAL ENGINEERING AND APPLIED SCIENCES CHAPTER 3
Engineering Mechanics
20. Which of the following collisions is an elastic collision? A.
Two bodies move towards each other, collide and then move away from each other. There is a rise in temperature B. Two bodies collide and the sound of collision is heard by a blind man C. Two steel balls collide such that their kinetic energy is conserved D. A man jumps on to a moving cart 21. A mass is revolving in a circle which is in the plane of paper. The direction of centripetal acceleration is along the radius: A. B. C. D.
away from the center radius toward the center at right angle to angular velocity none of the above
22. When a mass is rotating in a plane about a fixed point, its angular momentum is directed along: A. the radius C. a line at an angle of 45o to the plane of the rotation B. the tangent to orbit D. the axis of rotation 23. A solid iron sphere A rolls down an inclined plane, while an identical hollow 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 B C. more than that of B B. equal to that of B D. sometime more and sometimes less
131
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24. Which of the following has the highest moment of inertia when each of them has the same mass and the same radius? A. B. C. D.
A hollow sphere about one of its diameters. A solid sphere about one of its diameters. A disc about its central axis perpendicular to the plane of the disc All of the above have the same moment of inertia
25. When a planet moves around the sun, A. the angular momentum remains conserved B. the angular speed remains constant C. the linear velocity remains constant D. the linear momentum remains constant 26. What keeps an earth satellite moving on its orbit? A. Gravitational attraction between satellite and earth B. Ejection gases from the exhaust of the satellite C. burning of fuel D. Gravitational attraction of sun 27. The value of universal gravitational constant G depends upon: A. nature of material of two bodies B. heat content of two bodies C. acceleration of two bodies D. None of these 28. If the mass of an object could be doubled, then its inertia would be A. halved B. doubled C. unchanged D. quadrupled 29. It shows the forces acting on an isolated object. A. B. C. D.
force diagram schematic diagram free body diagram force polygon
132
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GENERAL ENGINEERING AND APPLIED SCIENCES CHAPTER 3
Engineering Mechanics
30. A wagon is uniformly accelerating from rest. The net force acting on the wagon is A. zero B. increasing C. constant D. decreasing 31. If the mass of an object were doubled, its acceleration due to gravity would be A. doubled also B. unchanged C. halved D. fivefold 32. Which of the following is not a vector quantity? A. B. C. D.
force energy weight velocity
33. Which of the following is not a scalar quantity? A. B. C. D.
time work temperature displacement
34. The resultant of two concurrent forces is minimum when the angle between them is A. B. C. D.
0 degree 90 degrees 45 degrees 180 degrees
133
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35. As the angle between two concurrent forces decreases from 180o, their resultant A. decreases B. increases C. unchanged D. cannot be determined 36. The maximum number of components that a single force may be resolved into is A. one B. two C. three D. four 37. The momentum of an object is the product of its A. B. C. D.
mass and displacement mass and velocity force and displacement force and time
38. If the direction of an object’s momentum is west, the direction of the velocity of the object is A. B. C. D.
east west north south
39. The direction of an object’s momentum is always the same as the direction of the object’s A. B. C. D.
inertia mass weight velocity
134
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GENERAL ENGINEERING AND APPLIED SCIENCES CHAPTER 3
Engineering Mechanics
40. The moment of inertia of a triangle with respect to the base how many time its moment of inertia with respect to its centroidal axis A. 1/2 B. 3 C. 1/4 D. 5 41. When to objects collide, which of the following is always true? A. the velocity of each object does not change B. there is no change in the displacement of each object C. 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. ballistics B. kinematics C. kinetics D. dynamics 43. What is the moment of inertia of a circle of radius r? A. πr4/4 B. πr4/12 C. πr4/15 D. πr4/16 44. The moment of inertia of a rectangle with respect to the base is how many times its moment of inertia with respect to the centroidal axis? A. B. C. D.
2 3 4 5
135
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CHAPTER 3 -MECHANICS
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45. The moment of inertia of a triangle with respect to its base “b” is A. B. C. D.
b2h2/12 bh3/6 bh3/12 bh3/3
46. The moment of inertia of a circle with respect to its tangent is how many times its moment of inertia with respect to its centroidal axis? A. B. C. D.
2 3 4 5
47. Moment of inertia is also called A. Moment of mass B. Moment of center C. second moment of area D. moment of volume 48. One newton is equivalent to A. kg-m/s B. kg – m/m/s C. kg – m/s/s D. m/s2 per kilogram 49. These are forces whose lines of action all pass through a common point. A. B. C. D.
collinear forces couple coplanar forces concurrent forces
50. The radial distance from the axis to the point of application of the force is called A. radius vector B. lever arm C. normal D. displacement 136
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GENERAL ENGINEERING AND APPLIED SCIENCES CHAPTER 3
Engineering Mechanics
Solved Problems
In 1.
Mechanics
A mango falls from a branch 5 meters above the ground. With what 2 speed in meters per second does it strike the ground? g=10m/s
Solution: From:
( v f )2 = ( v o )2 + 2gh Where: v o = 0 (free fall) Substitute: v f = 2gh = 2 (10 )( 5 ) v f = 10 m / s 2.
An automobile accelerates at a constant rate of 15mi/hr to 45 mi/hr in 15 seconds while traveling in a straight line. What is the average acceleration?
Solution:
a=
From:
vf − vo t
Where:
mi ⎛ 5280 ft ⎞ ⎛ 1hr ⎞ ⎜ ⎟ = 66 ft / s hr ⎜⎝ mi ⎟⎠ ⎝ 3600 s ⎠ mi ⎛ 5280 ft ⎞ ⎛ 1hr ⎞ v o = 15 ⎜ ⎜ ⎟ = 22 ft / s hr ⎝ mi ⎟⎠ ⎝ 3600 s ⎠ v f = 45
Then, a=
66 − 22 = 2.93 ft / s2 15 137
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CHAPTER 3 -MECHANICS
GEAS
GENERAL ENGINEERING & APPLIED SCIENCES
3.
A 50-kilogram block of wood rests on the top of the smooth plane whose length is 3m and whose altitude is 0.8 m. How long will it take for the block to slide to the bottom of the plane when released?
Solution: Given: vo = 0
m = 50 kg s=3m
v o = 0 (from rest)
mgsin θ
θ
h = 0.8 m
3m
N
0.8 m
W
Solving for the acceleration of the box: Take summation of forces along the incline equal to zero: mg sin θ = ma a = g sin θ ⎛ 0.8 ⎞ a = 9.81⎜ ⎟ ⎝ 3 ⎠ a = 2.62 m / s2
Solving for time, t: 1 s = v o t + at 2 2 1 3 = 0 + ( 2.62 ) t 2 2 t = 1.51 sec .
138
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GENERAL ENGINEERING AND APPLIED SCIENCES CHAPTER 3
4.
Engineering Mechanics
A baseball is thrown from a horizontal plane following a parabolic path with 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: v o = 100 m / s θ = 30°
From: R= R=
( vo )
2
sin 2θ g
(100 )
2
sin ⎣⎡2 ( 30 ) ⎦⎤ 9.81
R = 883 m 5.
A block weighing 500 KN rests on a ramp inclined at 25 degrees with the horizontal. The force tending to move the block down the ramp is
Solution: Given: W = 500 kN θ = 25°
F = W sin θ
W
25°
Let: F = the force along the incline that tends to move the block down the ramp Then, F = W sin θ
F = 500 sin 25 F = 211 kN
139
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CHAPTER 3 -MECHANICS
GEAS
GENERAL ENGINEERING & APPLIED SCIENCES
6.
How far does an automobile move while its speed increases uniformly from 15 kph to 45 kph in 20 sec?
Solution: ⎛ 15km ⎞ ⎛ 1000m ⎞ ⎛ 1hr ⎞ vo = ⎜ ⎟⎜ ⎟⎜ ⎟ = 4.17 m / s ⎝ hr ⎠ ⎝ km ⎠ ⎝ 3600s ⎠ ⎛ 45km ⎞ ⎛ 1000m ⎞ ⎛ 1hr ⎞ vf = ⎜ ⎟⎜ ⎟⎜ ⎟ = 12.5 m / s ⎝ hr ⎠ ⎝ km ⎠ ⎝ 3600s ⎠ From: v o = 15 kph ⎛ v + vo ⎞ S = vt = ⎜ f ⎟t ⎝ 2 ⎠ ⎛ 4.17 + 12.5 ⎞ S=⎜ ⎟ ( 20 ) 2 ⎝ ⎠ S = 166.7 m
7.
v f = 45 kph
S
A rotating wheel has a radius of 2 feet and 6 inches. A point on the rim of the wheel moves 30 feet in 2 seconds. Find the angular velocity of the wheel.
Solution: Solving for the tangential velocity: v = s t → tangential velocity 30 v= = 15 ft / s 2 From: v = rω 15 = 2.5ω ω = 6 rad / s
140
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GENERAL ENGINEERING AND APPLIED SCIENCES CHAPTER 3
8.
Engineering Mechanics
A 16 gm mass is moving at 30 cm/sec., while a 4gm mass is moving in an opposite direction at 50 cm/sec. They collide head on and stick together. Their velocity after collision is
Solution: Let: v c = combined velocity after collision v1
v2 Before Impact
m1
m2 vc
After Impact
(m1 + m2 )
From the Law of Conservation of Momentum: Momentum Before Impact = Momentum After Impact m1v1 + m2 v 2 = ( m1 + m2 ) v c 16 ( 0.30 ) + 4 ( −0.50 ) = (16 + 4 ) v c v c = 0.14 m / s
9.
A ball is dropped from a building 100 m high. If the mass of the ball is 10 gm, after what time will the ball strike the earth?
Solution: 1 2 gt 2 1 100 = ( 0 ) t + ( 9.81) t 2 2 t = 4.52 sec . h = vot +
141
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CHAPTER 3 -MECHANICS
GEAS
GENERAL ENGINEERING & APPLIED SCIENCES
10. Determine the angle of super elevation for a 200 m highway curve so that there will be no side thrust at a speed of 90 kph.
Solution: From:
tan θ =
v2 → banking angle fromula gr
Substitute given and solve for θ : ⎡⎛ 90 km ⎞ ⎛ 1000 m ⎞ ⎛ 1hr ⎞ ⎤ ⎢⎜ ⎟⎥ ⎟⎜ ⎟⎜ ⎝ hr ⎠ ⎝ km ⎠ ⎝ 3600 s ⎠ ⎦ tan θ = ⎣ [9.81 m / s] ( 200 m )
2
θ = 17.67° 11. 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.
Solution: CF =
mv 2 r
Where:
km ⎞ ⎛ 1000m ⎞ ⎛ 1hr ⎞ ⎛ v = ⎜ 150 ⎜ ⎟⎜ ⎟ hr ⎟⎠ ⎝ 1km ⎠ ⎝ 3600 s ⎠ ⎝ v = 41.67 m / s Thus, solving for FC: CF =
( 50,000 )( 41.67 ) ( 9.81)(150 )
2
CF = 59 kN
142
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GENERAL ENGINEERING AND APPLIED SCIENCES CHAPTER 3
Engineering Mechanics
12. A 40 gm rifle with a speed of 300 m/s is fired into a ballistic pendulum of mass 5 kg suspended from a chord 1 m long. Compute the vertical height through which the pendulum rises.
Solution:
Let: mb = mass of the bullet
; mB = mass of the block
v c = combined velocity of the block and the bullet mc = mb + mB = combined mass of the block and the bullet Using the Law of Conservation of Momentum: mb v b + mB v B = mc v c
( 0.04 )( 300 ) + 5 ( 0 ) = ( 0.4 + 5 ) v c v c = 2.38 m / s
From the law of conservation of energy: KEbottom = PE top 1 2 mc ( v c ) = mc gh 2 h=
( vc ) 2g
2
( 2.38 ) 2 ( 9.81) 2
=
h = 0.2887 m or 28.87 cm
143
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CHAPTER 3 -MECHANICS
GEAS
GENERAL ENGINEERING & APPLIED SCIENCES
13. A projectile is thrown with a speed of 100 ft/sec in a direction 30o above the horizontal. Determine the highest point to which it rises.
Solution: At maximum height, v fy = 0 :
( v ) − ( v ) = −2gh ( v ) = ( v sin θ) h = 2
fy
2
oy
max
2
oy
max
hmax
2
o
2g
2g
(100 sin30 ) =
2
2(32.2) hmax = 38.81 m
14. A missile is fired with a speed of 100 fps in a direction 30 degrees above the horizontal. Determine the maximum height to which it rises?
Solution: hmax =
( vo )
2
sin2 θ
2g
(100 ) ( sin 30 ) 2 ( 32.2 ) 2
hmax =
2
hmax = 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: ⎛ 1lb ⎞ W = 1.6 oz ⎜ ⎟ = 0.1lb ⎝ 16 oz ⎠ From: Impulse = change in momentum
I = m ( vf − vo ) 0.1 ( 225 − 0 ) 32 I = 0.703 slugs − ft / s I=
144
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GENERAL ENGINEERING AND APPLIED SCIENCES CHAPTER 3
Engineering Mechanics
16. A car accelerates from rest and reached a speed of 90 kph in 30 seconds. What is the acceleration in meter per second per second?
Solution: v o = 0 (from rest) ⎛ 1000 m ⎞ ⎛ 1hr ⎞ v f = 90kph ⎜ ⎟ ⎟⎜ ⎝ km ⎠ ⎝ 3600 s ⎠ v f = 25 m / s From: v f − v o = at Substitute and solve for a: 25 − 0 = a(30) a = 0.833 m / s2 17. A stone is thrown vertically upward at the rate of 20 m/s. It will return to the ground after how many seconds?
Solution: From: v f − v o = gt
v f = 0 (@ max imum height)
−v −20 t= o = g −9.81 t = 2.04 sec . Total time of flight: T = 2t = 2 ( 2.04 )
v o = 20 m / s
= 4.08 sec .
145
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CHAPTER 3 -MECHANICS
GEAS
GENERAL ENGINEERING & APPLIED SCIENCES
18. A car accelerates uniformly from standstill to 80 mi/hr in 5 seconds. What is its acceleration?
Solution:
v o = 0 (from rest) ⎛ 5280 ft ⎞ ⎛ 1hr ⎞ v f = 80 mph ⎜ ⎟ ⎟⎜ ⎝ mi ⎠ ⎝ 3600 s ⎠ v f = 117.33 ft / s From: v f − v o = at Substitute and solve for a: 117.33 − 0 = a(5) a = 23.47 ft / s2 19. A DC-9 jet with a takeoff mass of 120 tons has two engines producing average force of 80,000 N during takeoff. Determine the plane’s acceleration down the runway if the takeoff time is 10 seconds.
Solution:
Formula: F m Substitute: 2 ( 80,000 ) a= 120,000 a=
a = 1.33 m / s2
" Note:
1 ton = 1000kg
20. What is the acceleration of a point on a rim of a flywheel 0.8 m in diameter turning at the rate of 1400 rad/min?
Solution: a=
v2 →X r
But:
v = rω Substitute to equation n : a=
(rω) r
2
2
⎛ rad 1min ⎞ 2 = rω2 = ( 0.4 ) ⎜ 1400 × ⎟ = 217.77 m / s min 60 s ⎠ ⎝
146
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GENERAL ENGINEERING AND APPLIED SCIENCES CHAPTER 3
Engineering Mechanics
21. A bullet is fired vertically upward with a mass of 3 grams. If it reaches an altitude of 100 m, what is its initial velocity?
Solution: From:
( v f ) − ( vo ) 2
2
= ±2gh
Note: Use “+” , for going up Use “-“ , for going down Substitute values: @ maximum height, v f = 0 0 − ( v o ) = −2(9.81)100 2
v o = 44.29 m / s
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?
Solution: Converting to common units: ⎛ 2πrad ⎞ ⎛ 1min ⎞ ω = 1000 rpm ⎜ ⎟⎜ ⎟ ⎝ rev ⎠ ⎝ 60s ⎠ ω = 104.72 rad / s Also, ⎛ 1ft ⎞ r = 14 in ⎜ ⎟ = 1.167 ft ⎝ 12in ⎠ Solving for v: v = rω v = 1.167 ft (104.72rad / s ) v = 122.2 ft / s
147
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CHAPTER 3 -MECHANICS
GEAS
GENERAL ENGINEERING & APPLIED SCIENCES
23. A ball was thrown upward with an initial velocity of 50ft/s. How high does it go?
Solution:
vf = 0
From:
( v f ) − ( vo ) 2
2
= 2gh
Where: v f = 0; at max imum height Substitute:
hmax
0 − ( 50 ) = −2 ( 32 ) h 2
vo
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 the ground in m/sec?
Solution:
Solving for the time of flight to reach hmax : v o sin θ g 200 sin 30 t= 9.81 t = 10.19 sec t=
Considering free - fall from hmax :
v fy − v oy = gt ;(v oy = 0) t
v fy = 9.81(10.19) v fy = 100
t
30°
Solving for v fx :
v fx
v fx = v o cos θ
2t
v fx = 200 cos 30
v fy
vf
v fx = 173.2 Thus,
v=
(v ) + (v ) 2
fy
fx
2
=
(100 )
2
+ (173.2 )
2
v = 200 m / s
148
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GENERAL ENGINEERING AND APPLIED SCIENCES CHAPTER 3
Engineering Mechanics
25. A 40 kg block rests at the top of an inclined smooth plane whose length is 4 m and whose height is 0.5 m. How long will it take for the block to slide to the bottom of the plane when released?
Solution:
v o = 0 (from rest)
Given: vo = 0 m = 40 kg s=4m h = 0.5 m
mgsin θ
θ
4m
N
0.5 m
W
Solving for the acceleration of the box: Fnet = ma mgsin θ = ma a = g sin θ ⎛ 0.5 ⎞ a = 9.81⎜ ⎟ ⎝ 4 ⎠ a = 1.23 m / s2
Solving for time, t: 1 s = v o t + at 2 2 1 4 = 0 + (1.23 ) t 2 2 t = 2.55 sec
149
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CHAPTER 3 -MECHANICS
GEAS
GENERAL ENGINEERING & APPLIED SCIENCES
26. A block weighing 500 kN rests on a ramp inclined at 39 degrees with the horizontal. What is the force that tends to move the block down the ramp?
Solution: Given: W = 500 kN θ = 39°
F = W sin θ
W
Let:
25°
F = the force along the incline that tends to move the block down the ramp
Then, F = W sin θ F = 500 sin 39 F = 314.6 kN
27. A golf ball after being struck by a golf club at ground level, departs at an angle 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: v 2 sin 2θ R= o g R=
(100 )
2
θ = 45° R
sin 90
32.2 ⎛ 1yd ⎞ R = 310.6 ft ⎜ ⎟ ⎝ 3 ft ⎠ R = 103.5 yd
150
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GENERAL ENGINEERING AND APPLIED SCIENCES CHAPTER 3
Engineering Mechanics
28. An absentminded driver traveling at 75 mph suddenly sees a checkpoint ahead blocking the roadway. The driver applies the brakes following a 0.75 second delay. After the brakes are applied, the car decelerates at 4.2 m/sec2. Determine the stopping distance.
Solution:
v = 75 mph
v = 75 mph
s1
v=0
s2
⎞ ⎛ 1.609 km ⎞ ⎛ 1000m ⎞ ⎛ 1hr v = 75 mph ⎜ ⎜ ⎟ ⎟ ⎜ ⎟ mi ⎝ ⎠ ⎝ km ⎠ ⎝ 3600 sec ⎠ v = 33.52 m / s Solving for s1 : (distance travelled by the car at constant speed ) s1 = vt s1 = 33.52(.75) s1 = 25.14 m Solving for s2 : distance travelled with constant deceleration)
( v f ) − ( v o ) = 2as2 2 − ( 33.52 ) s2 = 2 ( −4.2 ) 2
2
s2 = 133.76 Thus, the total distance travelled is: S = s1 + s2 = 25.14 + 133.76 = 158.9 m
151
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CHAPTER 3 -MECHANICS
GEAS
GENERAL ENGINEERING & APPLIED SCIENCES
29. A compact disk 125 mm in diameter is rotating at 50 rpm. If it takes 10 complete revolutions before slowing to a stop, how fast is it decelerating?
Solution: Given: ⎛ 2πrad ⎞ ⎛ 1min ⎞ ; ω = 50 rpm ⎜ ⎟ = 5.236 rad / s ⎟⎜ ⎝ rev ⎠ ⎝ 60 s ⎠
d = 125 mm
⎛ 2πrad ⎞ θ = 10 rev ⎜ ⎟ = 62.83 rad ⎝ rev ⎠ From:
( ωf ) − ( ωo ) = 2αθ 2 2 ( ω ) − ( ωo ) α= f 2
2
ω
2θ Substitute values: α=
0 − ( 5.236 )
2
2 ( 62.83 )
= −0.22 rad / s2
30. A man driving a car at 45 mph suddenly saw a stalled vehicle on the road 100 ft ahead. What constant deceleration is required to stop the car 10 ft before the stalled vehicle? (Answer in ft per sec per sec).
Solution: ⎛ 5280 ft ⎞ ⎛ 1hr ⎞ v = 45 mph ⎜ ⎟⎜ ⎟ ⎝ 1mi ⎠ ⎝ 3600 s ⎠ v = 66 ft / s v o = 66 ft / s
−a
From: v 2f − v o2 = 2as a= a=
90
v 2f − v o2 2s 0 − ( 66 )
10
v f = 0 (stop)
2
2 ( 90 )
a = −24.20 ft / s2
152
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GENERAL ENGINEERING AND APPLIED SCIENCES CHAPTER 3
Engineering Mechanics
31. A wheel revolving at 300 rpm decelerates at 5 rad/sec^2. How long before the wheel stops?
Solution: ⎛ 2πrad ⎞ ⎛ 1min ⎞ ωo = 300 rpm ⎜ ⎟ ⎟⎜ ⎝ rev ⎠ ⎝ 60 s ⎠ ωo = 31.42 rad / s From: ωf − ωo = αt
ωf − ω 0 − 31.42 = α −5 t = 6.28 sec t=
32. An automobile tire is 30 inches in diameter. How fast in rpm does the wheel turn on the axle when the automobile maintains a speed of 50 mph?
Solution: Given: ⎛ 1ft ⎞ d = 30 in ⎜ ⎟ = 2.5 ft ⎝ 12in ⎠ ⎛ 5280 ft ⎞ ⎛ 1hr ⎞ v = 50 mph ⎜ ⎟ = 73.33 ft / s ⎟⎜ ⎝ mi ⎠ ⎝ 3600 s ⎠ From: v = rω v 73.33 ω= = 2.5 r 2
⎛ 1rev ⎞ ⎛ 60 s ⎞ ω = 58.664 rad / s ⎜ ⎟⎜ ⎟ ⎝ 2πrad ⎠ ⎝ min ⎠ ω = 560.2 rpm 153
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CHAPTER 3 -MECHANICS
GEAS
GENERAL ENGINEERING & APPLIED SCIENCES
33. A grenade is fired from a launcher with a velocity of 50 m/s at an angle of 30 degrees with the horizontal. Determine the maximum height that it could reach in meters.
Solution: hmax =
( vo )
2
sin2 θ
2g
→ formula for max imum height
( 50 ) ( sin 30 ) hmax = 2 ( 9.81) 2
2
hmax = 31.85 m 34. A 5g bullet is fired from a 5 kg gun with a speed of 400 m/sec. What is the speed of recoil of the gun?
Solution: Given: mb = 5g = 0.005kg mg = 5kg
v b = 400m / s vr = ?
From the law of conservation of momentum: Total momentum before = Total momentum after Before explosion, the velocity of the gun and the bullet is zero, thus the total momentum before explosion is zero. 0 = mb v b + mgv r vr =
−mb v b 0.005(400) =− mg 5
v r = −0.40 m / s
154
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