ENGINEERING MECHANICS OBJECTIVE QUESTIONS
1. What is the branch of engineering mechanics which refers to the study of stationary rigid
body?
A.
B.
C.
D.
Statics
Kinetics
Kinematics
Dynamics
2. What is the branch of engineering mechanics which refers to the study of rigid body in
motion under the action of forces?
A.
B.
C.
D.
Statics
Strength of Materials
Kinematics
Dynamics
3. What is the branch of engineering mechanics which refers to the study of rigid body in
motion without reference to the force that causes the motion?
A.
B.
C.
D.
Statics
Kinetics
Kinematics
Dynamics
4. What refers to the force that holds part to the rigid body together?
A.
B.
C.
D.
Natural force
External force
Internal force
Concentrated force
5. What refers to a pair of equal, opposite and parallel forces?
A.
B.
C.
D.
Couple
Moment
Torque
All of the above
6. What is a concurrent force system?
A.
B.
C.
D.
All forces act at the same point.
All forces have the same line of action.
All forces are parallel with one another
All forces are in the same plane.
7. When will a three- force member be considered in equilibrium?
A.
B.
C.
D.
When the sum of two forces is equal to the third force.
When they are concurrent or parallel.
When they are coplanar.
All of the above
8. A roller support has how many reactions?
A.
B.
C.
D.
None
1
2
3
9. A link or cable support has how many reactions?
A.
B.
C.
D.
None
1
2
3
10. A build-in, fixed support has how many reactions and moment?
A.
B.
C.
D.
1 reaction and 1 moment
2 reactions and 1 moment
1 reaction and 2 moments
2 reactions and no moment
11. Which support has one moment?
A.
B.
C.
D.
Frictionless guide
Pin connection
Fixed support
Roller
12. What is the science that describes and predicts the effect on bodies at rest or in motion by
forces acting on it?
A.
B.
C.
D.
Engineering Mechanics
Theory of Structures
Mechanics of Materials
Strength of Materials
13. What refers to a negligible body when compared to the distances involved regarding its
motion?
A.
B.
C.
D.
Particle
Atomic substance
Element
Quarts
14. The resulting force of a distributed load is always acting at:
A.
B.
C.
D.
the center of the beam subjected to the distributed load
the centroid of the area of the loading curve
the 1/3 point from the higher intensity side of the loading curve
the 2/3 point from the higher intensity side of the loading curve
15. The resultant force of a distributed load is always equal to:
A.
B.
C.
D.
twice the area under the loading curve
half the area under the loading curve
the area under the loading curve
one-fourth the area under the loading curve
16. When a body has more supports than are necessary to maintain equilibrium, the body is
said to be _________.
A.
B.
C.
D.
in static equilibrium
in dynamic equilibrium
statically determinate
statically indeterminate
17. When does an equation be considered “dimensionally homogenous”?
A. When it is unitless
B. When the dimensions of the various terms on the left side of the equation is not
the same as the dimensions of the various terms on the right side.
C. When the degree of the left side of the equation is the same as the right side.
D. When the dimensions of various terms on the left side of the equation is the
same as the dimensions of the various terms on the right side.
18. What refers to the branch of mathematics which deals with the dimensions of quantities?
A.
B.
C.
D.
Unit analysis
Dimensional analysis
System analysis
Homogeneity analysis
19. What is a “simple beam”?
A.
B.
C.
D.
A beam supported only at its end.
A beam supported with a fixed support at one-end and none on the other end.
A beam with more than two supports.
A beam with only one support at the midspan.
20. What assumption is used in the analysis of uniform flexible cable?
A. Cable is flexible.
B. Cable is inextensible.
C. The weight of the cable is very small when compared to the loads supported by
the cable.
D. All of the above
21. “The sum of individual moments about a point caused by multiple concurrent forces is
equal to the moment of the resultant force about the same point”. This statement is known
as ___________.
A.
B.
C.
D.
Pappus Proposition
D’Alembert’s Principle
Varignon’s Theorem
Newton’s Method
22. “Two forces acting on a particle may be replaced by a single force called resultant which
can be obtained by drawing diagonal of parallelogram, which has the sides equal to the
given forces”. This statement is known as ____________.
A.
B.
C.
D.
Pappus Proposition
Principle of Transmissibility
Parallelogram Law
Varignon’s Theorem
23. “The condition of equilibrium or motion of a rigid body remains unchanged if a force
acting at a given point of the rigid body is replaced by a force of same magnitude and
direction but acting at a different point provided that the two forces have the same line of
action”. This statement is known as
.
A.
B.
C.
D.
Pappus Propositions
Principle of Transmissibility
Parallelogram Law
Varignon’s Theorem
24. “If two forces acting simultaneously on a particle can be represented by the two sides of a
triangle taken in order that the third side represents the resultant in the opposite order”.
This statement is known as
.
A.
B.
C.
D.
Principle of Transmissibility
Parallelogram Law
Varignon’s theorem
Triangle Law of Forces
25. “If the number of concurrent forces acting simultaneously on a particle, are represented in
magnitude and direction by the sides of polygon taken in order, then the resultant of this
system of forces is represented by the closing side of the polygon in the opposite order”.
His statement is known as
.
A.
B.
C.
D.
Principle of Transmissibility
Parallelogram Law
Polygon Law
Triangle Law of Forces
26. A beam with more than one supports is called
A.
B.
C.
D.
Cantilever beam
Simple beam
Complex beam
Continuous beam
27. A truss consisting of coplanar number is called
A.
B.
C.
D.
.
Plane truss
Space truss
Ideal truss
Rigid truss
28. A truss consisting of non-coplanar member is called
A.
B.
C.
D.
.
.
Plane truss
Space truss
Ideal truss
Rigid truss
29. What method of determining the bar force of a truss if only few members are required?
A.
B.
C.
D.
Method of joints
Method of section
Maxwell’s diagram
Method of superposition
30. Which of the following statements about friction is FALSE?
A. The direction of frictional force on a surface is such as to oppose the tendency of
one surface to slide relative to the other.
B. The total frictional force is dependent on the area of contact between the two
surfaces.
C. The magnitude of the frictional force is equal to the force which tends to move the
body till the limiting value is reached.
D. Friction force is always less than the force required to prevent motion.
31. In the analysis of friction, the angle between the normal force and the resultant force
______________ the angle of friction.
A.
B.
C.
D.
may be greater than or less than
is greater than
is less than
is equal to
32. When a block is place on an inclined plane, its steepest inclination to which the block
will be in equilibrium is called _____________.
A.
B.
C.
D.
angle of friction
angle of reaction
angle of normal
angle of repose
33. What is usually used to move heavy loads by applying a force which is usually smaller
than the weight of the load?
A.
B.
C.
D.
Axle
Incline plane
Wedge
Belt
34. The angle of inclined plane of a jack screw is also known as ____________.
A.
B.
C.
D.
angle of thread
angle of lead
angle of friction
angle of pitch
35. center of gravity for a two-dimensional body is the point at which the entire
___________ acts regardless of the orientation of the body.
A.
B.
C.
D.
mass
weight
mass or weight
volume
36. Second moment of area is the product of:
A.
B.
C.
D.
area and square of the distance from the reference axis
area and distance from the reference axis
square of the area and distance from the reference axis
square of the area and square of the distance from the reference axis
37. Moment of inertia of an area about an axis is equal to the sum of the moment of inertia
about an axis passing through the centroid parallel to the given axis and __________.
A.
B.
C.
D.
area and square of the distance between two parallel axes
area and distance between two parallel axes
square of the area and distance between two parallel axes
square of the area and square of the distance between two parallel axes
38. What is the unit of mass moment of inertia?
A.
B.
C.
D.
kg-m4
kg-m3
kg-m
kg-m2
39. The number of independent degrees of freedom is:
A. Square root of the square of the difference of total degrees of freedom – number
of constrain equations
B. Square root of the total degrees of freedom – number of constrain equations
C. Total degrees of freedom – number of constrain equations
D. Total degrees of freedom – half the number of constrain equations
40. What velocity is normally referred to as the derivative of position vector with respect to
time?
A.
B.
C.
D.
Decreasing velocity
Average velocity
Instantaneous velocity
Increasing velocity
41. What refers to a force by which work done on a particle as it moves around any closed
path is zero?
A.
B.
C.
D.
Natural force
Virtual force
Conservative force
Non-conservative force
42. When a force causes a change in mechanical energy when it moves around a closed path,
it is said to be ____________ force.
A.
B.
C.
D.
natural
virtual
conservative
non-conservative
43. The following are quantities that describe motion and uses Newton’s law of motion and
d’Alembert’s principle except one. Which one?
A.
B.
C.
D.
Time
Mass
Acceleration
Force
44. Which of the following set of quantities that describe motion and uses the principle of
work and energy?
A.
B.
C.
D.
Force, mass, velocity, time
Force, mass, acceleration
Force, mass, distance, velocity
Force, weight, distance, time
45. Which of the following set of quantities that describe motion and uses the principle of
impulse and momentum?
A.
B.
C.
D.
Force, mass, velocity, time
Force, mass, acceleration
Force, mass, distance, velocity
Force, weight, distance, time
46. The principles of kinetics of particles are derived from which law?
A.
B.
C.
D.
Newton’s first law
Newton’s second law
Newton’s third law
d’Alembert’s principle
47. What type of impact is when the motion of one or both of the colliding bodies is not
directed along the line impact?
A.
B.
C.
D.
Central impact
Eccentric impact
Direct impact
Oblique impact
48. What type of impact is when the centers of mass of colliding bodies are not located on
the line of impact?
A.
B.
C.
D.
Central impact
Eccentric impact
Direct impact
Oblique impact
49. If the coefficient of restitution is zero, the impact is ______________.
A.
B.
C.
D.
partially plastic
perfectly inelastic
perfectly elastic
partially elastic
50. A uniform circular motion can be considered as a combination of ______________.
A.
B.
C.
D.
linear velocity and impulse
simple harmonic motion and momentum
two simple harmonic motion
rectilinear translation and curvilinear translation
ENGINEERING MECHANICS PROBLEMS (STATICS)
1. What is the magnitude of the resultant force of the two forces which are perpendicular to
each other? The two forces are 20 units and 30 units respectively.
A.
B.
C.
D.
36
42
25
40
2. A rope is stretched between two rigid walls 40 feet apart. At the midpoint, a load of 100
lbs was placed that caused it to sag 5 feet. Compute the approximate tension in the rope.
A.
B.
C.
D.
206 lbs
150 lbs
280 lbs
240 lbs
3. What is the effective component applied on the box that is being pulled by a 30 N force
inclined at 30° with horizontal?
A.
B.
C.
D.
36.21 N
25.98 N
15.32 N
20.62 N
4. A post is supported by a guy wire which exerts a pull of 100 N on the top of the post. If
the angle between the wire and the ground is 60°, what is the horizontal component of the
force supporting pole?
A.
B.
C.
D.
86.6 N
50.0 N
76.6 N
98.5 N
5. The resultant of two forces in a plane is 400 N at 120°. If one of the forces is 200 lbs at
20° what is the other force?
A.
B.
C.
D.
347.77 N at 114.85°
435.77 N at 104.37°
357.56 N at 114.24°
477.27 N at 144.38°
6. Determine the resultant of the following forces: A = 600 N at 40°, B = 800 N at 160° and
C = 200 N at 300°.
A.
B.
C.
D.
532.78 N, 55.32°
435.94 N, 235.12°
522.68 N, 111.57°
627.89 N, 225.81°
7. A collar, which may slide on a vertical rod is subjected three forces. Force A is 1200 N
vertically upward, Force B is 800 N at an angle of 60° from the vertical and a force F
which is vertically downward to the right. Find the direction of F if its magnitude is 2400
N and the resultant is horizontal.
A.
B.
C.
D.
41.61°
43.52°
40.13°
45.52°
8. In the system shown, a 5 kg block rests on a horizontal table top and is attached with
horizontal string to a second string as shown. What is the maximum value for the mass,
m, if the first block is to remain stationary?
5 kg
A.
B.
C.
D.
2.89 kg
1.98 kg
2.18 kg
1.89 kg
37°
m
9. Given the 3-dimensional vectors: A = i(xy) + j(2yz) + k(3zx) and B = i(yz) + j(2zx) +
k(3xy). Determine the scalar product at the point (1,2,3).
A.
B.
C.
D.
144
138
132
126
10. Determine the divergence of the vector: V = i(xz) + j(-xy) + k(xyz) at the point (3,2,1)
A.
B.
C.
D.
9.00
11.00
13.00
7.00
11. The three vectors described by 10 cm/ at 120k degrees, k = 0,1,2 encompass the sides of
an equilateral triangle. Determine the magnitude of the vector cross product: 0.5[(10/at 0
deg) x (10/ at 120 deg)].
A.
B.
C.
D.
86.8
25.0
50.0
43.3
12. The 5 vectors: 10 cm/ at 72k degrees, k = 0,1,2,3,4 encompass the sides of a regular
pentagon. Determine the magnitude of the vector cross product 2.5[(10/at 144 deg) x
(10/at 216 deg)].
A.
B.
C.
D.
198.1
237.7
285.2
165.1
13. What is the angle between two vectors A and B if A = 4i +12j + 6k and B = 24i – 8j +
6k?
A.
B.
C.
D.
168.45°
84.32°
86.32°
-84.64°
14. Given the 3-dimensional vectors: A = i(xy) + j(2yz) + k(3zx), B = i(yz) + j(2zx) + k(3xy).
Determine the magnitude of the vector sum |A + 𝐵| at coordinates (3,2,1).
A.
B.
C.
D.
32.92
29.88
27.20
24.73
15. What is the cross-product A x B of the vectors, A = I + 4j + 6k and B = 2i +3j +5k?
A.
B.
C.
D.
i–j–k
-i + j + k
2i + 7j – 5k
2i + 7j + 5k
16. Find the magnitude and location of the resultant of the loads 2000 N and 3000 N forces
acting vertically downward at 3 m and 8 m distance from the left end of a 13 m beam.
A.
B.
C.
D.
5 kN, 6 m from the left
6 kN, 3 m from the left
10 kN, 2 m from the right
3 kN, 4 m from the left
3000 N
2000 N
3m
5m
5m
A
17. A simply supported beam is five meters in length. It carries a uniformly distributed load
including its own weight of 300 N/m and a concentrated load of 100 N, 2 meters away
from the left end. Find the reactions of reaction A at the left end and reaction B at the
right end.
A.
B.
C.
D.
RA = 810 N, RB = 700 N
RA = 820 N, RB = 690 N
RA = 830 N, RB = 680 N
RA = 840 N, RB = 670 N
18. A beam is loaded as shown. Solve for RA and RB.
40 kN
20 kN
2m
A.
B.
C.
D.
140 & 40 kN
110 & 20 kN
130 & 20 kN
120 & 50 kN
W = 20 kN/m
2m
4m
RA
RB
19. A man can exert a maximum pull of 1,000 N but wishes to lift a new stone door for his
cave weighing 20,000 N. If he uses a lever how much closer must the fulcrum be to the
stone than to his hand?
A.
B.
C.
D.
10 times nearer
20 times farther
10 times farther
20 times nearer
20. A piece of metal plate is in shape of right triangle and is suspended both ends A and B.
Considering its weight 60 N with its side b parallel to the horizontal, what is the force on
the supporting string at A which is at the left end.
A
A.
B.
C.
D.
40 N
50 N
20 N
10 N
b
B
a
21. A block weighing 500 kN rest on a ramp inclined at 25° with the horizontal. What is the
force tending to move the block down the ramp?
A.
B.
C.
D.
121 kN
265 kN
211 kN
450 kN
22. Determine the reaction RB on a simply supported beam as shown.
A.
B.
C.
D.
510 N
520 N
530 N
540 N
23. A 300-N cylindrical tank is at rest as shown. Determine force F required to move the tank
up the higher-level surface.
A.
B.
C.
D.
120.5 N
165.5 N
173.2 N
203.4 N
24. A 300 N block is at rest on an inclined plane having a slope of 4 m vertical and 12 m
horizontal. If the coefficient of friction between the block and the inclined plane is 0.18,
solve the horizontal force of motion to impend?
A.
B.
C.
D.
143.5 N
231.4 N
163.8 N
204.8 N
25. A 400 N man climbs at the middle of a 150 N ladder leaned against the wall. The top
portion of the ladder is 4 m from the ground and its bottom is 2 m from the wall.
Assuming smooth wall and a stopper at the bottom of the ladder to prevent slipping, find
the reaction at the wall.
A.
B.
C.
D.
137.5 N
145.7 N
245.6 N
143.5 N
26. A homogeneous ladder 18 ft long and weighing 120 lbs rests against a smooth wall. The
angle between it and the floor is 70 degrees. The coefficient of friction between the floor
and the ladder is 0.25. How far up the ladder can a 180 lb man walk before the ladder
slips?
A.
B.
C.
D.
17.6 ft
12.0 ft
14.6 ft
13.2 ft
27. A sphere is weighing 300 N is tied to a smooth wall by a string which makes 20° with the
wall. Find the tension of the string supporting the sphere.
A.
B.
C.
D.
324.35 N
241.15 N
312.56 N
319.84 N
28. 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.
B.
C.
D.
503.21 ft
504.76 ft
505.12 ft
506.03 ft
29. The weight of a transmission cable is 1.5 kg/m distributed horizontally. If the maximum
safe tension of the cable is 6000 kg and the allowable sag is 30 m, determine the
horizontal distance between the electric posts supporting the transmission cable.
A.
B.
C.
D.
897 m
926 m
967 m
976 m
30. A cable 45.4 m long is carrying a uniformly distributed load along its span. If the cable is
strung between two posts at the same level, 40 m apart, compute the smallest value that
the cable may sag,
A.
B.
C.
D.
12.14 m
10.12 m
9.71 m
8.62 m
31. A pipeline crossing a river is suspended from a steel cable stretched between two posts
100 m apart. The weight of the pipe is 4 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.
B.
C.
D.
9047.28 kg
9404.95 kg
9545.88 kg
9245.37 kg
32. The distance between supports of a transmission cable is 20 m apart. The cable is loaded
with a uniformly distributed load of 20 kN/m throughout its span. The maximum sag of
the cable is 4 m. What is the maximum tension of the cable if one of the support is 2
meters above the other?
A.
B.
C.
D.
415.53 N
413.43 N
427.33 N
414.13 N
33. A cable weighing 0.4 pound per foot and 800 feet long is to be suspended with sag of 80
feet. Determine the maximum tension of the cable.
A.
B.
C.
D.
403 kg
456 kg
416 kg
425 kg
34. A cable is 200 m long weighs 50 N/m and is supported from two points at the same
elevation. Determine the required sag if the maximum tension that the cable can carry
shall not exceed 8000 N.
A.
B.
C.
D.
35.1 m
28.2 m
40.3 m
31.3 m
35. A transmission cable 300 m long, weighs 600 kg. The tension at the ends of the cable are
400 kg and 450 kg. find the distance of its lowest point to the ground.
A.
B.
C.
D.
145 m
148 m
150 m
153 m
36. A 250 kg block rests on a 30° plane. If the coefficient of kinetic friction is 0.20,
determine the horizontal force P applied on the block to start the block moving up the
plane.
A.
B.
C.
D.
59.10 kg
58.10 kg
219.71 kg
265.29 kg
37. Compute the number of turns of the rope to be wound around a pole in order to support a
man weighing 600 N with an input force of 10 N. Note: coefficient of friction is 0.30.
A. 2.172
B. 3.123
C. 1.234
D. 4.234
38. A block weighing 500 N is held by a rope that passes over a horizontal drum. The
coefficient of friction between the rope and the drum is 0.15. If the angle of contact is
150°, compute the force that will raise the object.
A.
B.
C.
D.
740.7 N
760.6 N
770.5 N
780.8
39. A circle has a diameter of 20 cm. Determine the moment of inertia of the circular area
relative to the axis perpendicular to the area through the center of the circle in cm4.
A.
B.
C.
D.
14,280
15,708
17,279
19,007
40. An isosceles triangle has a 10 cm base and a 10 cm altitude. Determine the moment of
inertia of the triangular area relative to a line parallel to the base and through the upper
vertex in cm4.
A.
B.
C.
D.
2,750 cm4
3,025 cm4
2,500 cm4
2,273 cm4
SOLUTIONS TO ENGINEERING MECHANICS (STATICS) PROBLEMS
Problem 1:
R = √(30)2 + (20)2
R = 36.05
Problem 2:
tan 𝜃 =
20
5
𝜃 = 75.963°
∑ 𝐹𝑦 = 0
100 = 𝑇 cos 𝜃 + 𝑇 cos 𝜃
100 = 2 𝑇 cos 𝜃
100
T = 2 cos 75.963°
T = 206 lbs
Problem 3:
𝐹𝑒𝑓𝑓𝑒𝑐𝑡𝑖𝑣𝑒 = 𝐹 cos 𝜃
𝐹𝑒𝑓𝑓𝑒𝑐𝑡𝑖𝑣𝑒 = 30 cos 30°
𝐹𝑒𝑓𝑓𝑒𝑐𝑡𝑖𝑣𝑒 = 25.98 𝑁
Problem 4:
𝐹𝑥 = 𝐹 sin 𝜃
𝐹𝑥 = 100 cos 60°
𝐹𝑥 = 50 𝑁
Problem 5:
Cosine Law:
𝐵 2 = 𝐴2 + 𝑅 2 − 2𝐴𝑅 cos 100°
𝐵 2 = 2002 + 4002 − 2(200)(400) cos 100°
𝐵 = 477.27 𝑁
Using sine law:
Sin 100°
477.27
+
sin 𝛼
400
𝛼 = 55.62°
𝜃 + (𝛼 − 20°) = 180°
𝜃 + (55.62° − 20°) = 180°
𝜃 = 144.38°
Problem 6:
∑ 𝐹𝑥 = 600 cos 40° − 800 cos 20° + 200 sin 30°
∑ 𝐹𝑥 = −192.13 𝑁 (𝑡𝑜 𝑡ℎ𝑒 𝑙𝑒𝑓𝑡)
∑ 𝐹𝑥 = 600 sin 40° + 800 sin 20° − 200 cos 30°
∑ 𝐹𝑥 = 486.08 𝑁 (𝑢𝑝𝑤𝑎𝑟𝑑)
The Magnitude of the Resultant (R):
R = √(∑ 𝐹𝑥 )2 + (∑ 𝐹𝑦 )
2
R = √(−192.13)2 + (486.08)2
R = 522.68
The Direction of the Resultant (𝜃𝑅 ):
𝜃 = tan−1
∑ 𝐹𝑦
∑ 𝐹𝑥
𝜃 = tan−1 (
486.08
)
192.13
𝜃 = 68.43° 𝑖𝑛 𝑡ℎ𝑒 𝑆𝑒𝑐𝑜𝑛𝑑 𝑄𝑢𝑎𝑑𝑟𝑎𝑛𝑡
Refer to the vector diagram:
𝜃𝑡 = 180° − 68.43°
𝜃𝑡 = 111.57°
Problem 7:
Since the resultant is horizontal, then ∑ 𝐹𝑦 = 0.
∑ 𝐹𝑦 = 800 cos 60 + 1200 − 2400 sin 𝜃 = 0
𝜃 = 41.61°
Problem 8:
The maximum static frictional force that acts on a 5-kg block is:
𝑓𝑠 = 𝜇𝑠 𝑚𝑔
𝑚
𝑓𝑠 = (0.50)(5 𝑘𝑔)(9.81 𝑠2 )
𝑓𝑠 = 24.5 𝑁
Solve for T1:
∑ 𝐹𝑥 = 0
∑ 𝐹𝑥 = 𝑇1 cos 37° − 𝑇2 = 0
To keep the block in static equilibrium:
(T2 = fs)
24.5
𝑇1 = cos 37° = 30.7 𝑁
∑ 𝐹𝑦 = 𝑇𝑡𝑦 − 𝑚𝑔 = 0
∑ 𝐹𝑦 = 𝑇1 sin 37° − 𝑚𝑔 = 0
Then:
𝑚=
(30.7)(sin 37°)
9.81
𝑚 = 1.89 𝑘𝑔
Problem 9:
𝐴 = 𝑖(𝑥𝑦) + 𝑗(2𝑦𝑧) + 𝑘(3𝑧𝑥)
𝐵 = 𝑖(𝑦𝑧) + 𝑗(2𝑧𝑥) + 𝑘(3𝑥𝑦)
𝐴 ∙ 𝐵 = (𝑥𝑦)(𝑦𝑧) + (2𝑦𝑧)(2𝑧𝑥) + (3𝑧𝑥)(3𝑥𝑦)
𝐴𝑡 (1,2,3) → 𝑥 = 1; 𝑦 = 2; 𝑧 = 3
𝐴 ∙ 𝐵 = (1)(2)(2)(3) + (2)(2)(3)(2)(3)(1) + (3)(3)(1)(3)(1)(2)
𝐴 ∙ 𝐵 = 138
Problem 10:
Divergence = ∇ ∙ V
𝜕
𝜕
𝜕
Divergence = [𝑖 𝜕𝑥 + 𝑗 𝜕𝑦 + 𝑘 𝜕𝑧] ∙ [𝑖(𝑥 2 ) + 𝑗(−𝑥𝑦) + 𝑘(𝑥𝑦𝑧)]
Divergence = [
𝜕(𝑥 2 )
𝜕𝑥
+
𝜕(−𝑥𝑦)
𝜕𝑦
+
𝜕(𝑥𝑦𝑧)
𝜕𝑧
]
Divergence =2𝑥 − 𝑥 + 𝑥𝑦
At (3,2,1) → 𝑥 = 3; 𝑦 = 2; 𝑧 = 1
Divergence = 2(3) – 3 + (3)(2)
Divergence = 9
Problem 11:
𝑖 𝑗
𝐴 𝑥 𝐵 = [1 4
2 3
𝑘
6]
5
𝐴 𝑥 𝐵 = (20𝑖 + 12𝑗 + 3𝑘) − (8𝑘 + 5𝑗 + 18𝑖)
𝐴 𝑥 𝐵 = 2𝑖 + 7𝑗 − 5𝑘
Change the given vectors in rectangular form using calculator:
𝐴 = 10∠0° = 10 + 𝑗0
𝐵 = 10∠120° = −5 + 𝑗8.66
Re-write the given vectors into three-dimensional (i,j,k) format:
𝐴 = 10𝑖 + 0𝑗 + 0𝑘
𝐵 = −5𝑖 + 8.66𝑗 + 0𝑘
𝑖
𝑗
𝑘
𝐴 𝑥 𝐵 = [ 10
0
0]
−5 8.66 0
𝐴 𝑥 𝐵 = [0 + 0 + 86.6𝑘] − [0 + 0 + 0]
𝐴 𝑥 𝐵 = 86.6 𝑘
Thus:
0.5|𝐴𝑥𝐵| = 0.5(86.6)
0.5|𝐴𝑥𝐵| = 43.3 𝑢𝑛𝑖𝑡𝑠
Problem 12:
Convert the given vector in rectangular form.
𝐴 = 10∠144° = −8.09 + 𝑗5.877
𝐵 = 10∠216° = −8.09 + 𝑗5.877
Re-write the given vectors into three-dimensional (i,j,k) format:
𝐴 = −8.09𝑖 + 5.877𝑗 + 0𝑘
𝐵 = −8.09𝑖 − 5.877𝑗 + 0𝑘
𝑖
𝑗
𝐴 𝑥 𝐵 = [−8.09 5.877
−9.09 −5.877
𝑘
0]
0
𝐴 𝑥 𝐵 = [0 + 0 + 45.545𝑘] − [−47.545𝑘 + 0 + 0]
𝐴 𝑥 𝐵 = 95.09𝑘
Thus:
2.5|𝐴𝑥𝐵| = 2.5(95.09)
2.5|𝐴𝑥𝐵| = 237.725 𝑢𝑛𝑖𝑡𝑠
Problem 13:
𝐴 = 4𝑖 + 12𝑗 + 6𝑘
𝐵 = 24𝑖 − 8𝑗 + 6𝑘
|𝐴| = √(4)2 +(12)2 + (6)2
|𝐴| = 14
|𝐵| = √(24)2 +(−8)2 + (6)2
|𝐵| = 26
𝐴 ∙ 𝐵 = 4(24) + 12(−8) + 6(6)
𝐴 ∙ 𝐵 = 36
𝐴 ∙ 𝐵 = |𝐴||𝐵| cos 𝜃 → 𝑓𝑜𝑟𝑚𝑢𝑙𝑎
36 = 14(26) cos 𝜃
𝜃 = cos −1
36
14(26)
𝜃 = 84.324°
Problem 14:
𝐴 + 𝐵 = 𝑖(𝑥𝑦 + 𝑧𝑦) + 𝑗(3𝑦𝑧 + 2𝑧𝑥) + 𝑘(3𝑧𝑥 + 3𝑥𝑦)
𝑤ℎ𝑒𝑟𝑒:
𝑥 = 3, 𝑦 = 2, 𝑧 = 1
𝐴 + 𝐵 = 𝑖(2 + 6) + 𝑗(4 + 6) + 𝑘(9 + 18)
𝐴 + 𝐵 = 8𝑖 + 10𝑗 + 27𝑘
|𝐴 + 𝐵| = √(8)2 +(10)2 + (27)2 = 29.88
Problem 15:
𝑖 𝑗
𝐴 𝑥 𝐵 = [1 4
2 3
𝑘
6]
5
𝐴 𝑥 𝐵 = [20𝑖 + 12𝑗 + 3𝑘] − [−8𝑘 + 5𝑗 + 18𝑖]
𝐴 𝑥 𝐵 = 2𝑖 + 7𝑗 − 5𝑘
Problem 16:
∑ 𝐹𝑦 = 0
2000 + 3000 = 𝑅
𝑅 = 5000 𝑁
∑ 𝑀𝐴 = 0
2000(3) + 3000(8) = 5000(𝑥)
𝑥 = 6 𝑚 𝑓𝑟𝑜𝑚 𝑡ℎ𝑒 𝑙𝑒𝑓𝑡 𝑒𝑛𝑑
Problem 17:
𝑅 = 300(5)
𝑅 = 1500 𝑁
∑ 𝑀𝐴 = 0
100(2) + 1500(2.5) − 𝑅𝐵 = 0
𝑅𝐵 = 700 𝑁
∑ 𝑀𝐵 = 0
𝑅𝐴 (5) + 100(3) − 1500(2.5) = 0
𝑅𝐴 = 810 𝑁
Problem 18:
𝑅 = 20(6)
𝑅 = 120 𝑘𝑁
∑ 𝑀𝐵 = 0
𝑅𝐴 (4) − 120(3) − 20(6) − 40(2) = 0
𝑅𝐴 = 140 𝑘𝑁
∑ 𝑀𝐴 = 0
40(2) + 120(1) − 𝑅2 (4) − 20(2) = 0
𝑅2 = 40 𝑘𝑁
Problem 19:
∑ 𝑀𝑓𝑢𝑙𝑐𝑟𝑢𝑚 = 0
𝑊(𝑥2 ) − 𝐹(𝑥1 ) = 0
𝑊(𝑥2 ) = 𝐹(𝑥1 )
(20,000)𝑥2 = (1,000)𝑥1
𝑥1 = 20𝑥2
Problem 20:
∑ 𝑀𝐵 = 0
2𝑏
𝑅𝐴 (𝑏) − 𝑊 ( ) = 0
3
𝑅𝐴 =
2𝑏𝑤
3𝑏
𝑅𝐴 =
2(60)
3
𝑅𝐴 = 40 𝑁
Problem 21:
𝑊𝑥 = 𝑊 sin 𝜃
𝑊𝑥 = 500 sin 25°
𝑊𝑥 = 211.31 𝑘𝑁
Problem 22:
𝑥1 =
1
(36)
3
𝑥1 = 12 𝑖𝑛
𝑥1 =
1
(36)
2
𝑥1 = 18 𝑖𝑛
𝑊1 =
1
(50 − 20)(36)
2
𝑊1 = 540 𝑁
𝑊2 = (20)(36)
𝑊2 = 720 𝑁
∑ 𝑀𝐴 = 0
𝑊1 𝑥1 + 𝑊2 𝑥2 − 𝑅𝐵 (36) = 0
540(12) + 720(18) = 36𝑅𝐵
𝑅𝐵 = 540 𝑁
Problem 23:
By Pythagorean theorem:
𝑥 = √(80)2 − (40)2
𝑥 = 69.28 𝑐𝑚
∑ 𝑀𝐴 = 0
𝑊𝑥 − 𝐹(120) = 0
300(69.28) − 120 𝐹 = 0
𝐹 = 173.2 𝑁
Problem 24:
∑ 𝐹𝑥 = 0
𝐹 cos 18.43° = 300 sin 18.43° + 0.18𝑁 → (1)
∑ 𝐹𝑦 = 0
𝑁 = 𝐹 sin 18.43° + 300 cos 18.43° → (2)
Substitute eq. 2 to eq. 1
𝐹 cos 18.43° = 300 sin 18.43° + 0.18(𝐹 sin 18.43° + 300 cos 18.43°
𝐹 = 163.8 𝑁
Problem 25:
∑ 𝑀𝐵 = 0
𝑅𝐴 (4) = 550(1)
𝑅𝐴 = 137.5 𝑁
Problem 26:
∑ 𝐹𝑦 = 0
𝑁 − 120 − 180 = 0
𝑁 = 300
∑ 𝑀𝐴 = 0
𝑁 (18cos 70°) − (0.25𝑁) (18sin 70°) − (120) (9cos 70°) − 180[(18 − 𝑥) cos 70°] = 0
6.156 𝑁 − 4.228 𝑁 − 369.38 − 1108.145 + 61.563𝑥 = 0
1.928 𝑁 − 1477.525 + 61.563𝑥 = 0
Substituting the value of N:
1.928(300) − 1477.525 + 61.563𝑥 = 0
𝑥 = 14.6 𝑓𝑡
Problem 27:
cos 20° =
300
𝑇
𝑇 = 319.25 𝑁
Problem 28:
From the given condition that the loading is horizontally distributed, the cable is PARABOLIC.
Using the approximate formula
𝑆=𝐿+
8𝑑2 32𝑑 4
−
→ 𝑓𝑜𝑟𝑚𝑢𝑙𝑎
3𝐿
5𝐿3
𝑆 = 500 +
8(30)2 32(30)4
−
3(500) 5(500)3
𝑆 = 504.758 𝑓𝑡
Problem 29:
𝑤𝐿 2
𝑇 = ( ) + 𝐻 2 → 𝑓𝑜𝑟𝑚𝑢𝑙𝑎
2
2
𝐻=
𝑤𝐿2
→ 𝑓𝑜𝑟𝑚𝑢𝑙𝑎
8𝑑
Substitute eq. 2 in eq. 1
𝑤𝐿 2
𝑤𝐿2
𝑇 =( ) +(
)
2
8𝑑
2
2
1.5 2 2
1.5 2 4
6000 = ( ) 𝐿 + (
) 𝐿
2
8(30)
2
60002 = 0.5625𝐿2 + 39.0625𝑥10−4 𝐿4
0 = 𝐿4 + 14400𝐿2 − 60002 (25600)
𝐿 = 976.128 𝑚
Problem 30:
Using the approximate formula
8𝑑2 32𝑑 4
𝑆=𝐿+
−
3𝐿
5𝐿3
45.4 = 40 +
8(𝑑)2 32(𝑑)4
−
3(40) 5(40)3
5.4 = 0.06667𝑑 2 − 0.0001𝑑4
54,000 = 666.67𝑑2 − 𝑑 4
0 = 𝑑4 − 666.67𝑑2 + 54,000
𝑑 = 9.714 𝑚
Problem 31:
𝑤 = 𝑤𝑝 + 𝑤𝑐
𝑤 = 14 + 1
𝑤 = 15 𝑘𝑔/𝑚
𝐻=
𝑤𝐿2
8𝑑
𝐻=
15(100)2
8(2)
𝐻 = 9375 𝑚
𝑇 = √(
𝑤𝐿 2
) + 𝐻2
2
𝑇 = √(
15(100) 2
) + (9375)2
2
𝑇 = 9404.95 𝑘𝑔
Problem 32:
𝑤𝑥1 2
𝐻=
2𝑑1
20𝑥1 2
𝐻=
2(4)
𝐻=
𝑤𝑥2 2
2𝑑2
𝐻=
20(20 − 𝑥1 )2
2(2)
→ 𝑒𝑞. 1
→ 𝑒𝑞. 2
Equating:
20𝑥1 2 20(20 − 𝑥1 )2
=
8
4
𝑥1 2 = 2(20 − 𝑥1 )2
𝑥1 2 = 800 − 80𝑥1 + 2𝑥1 2
0 = 𝑥1 2 − 80𝑥1 + 800
𝑥1 = 11.716 𝑚
Substituting in eq. 1
20(11.716)2
𝐻=
2(4)
𝐻 = 343.16 𝑘𝑁
𝑇 = √(𝑤𝑥1 )2 + 𝐻 2
𝑇 = 415.53 𝑁
Problem 33:
𝑇 = 𝑤𝑦
𝑇 = 0.4(80 + 𝑐)
→ 𝑒𝑞. 1
𝑦2 = 𝑆2 + 𝑐2
(80 + 𝑐)2 = 4002 + 𝑐 2
6400 + 160𝑐 + 𝑐 2 = 4002 + 𝑐 2
160𝑐 = 153600
𝑐 = 960 𝑓𝑡
Substituting c in eq. 1
𝑇 = 0.4(80 + 960)
𝑇 = 416 𝑘𝑔
Problem 34:
𝑇 = 𝑤𝑦
8000 = 50𝑦
𝑦 = 160 𝑚
𝑦2 = 𝑆2 + 𝑐2
1602 = 1002 + 𝑐 2
𝑐 = 124.9 𝑚
𝑑 =𝑦−𝑐
𝑑 = 160 − 124.9
𝑑 = 35.1 𝑚
Problem 35:
𝑤=
600
= 2 𝑘𝑔/𝑚
300
𝑇1 = 𝑤𝑦1
450 = 2𝑦1
𝑦1 = 225
𝑇2 = 𝑤𝑦2
400 = 2𝑦2
𝑦2 = 200
𝑦1 2 = 𝑆1 2 + 𝑐 2
2252 = 𝑆1 2 + 𝑐 2
→ 𝑒𝑞. 1
2002 = 𝑆1 2 + 𝑐 2
→ 𝑒𝑞. 2
Subtract eq. 2 from eq. 1:
2252 − 2002 = 𝑆1 2 + 𝑆2 2
𝑆1 2 = 10625 + 𝑆2 2
→ 𝑒𝑞. 3
𝑆1 + 𝑆2 = 300
𝑆1 = 300 − 𝑆2
𝑆1 2 = (300 − 𝑆2 )2
𝑆1 2 = 3002 − 600𝑆2 + 𝑆2
2
→ 𝑒𝑞. 4
Equate eq. 3 and eq. 4:
10625 + 𝑆2 2 = 3002 − 600𝑆2 + 𝑆2
𝑆2 = 132.29 𝑚
2
Substituting 𝑆2 in eq. 2:
2002 = 132.292 + 𝑐 2
𝑐 = 150 𝑚
Problem 36:
∑ 𝐹𝑦 = 0
𝑁 = 𝑊 cos 30° + 𝑃 sin 30°
𝑁 = 250 cos 30° + 𝑃 sin 30°
𝑁 = 216.506 + 0.5𝑃
∑ 𝐹𝑦 = 0
𝑃 cos 30° = 𝑊 sin 30° + 𝐹
𝑃 cos 30° = 𝑊 sin 30° + 𝜇𝑁
0.866𝑃 = 250(0.5) + 0.2(216.506 + 0.5𝑃)
0.866𝑃 = 125 + 43.3 + 0.1𝑃
𝑃 = 219.71 𝑘𝑔
Problem 37:
𝑇1
= 𝑒 𝜇𝛽
𝑇2
𝛽 = 𝑛 𝑡𝑢𝑟𝑛𝑠 𝑥
𝛽 = 2𝑛𝜋
2𝜋𝑟𝑎𝑑
𝑡𝑢𝑟𝑛
Substituting the values:
600
= 𝑒 𝜇(2𝑛𝜋)
10
60 = 𝑒 0.3(2𝑛𝜋)
ln 100 = 0.6𝑛𝜋
𝑛 = 2,172 𝑡𝑢𝑟𝑛𝑠
Problem 38:
𝛽 = 150° (
𝜋 𝑟𝑎𝑑
)
180°
𝛽 = 2.62 𝑟𝑎𝑑
The tight side, T1 is:
𝑇1
= 𝑒 0.15(2.62)
500
𝑇1 = 740.71 𝑁
Problem 39:
𝜋𝑑4
𝐽=
32
𝜋(20)4
𝐽=
32
𝐽 = 15708 𝑐𝑚4
Problem 40:
𝐼𝑥 = 𝐼𝑥𝑜 + 𝐴𝑑2
𝑏ℎ3
1
2 2
𝐼𝑥 =
+ ( 𝑏ℎ) ( ℎ)
36
2
3
𝑏ℎ3
𝐼𝑥 =
4
𝐼𝑥 =
(10)(10)4
4
𝐼𝑥 = 2500 𝑐𝑚4
ENGINEERING MECHANICS PROBLEMS (DYNAMICS)
1
1. The motion of a particle is defined by the relation 𝑥 = (3) 𝑡 3 − 3𝑡 2 + 8𝑡 + 2 where x is
the distance in meters and t is the time in seconds. What is the time when the velocity is
zero?
A.
B.
C.
D.
2 seconds
3 seconds
5 seconds
7 seconds
2. A particle moves along a straight line with the equation 𝑥 = 16𝑡 + 4𝑡 2 − 3𝑡 3 where x is
the distance in ft and t is the time in second. Compute the acceleration of the particle after
2 seconds?
A.
B.
C.
D.
-28 ft/s2
-30 ft/s2
-17 ft/s2
-24 ft/s2
3. Two cars A and B traveling in the same direction and stopped at a highway traffic sign.
As the signal turn green, car A accelerates at a constant rate of 1 m/s2. Two seconds later,
second car B accelerates at constant rate of 1.3 m/s2. When will the second car B
overtakes the first car A?
A.
B.
C.
D.
16.27 s
30.45 s
20.32 s
10.45 s
4. Two buses start at the same time towards each other from terminals A and B, 8 km apart.
The time needed for the first bus to travel from A to B is 8 minutes, and of the second bus
from B to A is 10 minutes. How much is the time needed by each bus to meet each if they
traveled at their respective uniform speeds?
A.
B.
C.
D.
5.45 min
10.7 min
4.44 min
2.54 min
5. A train changes its speed uniformly from 60 mph to 30 mph in distance of 1500 ft. what
is its acceleration?
A.
B.
C.
D.
-1.94 ft/s2
2.04 ft/s2
-2.04 ft/s2
1.94 ft/s2
6. A car starts from rest and has a constant acceleration of 3 ft/s2. Find the average velocity
during the first 10 seconds of motion.
A.
B.
C.
D.
13 ft/s2
15 ft/s2
14 ft/s2
20 ft/s2
7. A man aimed his rifle at the bull’s eye of a target 50 m away. If the speed of the bullet is
500 m/s, how far below the bull’s eye does the bullet strikes the target?
A.
B.
C.
D.
5.0 cm
6.8 cm
5.7 cm
6.0 cm
8. A man driving his car at a constant rate of 40 mph suddenly sees a sheep crossing the
road 60 feet ahead. Compute the constant deceleration (in ft/s2) required to avoid hitting
the sheep? Assume a reaction time of 0.5 seconds before the man applies brake.
A.
B.
C.
D.
34.65
45.54
55.65
61.87
9. A ball is thrown vertically into the air at 120 m/s. After 3 seconds, another ball is thrown
vertically. What is the velocity must the second ball have to [ass the first ball at 100 m
from the ground?
A.
B.
C.
D.
105.89 m/s
107.72 m/s
108.12 m/s
110.72 m/s
10. A ball is dropped from a height of 60 meters above the ground. How long does it take to
hit the ground?
A.
B.
C.
D.
2.1 sec
3.5 sec
5.5 sec
1.3 sec
11. A ball is thrown vertically upward from the ground and a student gazing out of the
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.
B.
C.
D.
15.25 m
14.87 m
9.97 m
11.28 m
12. A ball is thrown vertically upward with an initial velocity of 3 m/s from the window of a
tall building. The ball strikes the sidewalk at the ground level 4 seconds later. Determine
the velocity with which the ball strikes the ground.
A.
B.
C.
D.
39.25 m/s
38.50 m/s
37.75 m/s
36.24 m/s
13. A player throws a baseball upward with an initial velocity of 30 ft/sec and catches it with
a baseball glove. When will the ball strike the glove? Assume the glove is position in the
same elevation when the ball left his hand.
A.
B.
C.
D.
0.48 s
0.60 s
1.20 s
1.86 s
14. A highway curve has a super elevation of 7 degrees. What is the radius of the curve such
that there will be no lateral pressure between the tires and the roadway at a speed of 40
mph?
A.
B.
C.
D.
265.71 m
438.34 m
345.34 m
330.78 m
15. A baseball is thrown from a horizontal plane following parabolic path with an initial
velocity of 100 m/s at an angle of 30° above the horizontal. Solve the distance from the
throwing point that the ball attains its original level.
A.
B.
C.
D.
890 m
883 m
858 m
820 m
16. Compute the minimum distance that a truck slides on a horizontal asphalt road if it is
traveling at 20 m/s? the coefficient to sliding friction between asphalt and rubber tire is at
0.50. the weight of the truck is 8000 kg.
A.
B.
C.
D.
40.8
48.5
35.3
31.4
17. 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.
B.
C.
D.
15.74 km
14.54 km
12.31 km
20.43 km
18. A 25 g mass bullet was fired at the wall. The bullet’s speed upon hitting the wall is 350
m/s. what is the average force (in Newton) if the bullet penetrates 10 cm?
A.
B.
C.
D.
14,543.2 N
11,342.2 N
10,543.3 N
15,312.5 N
19. A girl tied 80-gram toy plane of a string which he rotated to form a vertical circular
motion with a diameter a 1000 mm. compute for the maximum pull exerted on the string
by the toy plane if it got loose leaving at the bottom of the circle at 25 m/s.
A.
B.
C.
D.
0.002 kN
0.05 kN
0.2 kN
0.1 kN
20. A gun is shot into 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.
B.
C.
D.
0.51 m
0.53 m
0.32 m
0.12 m
21. A train weighing 1000 kN is being pulled up a 2 grade. The train’s resistance is 5 N/kN.
The train’s velocity was increased from 6 to 12 m/s in a distance of 300 m. compute the
maximum power developed by the locomotive.
A.
B.
C.
D.
600 kW
450 kW
520 kW
320 kW
22. 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.
A.
B.
C.
D.
13.45°
12.71°
11.23°
10.45°
23. An airplane acquires a take-off velocity of 150 mph on a 2-mile runway. If the plane
started from rest and the acceleration remain constant, what is the time required to reach
take-off speed?
A.
B.
C.
D.
40 s
45 s
58 s
96 s
24. Water drops from a faucet at the rate of 4 drops per second. What is the distance between
two successive drops 1 second after the first drop has fallen.
A.
B.
C.
D.
5.32 ft
8.24 ft
7.04 ft
9.43 ft
25. A body which is 16.1 lb rests on a horizontal plane and acted upon by a 10-lb force. Find
the acceleration of the body if the coefficient of friction between the plane and the body
is 0.2. Note: 1 lbf = 32.2 lbm-ft/s2
A.
B.
C.
D.
12.34 ft/s2
11.57 ft/s2
15.57 ft/s2
13.56 ft/s2
26. A man on an elevator weighs 180 lbf. Compute the force exerted by the man on the floor
of the elevator if it is accelerating upward at 5 ft/s2.
A.
B.
C.
D.
207.95 lbf
210.45 lbf
190.56 lbf
205.54 lbf
27. A 10-lb stone is fastened to a 2-ft cord and is whirled in a vertical circle. Determine the
tension in the cord when it is rotated at 100 rpm.
A.
B.
C.
D.
47.95 lbf
58.08 lbf
19.56 lbf
20.54 lbf
28. An archer must split the apple atop his partner’s head from a distance of 30 m. the arrow
I horizontal when aimed directly to the apple. At what angle must he aim in order to hit
the apple with the arrow traveling at a speed of 35 m/s?
A.
B.
C.
D.
8.35°
10.55°
3.25°
6.95°
29. A hollow spherical shell has a radius of 5 units and mass of 10. What is its mass mpment
of inertia?
A.
B.
C.
D.
108.45
123.34
187.54
166.67
30. A coin 20 mg is place on the smooth edge of a 25 cm-radius phonograph record as the
record is brought up to its normal rotational speed of 45 rpm. What must be the
coefficient of friction between the coin and the record if the coin is not to slip off?
A.
B.
C.
D.
0.45
0.56
0.64
0.78
31. The acceleration due to gravity on the moon is 1.67 m/s2. If an astronaut can throw a ball
10 m straight upward on earth, how high should this man be able to throw the ball on the
moon? Assume that the throwing speeds are the same in the two cases.
A.
B.
C.
D.
58.67
50.84
65.67
45,67
32. A tennis ball is dropped into a certain height of 2 m. It rebounds to a height of 1.8 m.
What fraction of energy did it lose in the process of striking the floor?
A.
B.
C.
D.
One-tenth
One-fourth
One-third
One-seventh
33. A car is a rest on a sloping driveway. By experiment the driver releases the break of the
car and let the car move at a constant acceleration. How fast will the car be moving when
it reaches the street? Note: the street is 4 m below the original position of the car.
A.
B.
C.
D.
8.86 m/s
5.45 m/s
6.65 m/s
9.65 m/s
34. A solid sphere is placed at the top of a 45°incline. When released, it freely rolls down.
What will be its linear speed at the foot of the incline which is 2.0 m below the initial
position of the cylinder?
A.
B.
C.
D.
4.86 m/s
5.29 m/s
6.43 m/s
3.55 m/s
35. A ball is dropped from a height y above a smooth floor. How high will rebound if the
coefficient of restitution between the ball and the floor is 0.60?
A.
B.
C.
D.
0.45y
0.40y
0.60y
0.36y
36. A ball is thrown at na angle of 32.5° from the horizontal towards a smooth floor. At what
angle will it rebound if the coefficient of between the ball and the floor is 0.30?
A.
B.
C.
D.
11.33°
8.67°
9.12°
10.82°
37. A 1.62-ounce marble attains velocity of 170 mph (249.3 ft/s) in a hunting slingshot. The
contact with the sling is 1/15th second. What is the average force on the marble during
contact?
A.
B.
C.
D.
12.54 lbf
14.56 lbf
11.75 lbf
10.67 lbf
38. A man weighs 128 lb on the surface of the earth (radius = 3960 miles). At what distance
above the surface of the earth would he weight 80 lb?
A.
B.
C.
D.
3000 miles
2345 miles
7546 miles
1059 miles
39. A steel wheel 800 mm in diameter rolls on a horizontal steel rail. It carries a load of 700
N. the coefficient of rolling resistance is 0.250 mm. What is the force P necessary to roll
the wheel along the rail?
A.
B.
C.
D.
0.34 N
0.54 N
0.44 N
0.14 N
40. An electron strikes the screen of the cathode ray tube with a velocity of 10 to the 9th
power cm/s. Compute its kinetic energy in erg. The mass of an electron is 9 x 10-31 kg?
A.
B.
C.
D.
4.5 x 10-10 erg
3.0 x 10-10 erg
2.5 x 10-10 erg
1.5 x 10-10 erg
SOLUTIONS TO ENGINEERING MECHANICS (DYNAMICS) PROBLEMS
Problem 1:
1
𝑥 = ( ) 𝑡 3 − 3𝑡 2 + 8𝑡 + 2
3
𝑉=
𝑑𝑥
𝑑𝑡
= 𝑡 2 − 6𝑡 + 8
𝑊ℎ𝑒𝑛 𝑉 = 0
𝑡 2 − 6𝑡 + 8
(𝑡 − 4)(𝑡 − 2) = 0
𝑡=2
𝑡=4
Problem 2:
𝑥 = 16𝑡 + 4𝑡 2 − 3𝑡 3
𝑉=
𝑎=
𝑑𝑥
= 16 + 8𝑡 − 9𝑡 2
𝑑𝑡
𝑑2 𝑥
𝑑𝑡 2
= 8 − 18𝑡
𝐴𝑡 𝑡𝑖𝑚𝑒, 𝑡 = 2 𝑠𝑒𝑐𝑜𝑛𝑑𝑠:
𝑎 = 8 − 18(2)
𝑎 = −28 𝑓𝑡/𝑠 2
Problem 3:
1
𝑆𝐴 = 𝑉𝐴 𝑡𝐴 + 𝑎𝐴 𝑡𝐴 2
2
1
𝑆𝐴 = 0 + (1)𝑡 2
2
𝑆𝐴 = 0.5𝑡 2
→ 𝑒𝑞. 1
1
𝑆𝐵 = 𝑉𝐵 𝑡𝐵 + 𝑎𝐵 𝑡𝐵 2
2
1
𝑆𝐵 = 0 + (1.3)(𝑡 − 2)2
2
𝑆𝐵 = 0.65(𝑡 − 2)2
→ 𝑒𝑞. 2
Equate eq. 1 and eq. 2
0.5𝑡 2 = 0.65(𝑡 − 2)2
𝑡 = 16.27 𝑠
Problem 4:
𝑆1 + 𝑆2 = 8000
16.67𝑡 + 13.33𝑡 = 8000
𝑡 = 266.67 𝑠
Problem 5:
𝑉0 = 60
𝑚𝑖 5280𝑓𝑡 1ℎ𝑟
𝑥
𝑥
ℎ𝑟
1𝑚𝑖
3600𝑠
𝑉0 = 88
𝑓𝑡
𝑠
𝑉 = 30
𝑚𝑖 5380𝑓𝑡 1ℎ𝑟
𝑥
𝑥
ℎ𝑟
1𝑚𝑖
3600𝑠
𝑉 = 44
𝑓𝑡
𝑠
𝑉 2 = 𝑉0 2 + 2𝑎𝑆
(44)2 = (88)2 + 2𝑎(1500)
𝑎 = −1.936
𝑓𝑡
𝑠2
Problem 6:
1
𝑆 = 𝑉0 𝑡 + 𝑎𝑡 2
2
1
𝑆 = (0)(10) + (3)(10)2
2
𝑆 = 150 𝑓𝑡
𝑉𝑎𝑣𝑒 =
𝑆
𝑡
𝑉𝑎𝑣𝑒 =
150
10
𝑉𝑎𝑣𝑒 = 15
𝑓𝑡
𝑠
Problem 7:
𝑥 = 𝑉𝑥 𝑡
𝑡=
𝑥
𝑉𝑥
𝑡=
50
500
𝑡 = 0.1 𝑠
1
𝑦 = 𝑉𝑥 𝑡 + 𝑔𝑡 2
2
1
𝑦 = (0)(0.1) + (9.81)(0.1)2
2
𝑦 = 0.049 𝑚
𝑦 = 5 𝑐𝑚
Problem 8:
𝑉0 = 40
𝑚𝑖 5280𝑓𝑡 1ℎ𝑟
𝑥
𝑥
ℎ𝑟
1𝑚𝑖
3600𝑠
𝑉0 = 58.67
𝑓𝑡
𝑠
𝑆 = 𝑉0 𝑡
𝑆 = 58.67(0.5)
𝑆 = 29.33 𝑓𝑡
𝑉 2 = 𝑉1 2 − 2𝑎(60 − 𝑆)
02 = (61.6)2 − 2𝑎(60 − 29.33)
𝑎 = −61.87
𝑓𝑡
𝑠2
Problem 9:
The first ball must be on its way down as it was passed by the first ball 100 m from the ground.
1
ℎ = 𝑉0 𝑡 + 𝑔𝑡 2
2
1
100 = (120)𝑡 + (9.81)(𝑡)2
2
0 = 4.905𝑡 2 − 120𝑡 + 100
𝑡 = 23.6 𝑠
𝑡 = 0.863 𝑠
Considering quadratic equation, for one height(h) from the ground there are two values od “t” (as
solved). After 0/863 s the ball is 100 m from the ground on its way upward until it reached the
highest point. After 23.6 seconds the ball is again 100 m from the ground but on its way
downward.
Problem 16:
∑ 𝐹𝑥 = 0
𝜇𝑁 = 𝑚𝑎
→ 𝑒𝑞. 1
∑ 𝐹𝑦 = 0
𝑊 = 𝑁 = 𝑚𝑔 → 𝑒𝑞. 2
Substitute eq. 2 in eq. 1:
𝑎=
𝜇𝑊 𝜇𝑚𝑔
=
𝑚
𝑚
𝑎 = 𝜇𝑔
𝑎 = 0.5(9.81)
𝑎 = 4.905
𝑚
𝑠2
𝑉 2 − 𝑉0 2
𝑆=
2𝑎
0 − 202
𝑆=
2(−4.905)
𝑆 = 40.77 𝑚
Problem 17:
𝑦 = 𝑥 tan 𝜃 −
𝑔𝑥 2
2𝑉𝑜 2 cos2 𝜃
−300 = 𝑥 tan 35° −
9.81𝑥 2
2(400)2 cos2 35°
𝑥 = 15743.4 𝑚
𝑥 ≈ 15.7 𝑘𝑚
Problem 18:
Impulse-Momentum Theorem:
𝐹 ∙ Δ𝑡 = 𝑚 ∙ ∆𝑉
𝐹 = 𝑚𝑎
0.025(02 − 3502 )
𝐹=
2(0.10)
𝐹 = 15,312.5 𝑁
Problem 19:
∑ 𝐹𝑦 = 0
𝑇=𝑊+
𝑚𝑉 2
𝑅
𝑇 = 0.08(9.81) +
(0.08)(25)2
(0.5)
𝑇 = 100.78 𝑁
Problem 20:
Law of Conservation of Momentum
∑ 𝑃𝑏𝑒𝑓𝑜𝑟𝑒 = ∑ 𝑃𝑎𝑓𝑡𝑒𝑟
𝑚𝑏 𝑉𝑏 + 𝑚𝐵 𝑉𝐵 = (𝑚𝑏 + 𝑚𝐵 )𝑉
(5)(320) + (500)(0) = (5 + 500)𝑉
𝑉 = 3.16
𝑚
𝑠
Law of Conservation of Energy:
𝐾𝐸 = 𝑃𝐸
1
(𝑚 + 𝑚𝐵 )𝑉 2 = (𝑚𝑏 + 𝑚𝐵 )𝑔ℎ
2 𝑏
ℎ=
𝑉2
2𝑔
ℎ=
3.162
2(9.81)
ℎ = 0.51 𝑚
Problem 21:
The 2% upgrade is:
tan 𝜃 = 0.02
For very small angle (𝜃):
sin 𝜃 = tan 𝜃 = 0.02
Work – Energy Theorem:
𝑊𝑝𝑜𝑠𝑖𝑡𝑖𝑣𝑒 − 𝑊𝑛𝑒𝑔𝑎𝑡𝑖𝑣𝑒 = ∆𝐾𝐸
𝑊
𝑃(300) − (𝑊 sin 𝜃)(300) − 𝑅(300) = 2𝑔 (𝑉 2 − 𝑉0 2 )
1000
300𝑃 − 1000(0.02)(300) − 0.005(1000)(300) = 2(9.81) [122 − 62 ]
300𝑃 = 13004.6
𝑃 = 43.34 𝑘𝑁
The maximum power developed:
𝑚
𝑃𝑜𝑤𝑒𝑟 = (43.34 𝑘𝑁) (12 𝑠 )
𝑃𝑜𝑤𝑒𝑟 = 520.2 𝑘𝑊
Problem 22:
𝑉2
tan 𝜃 =
𝑔𝑅
2
tan 𝜃 =
5280
(45 (3600))
32.2(600)
𝜃 = 12.71°
Problem 23:
𝑉𝑓 2 − 𝑉𝑜 2 = 2𝑎𝑆
𝑎=
(220)2 − 0
2(10,560)
𝑎 = 2.29
𝑓𝑡
𝑠2
𝑉𝑓 − 𝑉𝑜 = 𝑎𝑡
𝑡=
220 − 0
2.29
𝑡 = 96 𝑠
Problem 24:
1
𝑆 = 𝑉𝑜 𝑡 + 𝑔𝑡 2
2
→ 𝑓𝑜𝑟𝑚𝑢𝑙𝑎
1
𝑆1 = 0 + (32.2)(1)2 = 16.1 𝑓𝑡
2
1
3 2
𝑆2 = 0 + (32.2) ( )
2
4
𝑆2 = 9.056 𝑓𝑡
𝑑 = 𝑆1 − 𝑆2
𝑑 = 16.1 − 9.056
𝑑 = 7.044 𝑓𝑡
Problem 25:
D’Alembert’s Principle:
∑ 𝐹𝑥 = 0
10 − 𝜇𝑊 − 𝑅𝐸𝐹 = 0
𝑎=
10 − 0.2(16.1)
16.1
(32.2)
𝑎 = 13.56
𝑓𝑡
𝑠2
Problem 26:
D’Alembert’s principle:
∑ 𝐹𝑦 = 0
𝑅 − 𝑊 − 𝑅𝐸𝐹 = 0
𝑅 = 180 + (
180
) (5)
32.2
𝑅 = 207.95 𝑙𝑏𝑓
Problem 27:
∑ 𝐹𝑦 = 0
𝑇 = 𝑚𝑅𝜔2 − 𝑊
10
𝑇=(
) (2)(10.47)2 − 10
32.2
𝑇 = 58.08 𝑙𝑏𝑓
Problem 28:
𝑉𝑜 2 sin 2𝜃
𝑥=
𝑔
30 =
(35)2 sin 2𝜃
9.81
𝜃 = 6.95°
Problem 29:
2
𝐼 = 𝑚𝑅 2
3
2
𝐼 = (10)(5)2
3
𝐼 = 166.67
Problem 30:
friction = centrifugal force
𝑊
𝜇𝑁 = ( ) 𝑅𝜔2
𝑔
2𝜋 2
25 (45𝑥 60 )
𝜇=
981
𝜇 = 0.56
Problem 31:
On earth:
𝑉𝑒 2 = 2𝑔𝑒 (10)
On moon:
𝑉𝑚 2 = 2𝑔𝑚 ℎ𝑚
The throwing velocities are the same, so the second expression can be divided by the first and
will arrive:
ℎ𝑚𝑜𝑜𝑛 = (10) (
𝑔𝑒
9.81
) = (10) (
)
𝑔𝑚
1.67
ℎ𝑚𝑜𝑜𝑛 = 58.67 𝑚
Problem 32:
Fraction of Energy Lost =
𝑃𝐸2 − 𝑃𝐸1
𝑃𝐸1
Fraction of Energy Lost =
𝑚𝑔ℎ2 − 𝑚𝑔ℎ1
𝑚𝑔ℎ1
Fraction of Energy Lost =
ℎ2 − ℎ1
ℎ1
Fraction of Energy Lost =
2.0 − 1.8
2.0
Fraction of Energy Lost = 0.10
Problem 33:
PE = KE
1
mgh = 𝑚𝑉 2
2
V = √2𝑔ℎ
V = √2(9.81)(4)
V = 8.86
𝑚
𝑠
Problem 34:
PE = KE𝑡𝑜𝑡𝑎𝑙
1
1
mgh = 𝑚𝑉 2 + 𝐼𝜔2
2
2
1
1 2
𝑉 2
2
2
mgh = 𝑚𝑉 + ( 𝑚𝑅 ) ( )
2
2 5
𝑅
gh =
1 2 1 2
𝑉 + 𝑉
2
5
9.81(2.0) = 0.7𝑉 2
𝑉 = 5.29
𝑚
𝑠
Problem 35:
𝑒=√
h𝑔
h𝑑
h𝑅
0.60 = √
y
h𝑅 = 0.36𝑦
Problem 36:
𝑒 = cot 𝜃𝑖𝑛𝑐𝑖𝑑𝑒𝑛𝑡 tan 𝜃𝑟𝑒𝑏𝑜𝑢𝑛𝑑
0.30 = tan 𝜃𝑅 (
1
)
tan32.5°
𝜃𝑅 = 10.82°
Problem 37:
Impulse-Momentum Theorem:
𝐹 ∙ ∆𝑡 = 𝑚 ∙ ∆𝑉
𝑓𝑡
1.62 𝑜𝑧
𝑜𝑧 ) (249.3 𝑠 )
16
𝑙𝑏𝑚
𝐹=
𝑓𝑡
𝑙𝑏𝑚 − 2
𝑠 ) ( 1 𝑠)
(32.2
𝑙𝑏𝑓
15
(
𝐹 = 11.75 𝑙𝑏𝑓
Problem 38:
From Newton’s Law of Universal Gravitation:
On Earth:
𝑊𝑒 =
𝐺𝑚𝑚 𝑚𝑒
𝑅𝑒 2
→ 𝑒𝑞. 1
Above the earth:
𝑊𝑚 =
𝐺𝑚𝑚 𝑚𝑒
𝑅2
→ 𝑒𝑞. 2
Dividing eq. 2 by eq. 1:
𝐺𝑚𝑚 𝑚𝑒
𝑅2
= 𝑊/𝑊𝑒
𝐺𝑚𝑚 𝑚𝑒
𝑅𝑒 2
𝑊𝑒 𝑅𝑒 2 = 𝑊𝑅 2
128(3960)2 = 80(𝑅)2
𝑅 = 5009 𝑚𝑖𝑙𝑒𝑠
𝑑 = 5009 − 3960 = 1049 𝑚𝑖𝑙𝑒𝑠
Problem 39:
𝑃=
𝜇𝑓 𝑁
𝑅
𝑃=
(0.250)700
400
𝑃 = 0.14𝑁
Problem 40:
1
𝐾𝐸 = 𝑚𝑉 2
2
1
𝑐𝑚 2
𝐾𝐸 = (9𝑥10−28 𝑔)(1𝑥109
)
2
𝑠
𝐾𝐸 = 4.5𝑥10−10 𝑔 ∙ 𝑐𝑚/𝑠 2
𝐾𝐸 = 4.5𝑥10−10 𝑑𝑦𝑛𝑒 − 𝑐𝑚
𝐾𝐸 = 4.5𝑥10−10 𝑒𝑟𝑔