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Engineering Mechanics: Newton's Laws of Motion, Friction, and Rotary Motion, Study notes of Mechanics

Liceo de Cagayan University (LDCU)Mechanics

Engineering Mechanics is the branch of physical science that deals with the study of bodies and systems of and the forces acting on them.

Typology: Study notes

2020/2021

Available from 01/13/2022

Jess_Rangcasajo
Jess_Rangcasajo 🇵🇭

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ENGINEERING MECHANICS (Part 1)
Content:
1. Branches of Mechanics
2. Vector and Scalar Quantities
3. Resultant of two or more vector
4. Forces in equilibrium
5. Rectilinear motion
6. Free fall
7. Curvilinear motion
8. Newton’s law of motion
9. Friction
Engineering Mechanics is the branch of physical science
that deals with the study of bodies and systems of and
the forces acting on them.
Branches of Mechanics
1. Statics : deals with bodies in the state of rest
2. Dynamics : deals with bodies in motion
a. Kinematics: study of motion without reference
to the forces that causes motion
b. Kinetics: study of motion with reference to
the forces that causes motion
Vector and Scalar Quantities
Scalar quantity : a physical quantity, which has
magnitude only
Example: temperature, height, mass, age, speed,
distance, etc.
Vector quantity : a physical quantity which has both
magnitude and direction
Resultant of two or more vectors
For parallel vectors:
For vectors parallel horizontally:
R=F1± F2± F3± … ..± Fn
For vectors parallel horizontally:
R=F1± F2± F3± … ..± Fn
For Perpendicular vectors:
R=
√
(F1)2+(F2)2
Vectors acting at an angle other than 90º
Parallelogram Method:
R=
√
(F1)2+(F2)2−2F1F2cos θ
Component method:
R=
√
(
∑Fx
)
2+
(
∑Fy
)
2
θ=arctan
(
∑Fy
∑Fx
)
FORCES OF EQUILIBRIUM
TYPES OF EQUILIBRIUM:
Static equilibrium: is the condition of a body at rest
and remains at rest under the action of concurrent
forces.
Translational Equilibrium: is the condition of a body in
motion with constant motion
Condition of Equilibrium:
1. The vector sum of all forces acting on the body
must be equal to zero.
∑Fx=0
∑Fy=0
1 --- GEAS MECHANICS by EJBR
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ENGINEERING MECHANICS (Part 1)

Content:

  1. Branches of Mechanics
  2. Vector and Scalar Quantities
  3. Resultant of two or more vector
  4. Forces in equilibrium
  5. Rectilinear motion
  6. Free fall
  7. Curvilinear motion
  8. Newton’s law of motion
  9. Friction Engineering Mechanics is the branch of physical science that deals with the study of bodies and systems of and the forces acting on them. Branches of Mechanics
  10. Statics : deals with bodies in the state of rest
  11. Dynamics : deals with bodies in motion a. Kinematics: study of motion without reference to the forces that causes motion b. Kinetics: study of motion with reference to the forces that causes motion Vector and Scalar QuantitiesScalar quantity : a physical quantity, which has magnitude only Example: temperature, height, mass, age, speed, distance, etc.  Vector quantity : a physical quantity which has both magnitude and direction Resultant of two or more vectors For parallel vectors: For vectors parallel horizontally: R = F 1 ± F 2 ± F 3 ± … .. ± Fn For vectors parallel horizontally: R = F 1 ± F 2 ± F 3 ± … .. ± Fn For Perpendicular vectors: R =√( F 1 ) 2 +( F 2 ) 2 Vectors acting at an angle other than 90º Parallelogram Method: R =√( F 1 ) 2 +( F 2 ) 2 − 2 F 1 F 2 cos θ Component method: R =√(∑ Fx ) 2 +(∑ F (^) y ) 2 θ =arctan ( ∑ F^ y ∑ Fx )^ FORCES OF EQUILIBRIUM TYPES OF EQUILIBRIUM:Static equilibrium: is the condition of a body at rest and remains at rest under the action of concurrent forces.  Translational Equilibrium: is the condition of a body in motion with constant motion Condition of Equilibrium:
  12. The vector sum of all forces acting on the body must be equal to zero. ∑ F^ x =^0 ∑ F^ y =^0
  1. The sum of all torques acting on a body must be equal to zero. ∑ M^ 0 =^0 RECTILINEAR MOTION Uniform Motion: is a motion by which there is no change in the magnitude of the velocity. S = v x t Where: S – distance v – uniform speed or velocity t - time Uniformly Accelerated Motion Uniformly accelerated motion is a motion by which velocity increases uniformly  vf^2 − vo^2 = ± 2 ass = vo t ± 1 2 a t 2  vf − vo = ±ats = ( vf + vo ) t 2

Free fall

A free fall is a motion under gravity.  v f − vo = ± >¿  vf^2 − vo^2 = ± ghh = vo t ± 1 2 g t 2 CURVILINEAR MOTION Projectile: is a body which after being given an initial velocity with an angle of release is allowed to travel under the action of gravity only. At any point: Maximum Height: Time to reach the highest point: Range: Total time of flight:

ENGINEERING MECHANICS (Part 2)

Content:

  1. Impulse & Momentum
  2. Rotary Motion
  3. Circular Motion
  4. Rotation of Rigid Bodies
  5. Banking of Curves
  6. Cables Impulse and Momentum Impulse (I) is the product of force and the time acts. I = F x t Momentum (P) is the product of the mass and the velocity of the body P = m x v Impulse-Momentum Theorem  Impulse is equal to the change in momentum I = ∆ P Ft = m v 2 − m v 1 Law of conservation of Momentum When two bodies of masses m1 and m2 collide, the total momentum before the impact is equal to the total momentum after impact Pbefore = Pafter m 1 v 1 + m 2 v 2 = m 1 v 1 '
  • m 2 v 2 ' Types of Collision Collision: refers to the mutual action of the molecules, atoms, etc. whenever they encounter one another
  1. Elastic Collision: is a collision that conserves kinetic energy
  2. Inelastic Collision: is a collision that does not conserve energy
  3. Perfectly Inelastic Collision: is the collision where the objects stick together after the impact. In this type of collision, Kinetic energy loss is maximum. Coefficient of Restitution (e) e = relative velocity of recession relative velocity of approach = v 2 − v 1 v 1 − v 2 Note: e = 0, for perfectly inelastic collision e = 1, for perfectly elastic collision Special Cases:
  4. If a ball is dropped from a height ho and rebounds to a height hr , the coefficient of restitution between the ball and the floor is: e =

hr ho Where: ho – original height hr – height of rebound

  1. If a ball is thrown at an angle θ 1 with the normal to a smooth surface and rebounds at an angle θ 2 e = tan θ 1 tan θ 2 ROTARY MOTION

Uniform Motion: θ = ω x t Uniformly Accelerated Motion θ = ωo t ± 1 2 α t 2 ωf = ωo ± αt θ = ( ωf + ωo ) t 2 Relationship between angular and tangential quantities: S = rθ v = rω a = rα Note: θ , ω , ∧ α Must be in rad, rad/s, and rad/s^2 respectively Uniform Circular motion Centripetal force: Fc = mv 2 r = mr ω 2 Centripetal acceleration: ac = v 2 r = r ω 2 Note: Centripetal force must be directed toward the center of the circular path ROTATION OF RIGID BODIES The kinetic energy of rotation: KEr = 1 2 l ω 2 Total Kinetic Energy KET = KEr + KEt KET = 1 2 lω 2

1 2 mv 2 Where: KEr – kinetic energy of rotation KEt – kinetic energy of translation KET – total kinetic energy l – moment of inertia ω – angular velocity, (rad/s) BANKING OF CURVES Ideal angle of banking θ =tan − 1 v 2 gr For maximum velocity of the car without skidding tan (^ θ + ∅ )= v 2 gr Where: θ – angle of banking ∅ - angle of friction v -velocity r – radius of curvature CABLES Parabolic Cables:

a. internal force b. external force c. body force d. surface force

  1. The moment of a force about a point P is equal to the sum of the moments of its components about P. a. Cavalieri’s Theorem b. Pascal’s Theorem c. Varignon’s Theorem d. Torricelli’s Theorem
  2. A type of force acting on a body due to the acceleration of gravity. a. load b. shear c. bear d. mass
  3. A type of force acting on a body caused by the friction between the body and the ground. a. load b. shear c. bear d. mass
  4. The gravity in the moon is about a. 1.6 m/s^2 b. 2.6 m/s^2 c. 3.6 m/s^2 d. 0.6 m/s^2
  5. If the force is moved in the direction parallel to the direction of the force, the moment exerted by the force ___________. a. increases b. decreases c. is unchanged d. becomes zero
  6. He is the father of the modern engineering mechanics a. Gilbert Lewis b. Stephen Timoshenko c. J. Gordon d. A. Cotrell
  7. It is a method of applying mechanics that assumes all objects are continuous. a. Discrete Mechanics b. Finite Element Method c. Continuum Mechanics d. Contact Mechanics
  8. Which of the following is an example of contact force? a. gravitational force b. magnetic force c. air resistance force d. electric force
  9. In fluids, _________ is used to assess to what extent the approximation of continuity can be made. a. Brayton Number b. Knudsen Number c. Reynolds Number d. Prandtl Number
  10. It is a mathematical property of a section concerned with a surface area and how that area is distributed about the reference axis. a. moment of area b. second moment of area c. third moment of area d. fourth moment of area
  11. It is the material’s ability to resist twisting a. mass moment of inertia b. moment of area c. second moment of area d. polar moment of area
  12. It is the upward force on an object produced by the surrounding fluid in which it is fully or partially immersed. a. Archimedes’ force b. fluid pressure c. buoyancy d. weight reaction
  13. A rock of weight 10 N suspended by a string is lowered into water, displacing water of weight 3 N. Determine the tension in the string. a. 13 N b. 7 N c. 10 N d. 3 N
  14. Determine the magnitude of the force vector F = 20i + 60j – 90k (N). a. 130 N b. 120 N c. 100 N d. 110 N
  15. Determine the dot product of the two vectors U = 8i – 6j + 4k and V = 3i + 7j + 9k. a. 18 b. 16 c. 14 d. 12
  1. Two perpendicular vectors are given in terms of their components by U = Uxi – 4j + 6k and V = 3i + 2j – 3k. Determine the component Ux. a. 5. b. 6. c. 7. d. 8.
  2. The motion of a particle is defined by the relation x = (1/3)t^3 – 3t^2 + 8t + 2 where x is the distance in meters and is the time in seconds. What is the time when the velocity is zero? a. 2 seconds b. 3 seconds c. 5 seconds d. 7 seconds
  3. In ____ the members are subjected to bending action a. Forces b. Trusses c. Frames d. Structure
  4. In ____ the internal force in a bar is directed along the axis of bars. a. Forces b. Trusses c. Frames d. Structure
  5. An axis passing through the centroid of an area is known as a___. a. Parallel axis b. Centroidal axis c. Radius d. Perpendicular axis
  6. The ball is thrown vertically upward with an initial velocity of 3m/s from a window of a tall bldg. The ball strikes the ground level 4 seconds later. Determine the height of window above the ground. a. 66.331m b. 66.450m c. 67.239m d. 67.492m
  7. A projectile leaves a velocity of 50m/s at an angle of 30º with the horizontal. Find the time it would take for the projectile to reach the maximum height. a. 2.55s b. 2.60s c. 3.10s d. 2.89s TAKE HOME ACTIVITYAnswer must be submitted in the next meeting!
  8. These equations state that changes in momentum of fluid particles depend only on the external pressure and internal viscous forces acting on the fluid. a. Navier – Stokes Equations b. Torricelli Equations c. Reynolds Equations d. Lagrangian Equations
  9. The equilibrant of the forces 10 N at 10° and 15 N at 100° is a. A.18 N at 246° b. B. 18 N at 66° c. C. 25 N at -114° d. D. 25 N at 66°
  10. The second moment of area is an important value which is used to __________. It can also be called moment of inertia. a. determine the state of stress in a section b. calculate the resistance to buckling c. determine the amount of deflection in a beam d. all of the above
  11. If a 10-kg object experiences a 20-N force for a duration of 0.05-second, then what is the momentum change of the object? a. 1 N-s b. 400 N-s c. 0.5 N-s d. 200 N-s
  12. When hit, the velocity of a 0.2 kg baseball changes from +25 m/s to -25 m/s. What is the magnitude of the impulse delivered by the bat to the ball? a. 1 N-s b. 5 N-s c. 10 N-s d. 20 N-s
  13. When a block is place on an inclined plane, its steepest inclination to which the block will be in equilibrium is called _____. a. angle of friction b. angle of reaction c. angle of normal d. angle of repose
  1. A 0.324kg ball is stuck 0.54m from the center of a disk spinning at 5.55rad/s. what is its angular momentum? a. 0.5243Js* b. 0.6321Js c. 1.021Js d. 1.1524J*s
  2. A 5.2 kg box sits on a ramp inclined at 42.4º to horizontal. What is the normal force on the box from the ramp? a. 37.64N b. 37.53N c. 37.54N d. 37.63N
  3. What height above the water does a 133.9kg diver jumps need to jump from for gravity to do 6062J of work on him/her? a. 4.61m b. 4.60m c. 4.62m d. 6.43m
  4. A lady is pulling a cart (total 55.7kg) with a force of 395N. Neglecting friction, what is the acceleration of the cart? a. 7.1 (^) m / s^2 b. 7.092 (^) m / s^2 c. 7.091 m / s^2 d. 7.093 (^) m / s^2
  5. How much work must be done to move a 34.6kg box 3.66m across the floor if the coefficient of kinetic friction between the box and the floor is 0.3. a. 372.3J b. -321.9J c. -372.3 J d. 3.234 J --- end ---