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This comprehensive overview explores the branches of mechanics: kinematics and dynamics, emphasizing Newton's Laws of Motion. Kinematics focuses on motion analysis without considering force, examining concepts like speed, velocity, and acceleration through various motion examples. Dynamics, in contrast, investigates the forces causing motion. Key concepts include types of forces, static and kinetic friction, and the mathematical relationships governing motion. Practical problems are presented to deepen understanding of these principles in real-world applications.
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DYNAMICS & NEWTON’S LAW Topics: Force Newton’s Law of Motion Application of Newton’s Law
Kinematics: • is the branch of mechanics which studies a motion that is observed based on physical quantities such as position, distance, displacement, speed, velocity and acceleration with no consider on what causes the motion (force). • Examples: • Uniform rectilinear motion of a particle • Accelerated uniform rectilinear motion • Vertical motion • Parabolic motion • etc
Dynamics: • is the branch of mechanics concernings motion of object with regard to what causes the motion (force). • Examples: • Motion of object on a smooth plane • Motion of object on a rough plane • Motion of object on an inclined plane • Motion of object that connected by a pulley • Lift • etc
Force • is a quantity that can change the state of motions(e.g., golf game), forms(e.g., bread batter making) and sizes(e.g.,spring) of an object. • force can be a push or a pull upon an object.
The first Newton’s Law In the absence of external forces, an object at rest remains at rest and an object in motion continues in motion with a constant velocity (that is, with a constant speed in a straight line). V= 0 At rest a = 0 GLB
The 2nd Newton’s Law So, What will happen , if the resultant of Force is not ZERO ?
a F a = ? a~F a ~ 1/m F m
Sample problem What is the acceleration resulted when a force resultant of 12 N is applied to an object which is 6 kg in mass? Solution:
3. Newton’s Third Law If two objects interact, the force F12 exerted by object 1 on object 2 is equal in magnitude and opposite in direction to the force F21 exerted by object 2 on object 1: Faction = - F reaction
application Normal force On horisontal plane : N = W N Weight θ w sinθ θ On inclined plane : N = W cosθ w cosθ w
N W = mg
T T = W F T fk w w’
There two kinds of frictional force • Static frictional force: is a frictional force exerts on the object which is at rest. • Kinetic frictional force: is a frictional force works on the moving object. μs > μk, So … fs> fk fs = static frictional force (N) fk = kinetic frictional force (N) μs = coefficient of static friction μk = coefficient of kinetic friction μ = 0 (smooth) 0 ≤ μ≤ 1 μ > 0 (rough)
Friction force F fs maks
F fs maks Friction force fsmaks = μs N If F < fsmaksat rest If F = fsmaks Exactly will be move If F > fsmaksObject is moving
Sample Problem Calculate the acceleration of system , if Force applied is : (g=10m/s2 ) a. 50 N b. 52 N c. 60 N d. 70 N e. 61 N F 12 kg μS= 0,5 μk = 0,3
Sample problem A car is 800 kg in mass, from rest it is accelerated with a constant acceleration, after 2 seconds the car travels at a distance of 20 m. Determine the force resulted by the car if during the motion it experiences the frictional force of 200 N! Solution: Because a = constant, then the car moves in accelerated uniform rectilinear motion, therefore the acceleration (a) can be determined by;
Sample problem: Determine the normal force of cube A, B, and C, if m = 10 kg and g = 9.8 m/s2! 20 N 15 N N=? N=? N=? m m m A B C w w w
Cube A Solution: Because on cube A exerts no outside force, N = w = m.g = (10kg)(9.8m/s2) = 98 N • Cube B Because on cube B works an outside force of 20 N, N = ƩF = w + F = m.g + F = (10kg)(9.8m/s2) + 20 N = 118 N
Cube C Because on cube C works an outside force of 15 N, N = ƩF = w - F = m.g - F = (10kg)(9.8m/s2) - 15 N = 83 N
Problem 1 Calculate the acceleration of system and stress of string (percepatan system dan tegangan tali ) if flour is smooth ( jika lantai licin ) g = 10 m/s2 7 kg 3 kg
7 kg 3 kg Answer N Because the floor is smooth , μ=0, fs =0 W2=70N w=mg = 30N
Problem Calculate the acceleration of system and stress of string (percepatan system dan tegangan tali ) if flour is rough ( jika lantai kasar ) g = 10 m/s2 7 kg μs=0,4 μk=0,3 3 kg
7 kg 3 kg Answer fk=μk.N = 0,3.70=21N μs=0,4, fsmaks =0,4.70 = 28N N W2=70N μs=0,4 μk=0,3 w=mg = 30N
Problem 2 Determine the acceleration (a) and tension of each string ! m1= 4 Kg m2 = 6 Kg There isn’t Friction between string and pulley
2Kg 3Kg 30o Determine the acceleration (a) and tension of each string ! If the inclined plane is smooth. ( g = 10 m/s2)
2Kg 3Kg 30o Answer T T w2sin30o w2cos30o W2 W1
T T 2Kg w2sin30o 3Kg w2cos30o W2 W1 30o Tension of string (T)