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Newton’s Law of Motion

Newton’s Law of Motion. Terms and Definition.

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Newton’s Law of Motion

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  1. Newton’s Law of Motion

  2. Terms and Definition Mass of an object is a measure of the inertia of the object. Inertia is the tendency of a body at rest to remain atrest, and a body in motion to continue moving with unchanged velocity. Mass is as a representation of the amount of or quantity-of-matter Massa adalah suatu ukuran inersia dari sebuah obyek. Inersia adalah kecenderungan suatu benda untuk tetap diam bila dalam keadaan diam dan atau untuk tetap bergerak dengan memiliki kecepatan konstan. Massa adalah ukuran dari jumlah keseluruhan suatu dari suatu zat

  3. Terms and Definition Force, in mechanics, is that which changes the velocity of an object. It is a vector quantity, having magnitude and direction Gaya, dalam pengertian mekanik, adalah besaran vektor yang mengubah kecepatan dari suatu benda yang bergerak

  4. From the figure: Mass is contained in each body of the person on roller coaster, the roller coaster itself, the rail structures. Dari gambar di samping: Massa terdapat dalam setiap individu yang menaiki roller coaster, termasuk roller coaster dan juga struktur rel untuk roller coaster

  5. From the figure: Force is a vector that causes the roller coaster moving in changed velocity Dari gambar di samping: Gaya adalah besaran vektor yang menyebabkan roller coaster bergerak dengan kecepatan berubah-ubah

  6. Newton’s First Law – Law of Inertia “ An object at rest will remain at rest: an object in motion will in motion with a constant velocity, except insofar as it is acted upon by an external force” Hukum Newton Pertama – Hukum Inersia “ Sebuah benda akan tetap diam dan akan tetap bergerak beraturan jika gaya total yang bekerja pada sistem adalah nol”

  7. ΣF = 0, F1 + F2 = 0, at rest

  8. When the system is at constant velocity and net force is null, the system remains at constant velocity

  9. Example: Forces acted on a point as follows: F1 = 29i + 37j – 77k F2 = 53i -100j -5k F3 = -82i + 63j + 82k Fnet = F1 + F2 + F3 = 0 z F3 ß y F2 x F1

  10. Newton’s Second Law “ A particle acted upon by an unbalanced force F experiences an acceleration a that has the same direction as the force and a magnitude that is directly proportional to the force ” Hukum Newton Kedua “ Sebuah benda yang dibebani gaya total sebesar F akan mengalami percepatan sebesar a yang memiliki arah yang sama dengan gaya total dan besarnya berbanding lurus terhadap gaya total tersebut”

  11. From the figure: Three members are pushing the bobsled resulting it to accelerate on the direction the same with the force exerted on it Dari gambar di samping: Percepatan dialami pada mobil luncur dengan arah yang sama dengan gaya total

  12. Gaya gesek Gaya normal Gaya pegas Gaya tarik Gaya dorong roket, mobil, pesawat Gaya berat

  13. Formula used in Newton’s second law Fnet = m a dv Fnet = m dt For component vector v = vx + vy + vz dvx dvz Fx, net = m Fz, net = m dt dt dvy Fy, net = m dt

  14. Example 1: Given: an object with a mass of 5.0 kg, to accelerate at 0.3 m/s2 Required: tension of the rope Solution: ΣFy = m ay ΣFy = FT - FW Solve for FT

  15. Problem 1: The only force acting on a 5.0 kg object has components Fx = 20 N and Fy = 30 N. Find the acceleration of the object (magnitude and direction)

  16. Problem 2: Jika Fg = 27 N, berapakah besarnya Fg’ untuk setimbang dan untuk memiliki percepatan 0.5 m/dt2 (massa pulley dan tali diabaikan)

  17. Newton’s Third Law “ The mutual forces of action and reaction between two particles are equal, opposite, and collinear” Hukum Newton Ketiga “ Gaya aksi dan reaksi memiliki besaran yang sama dengan arah yang berlawanan dan segaris”

  18. Application • Friction Force – Gaya gesek • Normal Force – Gaya normal

  19. Application • Spring Force – Gaya pegas

  20. Application • Tension Force – Gaya Tarik • Weight/Gravitational Force – Gaya berat

  21. Application • Thrust Force – Gaya dorong roket

  22. Normal Force The normal force is the contact force that is always perpendicular to the surface or plane on which an object lies

  23. FN = mg cosθ Normal force is always perpendicular to the plane on which the object lies

  24. Friction Force When an object slides along a rough surface, the force of kinetic friction acts opposite to the direction of the velocity of the object. The amount of friction force depends on the normal force of the object and its coefficient of friction Static Friction (fs) is a friction force that happens on an object between interval of at rest until about to move Kinetic Friction (fk) is a friction force which occurs during the movement of an object

  25. Example of coefficient of friction value μs > μk μs is coefficient of static friction μk is coefficient of kinetic friction

  26. fs = μs FN Static friction force fk = μk FN Kinetic friction force

  27. Example: Given: μs = 0.4 (wood on wood) μk = 0.2 (wood on wood) m = 3.5 kg Required: static friction force and kinetic friction force Solution: W = m.g = (3.5) (9.81) = 34.34 N FN = W = 34.34 N fs = μs FN = (0.4) (34.34) = 13.74 N fk = μk FN = (0.2) (34.34) = 6.87 N

  28. Tension Force Tension force is the force that is acted on a massless and unstretchable cord when it is used to pull an object. It is so called tension force because in the system the cord is in the state of tension

  29. Problems: • From the figure, find the acceleration of the box μk= 0.5

  30. Dari gambar di bawah, berapakah koefisien gesek kinetik antara kotak dengan bidang? a= 0.5 m/dt2

  31. Dari gambar di bawah, masing-masing massa blok adalah 5 kg dengan koefisien gesek kinetik antara blok dengan bidang adalah μk = 0.1. Hitung percepatan masing-masing blok?

  32. Dari gambar di bawah, blok A memiliki massa 5 kg bergerak dengan kecepatan 0.5 m/dt pada saat gambar diambil. Jika koefisien gesek kinetik antara blok dengan bidang adalah μk = 0.2. Tentukan kecepatan blok A setelah bergerak sejauh 1.2 m. Massa blok B adalah 10 kg.

  33. Dari gambar di bawah, kotak brankas memiliki massa 100 kg. Di ujung tali, seorang pemuda dengan massa 45 kg memegang untuk menahan. Karena dalam keadaan panik dan tidak mau melepaskan tali, berapakah percepatan yang akan dia alami? Massa tali dan pulley diabaikan

  34. Dari gambar di bawah, massa A = 25 kg, massa B = 15 kg. Koefisien gesek antara blok dengan meja adalah 0.20. seberapa jauhkah blok B akan turun setelah 3 detik sesudah blok B dilepas? μk= 0.2

  35. Dari sistem di bawah, tentukan percepatan dari sistem tersebut dan gaya tarik yang dialami oleh masing-masing tali.

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