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Dynamics: Newton’s Laws of Motion

Dynamics: Newton’s Laws of Motion. m. Two basic concepts: mass m and force Mass m is a scalar quantity that measures the property of inertia. m = ρ V ~ amount of matter Force is a vector quantity that, being divided by the mass

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Dynamics: Newton’s Laws of Motion

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  1. Dynamics: Newton’s Laws of Motion m Two basic concepts: mass m and force Mass m is a scalar quantity that measures the property of inertia. m = ρV ~ amount of matter Force is a vector quantity that, being divided by the mass of a body, evaluates its acceleration due to a particular interaction. V Typical masses: standard 1kg, electron me=9·10-31 kg, car 2000 kg, Earth 6·1024 kg, sun 2·1030 kg Interactions and various long-range and contact forces Superposition of forces: net force

  2. Newton’s First Law (Law of Inertia) An object continues in a state of rest or in a state of motion at a constant velocity along a straight line, unless compelled to change that state by a net force. Contrary to ancient Greeks, Aristotle (384 – 322 B.C.E., Athens, a student of Socrates and a tutor of Alexander the Great) Note:The First Law is valid only in the inertial frames of reference and states that such frames exist. 1 Mps Hubble Law of the Universe Expansion: r Hubble constant Age of the Universe Parallax second: 1 ps = 3.26 light years

  3. Newton’s Second Law of Motion Acceleration is equal to a net external force divided by the mass of an object: m Units: [F] = [M] [a] = [M] [L] / [T]2→ 1 newton = 1 kg·m / s2 Free-body diagram represents the object and the forces that act on it. y Only external forces Internal forces and the forces that the object exerts on its environment are not included!!! x

  4. Example of free-body diagrams (problem 4.54) Data: F=200 N, m1=6 kg, m2=4 kg, m3=5 kg Find: (a) free-body diagrams; (b) acceleration ay; (c) tension force T1 ; (d) tension force T2 . m1= y -m1g T1 -T1 Solution:Newton’s second law (b) for the whole system (m1+m2+m3) ay = F – (m1+m2+m3) g , ay = F/(m1+m2+m3) – g = 3.53 m/s2 (c) for the upper block T1= F –m1g –m1ay= =F[1-m1/(m1+m2+m3)]= F·9/15 = 120 N (d) for the “upper block + half of rope” similar to (c) → T2= F – (m1+m2/2)(g+ay)= =F[1-(m1+m2/2)/(m1+m2+m3)]= F·7/15= 93 N m2= -m2g T3 -T3 m3= -m3g

  5. Newton’s Third Law Whenever one body exerts a force on a second body, the second body exerts an oppositely directed force of equal magnitude on the first body. “Action-reaction” law: For every action (force) there is an equal, but opposite, reaction. Sun Mearth=6·1024 kg Msun=2·1030 kg

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