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Chapter 4 Forces and Newton’s Laws of Motion F=ma; gravity 0) Background • Galileo – inertia (horizontal motion) – constant acceleration (vertical motion) • Descartes & Huygens – Conservation of momentum: mass x velocity = constant • Kepler & Braha – laws of planetary motion (kinematics only) Question of the day: Explain planetary motion 1) Newton’s first law: the law of inertia A free object moves with constant velocity a) Free object --> no forces acting on it b) Constant velocity --> at rest or motion in a straight line with constant speed c) Natural state is motion with constant velocity - Aristotle: rest is natural state Galileo: circular motion (orbits) is natural state d) Inertial reference frames - A reference frame in which the law of inertia holds - does not hold on a carousal, or an accelerating car Requires ability to identify a free object: If no force acts on a body, a reference frame in which it has no acceleration is an inertial frame. 1) Newton’s first law: the law of inertia A free object moves with constant velocity e) Velocity is relative - - All frames moving at constant velocity with respect to an inertial frame are also inertial frames No local experiment can determine the state of uniform motion Cannot define absolute rest: No preferred reference frame - (Principle of Relativity) 2) Newton’s second law: F=ma a) Mass (i) (ii) quantity of matter (determined with a balance) quantity that resists acceleration (inertial mass) Define 1 kg as mass of a standard cylinder Addition of masses (scalar): m = m1 + m2 - in particular two identical masses have twice the mass, to satisfy quantity of matter definition (iii) Observe acceleration vs mass for a given force: mass acceleration 1 kg 1 m/s2 2 kg 1/2 m/s2 3 kg 1/3 m/s2 mass is inversely proportional to acceleration 2) Newton’s second law: F=ma b) Force (i) (ii) push or pull disturbs “natural” state: causes acceleration 2 Define 1 N (newton) as force required rto accelerate 1 kg by 1 m/s r Addition of forces (vector): Fnet F1 F2 Identical forces in opposite direction produce no acceleration Two identical forces at 60º produce the same acceleration as a third identical force at 0º (cos(60º)=1/2) Two identical parallel forces corresponds to twice the force. 2) Newton’s second law: F=ma (iii) Observe acceleration vs. force for a given object Force 1N 2N 3N Acceleration 1 m/s2 2 m/s2 3 m/s2 Force is proportional to acceleration (iv) Types of force: - gravity - electromagnetic - weak nuclear -strong nuclear electroweak 2) Newton’s second law: F=ma c) Second Law Combine 1 F a for a given mass, and m for a given force a to give F ma or F (const)ma Define proportionality constant =1. Then, r F ma For m = 1 kg, and a = 1 m/s2, F = 1 N by definition, and F = ma gives F = 1 kg m /s2, so 1 N = 1 kg m/s2 2) Newton’s second law: F=ma • F = ma can be used as the defining equation for force and inertial mass, but only because of the physical observation that force is proportional to acceleration (for a given mass), and mass is inversely proportional to acceleration (for a given force). • Inertia is the tendency of an object not to accelerate • Newton’s second law formally refers to the rate of change of r momentum: (mv) F t r v r For constant mass, Fm ma t • Special case: r F 0 0 ma r a0 velocity is constant (1st law) 2) Newton’s second law: F=ma d) Free-body diagrams Replace object(s) by dot(s). Represent all forces from the dot. Solve F=ma for each object F2 FN F1 F2 F1 m mg 2) Newton’s second law: F=ma d) Free-body diagrams 10 N 10 N 10 N scale m ?N 10 N 10 N scale m m 2) Newton’s second law: F=ma e) Components of force r F ma means F x max & sum of all forces F y may 2) Newton’s second law: F=ma e) Components of force Example: F1 = 15 N m = 1300 kg m º Find acceleration. F2 = 17 N y F1 Fx F2 F1 cos 23 N F y x F2 F1 sin 14 N ax F ay x m Fy m 0.018 m/s 2 0.011 m/s 2 3) Newton’s third law For every action, there is an equal and opposite reaction FAB A Conservation of momentum: B FBA FAB = -FBA r (mB vB ) FAB t r r (mA vA ) FBA t r r r r (mB vB ) (mA vA ) FAB FBA t t r r (mB vB mA vA ) 0 t 4) Gravity 5) Normal Force 6) Friction 7) Tension and pulleys • Tension: force exerted by rope or cable – For an ideal (massless, inextensible) line, the same force is exerted at both ends (in opposite directions) – objects connected by a line (no slack) have the same acceleration • Pulley: changes direction of force – For an ideal pulley (massless, frictionless) the magnitude of the tension is the same on both sides – magnitude of acceleration of connected objects is the same 7) Tension and pulleys +a T1 m1 • T 1 = T2 = T • a1 = a 2 = a • For the example, a1y = -a2y T2 m2 • Simplify problem, by choosing sign for a sense of the motion 7) Tension and pulleys T +a T1 m1g m1 T2 m2 T m2g Equations of motion: T m1g m1a m 2 g T m2 a Solve for a (eliminate T ): Solve for T (eliminate a): m2 m1 a g m m 2m1m2 T g m1 m2 1 2 7) Tension and pulleys e.g. m1 = 5 kg; m2 = 10 kg +a T1 m1 T2 g 10 5 a g 10 5 3 100kg T g 65N 15 m2 Solve for a (eliminate T ): Solve for T (eliminate a): m2 m1 a g m m 2m1m2 T g m1 m2 1 2 7) Tension and pulleys +a Acceleration can be determined by considering external forces (tension is an internal force holding objects together) F ext () m1 m1g m a m2 g m1g (m1 m2 )a m2 m1 a g m m m2 1 m2g 2 Example If m1 = m2, and rope and pulley are ideal, what happens when the monkey climbs the rope? T1 T2 m1g m2g T1 m1 T2 Quick Time™a nd a TIFF ( Unco mpre ssed ) dec ompr esso r ar e nee ded to see this pictur e. m2 Since T1 = T2, any change in T2 to cause the monkey to ascend, results in a change in T1, causing the bananas to ascend at the same rate. Example If m1 = m2, and rope and pulley are ideal, what happens when the monkey climbs the rope? T1 T2 m1g m2g Since T1 = T2, any change in T2 to cause the monkey to ascend, results in a change in T1, causing the bananas to ascend at the same rate. Example If m1 = m2, and rope and pulley are ideal, what happens when the monkey climbs the rope? T1 T2 m1g m2g Since T1 = T2, any change in T2 to cause the monkey to ascend, results in a change in T1, causing the bananas to ascend at the same rate. Example If m1 = m2, and rope and pulley are ideal, what happens when the monkey climbs the rope? T1 T2 m1g m2g Since T1 = T2, any change in T2 to cause the monkey to ascend, results in a change in T1, causing the bananas to ascend at the same rate. Example If m1 = m2, and rope and pulley are ideal, what happens when the monkey climbs the rope? T1 T2 m1g m2g Since T1 = T2, any change in T2 to cause the monkey to ascend, results in a change in T1, causing the bananas to ascend at the same rate. 8) Equilibrium applications • Equilibrium means zero acceleration • Balance forces in x and y directions F F x 0 y 0 8) Equilibrium applications Example: Find tension on leg (F) Free body diagram for pulley: T Free body diagram for weight: T T=mg F T1 cos 35º T2 cos 35º mg Since T T1 T2 , F 2T cos 35º 36 N 9) Non-equilibrium applications • Non-equilibrium means non-zero acceleration • Determine acceleration from 2nd law: F F x max y may • Solve kinematic equations Example: Apparent weight Apparent weight (measured by scale) is the normal force At rest or moving with constant velocity FN W FN W ma 0 FN W 700N Example: Apparent weight Apparent weight (measured by scale) is the normal force Accelerating up FN FN W may ma FN W ma 1000N W FN W FN W FN a g 1 g W m W If FN 2W , a g For FN 1000N, and W 700N, a 3 g 7 Example: Apparent weight Apparent weight (measured by scale) is the normal force Accelerating down FN FN W may ma FN W ma 400N W W FN W FN FN a g 1 g m W W If FN 0, a g For FN 400N, and W 700N, a 3 g 7 Example: Apparent weight Apparent weight (measured by scale) is the normal force Free fall FN = 0 FN W may ma mg FN W mg 0 W weightlessness