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SECOND LAW OF MOTION If there is a net force acting on an object, the object will have an acceleration and the object’s velocity will change. Newton's Second Law of Motion states that for a particular force, the acceleration of an object is directly proportional to the net force and inversely proportional to the mass of the object. The direction of the force is the same as that of the acceleration. In equation form: F a m F ma SECOND LAW OF MOTION In the SI system, the unit for force is the newton (N): A newton is that net force which, when applied to a 1-kg mass, gives it an acceleration of 1 m/s2. Acceleration and Force With Zero Friction Forces Pushing the cart with twice the force produces twice the acceleration. Three times the force triples the acceleration. Acceleration and Mass Again With Zero Friction F F a/2 a Pushing two carts with same force F produces one-half the acceleration. The acceleration varies inversely with the mass. THIRD LAW OF MOTION According to Newton's third law of motion, when one body exerts a force on another body, the second body exerts on the first an equal force in opposite direction. The Third Law of Motion applies to two different forces on two different objects: " For every action force, there must be an equal and opposite reaction force. " Action and reaction forces never balance out because they act on different objects. Newton’s Third Law • Third Law: For every action force, there must be an equal and opposite reaction force. Forces occur in pairs. Action Reaction Reaction Action Acting and Reacting Forces • Use the words by and on to study action/reaction forces below as they relate to the hand and the bar: Action The action force is exerted by bar the hands _____ on the _____. The reaction force is exerted by the _____ bar on the _____. hands Reaction MASS The property that a body has of resisting any change in its state of rest or of uniform motion is called inertia. The inertia of a body is related to the amount of matter it contains. A quantitative measure of inertia is mass. The unit of mass is the kilogram (kg). WEIGHT (FG) Weight (a vector quantity) is different from mass (a scalar quantity). The weight of a body varies with its location near the Earth (or other astronomical body), whereas its mass is the same everywhere in the universe. The weight of a body is the force that causes it to be accelerated downward with the acceleration of gravity g. FG = mg Units: Newtons (N) g = acceleration due to gravity = 9.8 m/s2 THE NORMAL FORCE A normal force is a force exerted by one surface on another in a direction perpendicular to the surface of contact. Note: The gravitational force and the normal force are not an action-reaction pair. Applying Newton’s Second Law • Read the problem and draw a sketch. • Write down the data in correct units: m = kg FG = N • Draw a free-body diagram for each object. • Choose x or y-axis along motion and choose direction of motion as positive. • Write Newton’s law for both axes: SFx = m ax SFy = m ay • Solve for unknown quantities. 4.1 A 5.0-kg object is to be given an upward acceleration of 0.3 m/s2 by a rope pulling straight upward on it. What must be the tension in the rope? m = 5 kg a = 0.3 m/s2 N2L ΣFy = FT - FG = ma FT a (+) FG FT = m(a + g) = 5(0.3 + 9.8) = 50.5 N 5 kg 4.2 A cord passing over a frictionless pulley has a 7.0 kg mass hanging from one end and a 9.0-kg mass hanging from the other. (This arrangement is called Atwood's machine). a. Find the acceleration of the masses. m1 = 7 kg m2 = 9 kg N2L FT 7kg 9 kg FT a (+) a (+) FG1 FG2 FT FT a (+) a (+) FG1 FG2 FG 2 FG1 a mT ΣF = FT - FG1 - FT + FG2 = mTOTALa 9(9.8) 7(9.8) = 1.22 m/s2 97 FT FT a (+) a (+) FG1 FG2 b. Find the tension of the cord Using either side of the pulley yields the same answer! FT – FG1 = m1a FT = m1a + FG1 = m1 (a + g) = 7(1.22 + 9.8) = 77.1 N APPARENT WEIGHT The actual weight of a body is the gravitational force that acts on it. The body's apparent weight is the force the body exerts on whatever it rests on. Apparent weight can be thought of as the reading on a scale a body is placed on. scale 4.3 What will a spring scale read for the weight of a 75 kg man in an elevator that moves: m = 75 kg FG = 75(9.8) = 735 N a. With constant upward speed of 5 m/s N1L m = 75 kg FG = 735 N FN ΣFy = FN - FG = 0 F N = FG = 735 N FG b. With constant downward speed of 5 m/s FN = 735 N FN b. Going up with an acceleration of 0.25 g a = 2.45 m/s2 N2L FG ΣF = FN - FG = ma FN = ma + FG = 75 (2.45) + 735 = 919 N FN c. Going down with an acceleration of 0.25 g a = 2.45 m/s2 N2L FG ΣF = FG - FN = ma FN = FG - ma = 735 - 75 (2.45) = 551 N For free fall the only force acting is FG e. In free fall? so FN = 0 N N2L FG “Weightlessness” More properly, this effect is called apparent weightlessness, because the gravitational force still exists. It can be experienced on Earth, but only briefly: FRICTION: STATIC AND KINETIC FRICTION Frictional forces act to oppose relative motion between surfaces that are in contact. Such forces act parallel to the surfaces. Static friction occurs between surfaces at rest relative to each other. When an increasing force is applied to a book resting on a table, for instance, the force of static friction at first increases as well to prevent motion. In a given situation, static friction has a certain maximum value called starting friction. When the force applied to the book is greater than the starting friction, the book begins to move across the table. The origin of friction: on a microscopic scale, most surfaces are rough. FRICTION: STATIC AND KINETIC FRICTION The kinetic friction (or sliding friction) that occurs afterward is usually less than the starting friction, so less force is needed to keep the book moving than to start it moving. COEFFICIENT OF FRICTION The frictional force between two surfaces depends on the normal force FN pressing them together and on the natures of the surfaces. The latter factor is expressed quantitatively in the coefficient of friction (mu) whose value depends on the materials in contact. The frictional force is experimentally found to be: Ff s FN Static friction Ff k FN Kinetic friction 4.4 A horizontal force of 140 N is needed to pull a 60.0 kg box across the horizontal floor at constant speed. What is the coefficient of friction between floor and box? N1L FN Fa = 140 N m = 60 kg Fa Ff FG ΣFx = Fa – Ff = 0 Fa = Ff = μFN ΣFy = FN – FG = 0 FN = FG =mg = 60(9.8) = 588 N Fa 140 FN 588 = 0.24 4.5 A 70-kg box is slid along the floor by a horizontal 400-N force. Find the acceleration of the box if the value of the coefficient of friction between the box and the floor is 0.50. N2L m = 70 kg Fa = 400 N μ = 0.5 ΣFy = FN – FG = 0 FN = FG =mg = 70(9.8) = 686 N FN Ff Fa FG ΣFx = Fa – Ff = ma Ff = μFN = (0.5)(686) = 343 N m = 70 kg Fa= 400 N μ = 0.5 ΣFx = Fa – Ff = ma a FN Ff Fa FG Fa F f m 400 343 = 0.81 m/s2 70 4.6 A 70-kg box is pulled by a rope with a 400-N force at an angle of 30 to the horizontal. Find the acceleration of the box if the coefficient of friction is 0.50 N2L FN Fa Ff FG m = 70 kg Fa = 400 N, 30 μ = 0.5 Fax = 400 cos 30 = 346.4 N Fay = 400 sin 30 = 200 N ΣFy = FN + Fay - Fg = 0 FN = FG - Fay = 70(9.8) - 200 = 486 N Ff = μFN = (0.5)(486) = 243 N m = 70 kg Fa = 400 N, 30 μ = 0.5 FN Fa Ff ΣFx = Fax – Ff = ma FG a Fax F f m 346 .4 243 = 1.47 m/s2 70 4.7 A box sits on an incline that makes an angle of 30˚ with the horizontal. Find the acceleration of the box down the incline if the coefficient of friction is 0.30. N2L FN Ff ΣFy = FN - FGy = 0 FN = FGy ΣFx = FGx – Ff = ma θ FG θ = 30 μ = 0.3 FGy FGx Ff = μFN ΣFx = FGx –μ FGy = ma FN Ff ΣFx = FGx – μ FGy = ma mg sin θ - μ mg cos θ = ma θ FG FGx FGy a g (sin cos ) 9.8(sin 30 cos30 ) = 2.35 m/s2 4.8 Two blocks m1 (300 g) and m2 (500 g), are pushed by a force F. If the coefficient of friction is 0.40. a. What must be the value of F if the blocks are to have an N2L 2 acceleration of 200 cm/s ? m1 = 0.3 kg m2 = 0.5 kg μ = 0.4 a = 2 m/s2 FN1 Ff1 F Ff2 FG1 FG1 = FN1 FG2 = FN2 FN2 FG2 ΣFx = F - Ff1 - Ff2 = mT a F = mTa + Ff1 + Ff2 = mTa + μ FNT = mTa + μ mT g = 0.8(2) + (0.4) 0.8 (9.8) = 4.7 N b. How large a force does m1 then exert on m2? m2 alone FN2 Ff2 F12 FG2 ΣFx = m2a F12 - Ff2 = m2a F12 = Ff2 + m2a = μ m2 g + m2a = 0.4(0.5) (9.8) + 0.5 (2) = 2.96 N 4.9 An object mA = 25 kg rests on a tabletop. A rope attached to it passes over a light frictionless pulley and is attached to a mass mB = 15 kg. The coefficient of friction between the table and block A is 0.20. a. What is the acceleration of the 25 kg block? FN Ff 25 kg FT FGA FT 15 kg FGB mA = 25 kg mB = 15 kg μ = 0.2 FN = FGA = 25(9.8) = 245 N FfA = μFN = 0.2 (245) = 49 N N2L ΣF = FT - FfA + FGB - FT = mTa FN Ff a FT FT FGA FGB mA = 25 kg mB = 15 kg μ = 0.2 FGB FfA mT 15(9.8) 49 25 15 = 2.45 m/s2 b. What is the tension on the string? ΣF = FT - Ff = mAa FN Ff FT FT FGA FGB F T = m Aa + F f = 25 (2.45) + 49 = 110.25 N 4.10 Blocks 1 and 2 of masses ml and m2, respectively, are connected by a light string, as shown above. These blocks are further connected to a block of mass M by another light string that passes over a pulley of negligible mass and friction. Blocks l and 2 move with a constant velocity v down the inclined plane, which makes an angle with the horizontal. The kinetic frictional force on block 1 is f and that on block 2 is 2f. a. On the figure below, draw and label all the forces on block ml. FN FT FG Ff Express your answers to each of the following in terms of ml, m2, g, , and f. b. Determine the coefficient of kinetic friction between the inclined plane and block 1. f N N m1 g cos f m1 g cos c. Determine the value of the suspended mass M that allows blocks 1 and 2 to move with constant velocity down the plane. FN2 FN1 FT N1L 2f FT FT f FT m2 g m1 g Mg m1g sin f m2 g sin 2 f Mg 0 Mg m1g sin f m2 g sin 2 f 3f M (m1 m2 ) sin g d. The string between blocks 1 and 2 is now cut. Determine the acceleration of block 1 while it is on the inclined plane. m1 g sin f m1a FN1 f m1 g N2L f a g sin m1 Summary Newton’s First Law: An object at rest or an object in motion at constant speed will remain at rest or at constant speed in the absence of a net force. Newton’s Second Law: A net force produces an acceleration in the direction of the force that is directly proportional to the force and inversely proportional to the mass. Newton’s Third Law: For every action force, there must be an equal and opposite reaction force. Forces occur in pairs. Summary: Procedure FR ma; FR a m N = (kg)(m/s2) • Read and write data with units. • Draw free-body diagram for each object. • Choose x or y-axis along motion and choose direction of motion as positive. • Write Newton’s law for both axes: SFx = m ax SFy = m ay • Solve for unknown quantities.