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Transcript
5.5 The Gravitational Force and Weight The Gravitational Force and Weight The gravitational force: Fg ≡ Force that The Earth exerts on an object This force is directed toward the center of the earth. Its magnitude is called THE WEIGHT of the object Weight ≡ |Fg| ≡ mg (5.6) Because it is dependent on g, the weight varies with location g, and therefore the weight, is less at higher altitudes Weight is not an inherent property of the object Gravitational Force, 2 Object in FREE FALL. Newton’s 2nd Law: ∑F = m a If no other forces are acting, only Fg acts (in vertical direction). ∑Fy = m ay Or: Fg = mg (down, of course) (5.6) Where: g = 9.8 m/s2 SI Units: Newton (just like any force!). 2 If m = 1 kg Fg = (1kg)(9.8m/s ) = 9.8N Gravitational Force, final REMARKS!!! Fg Depends on g, then it varies with geographic location. Fg Decreases from Sea level to a higher altitude. You want lose Weight without diet climb a high mountain. m in Equation 5.6 is called the gravitational mass The kilogram is not a unit of Weight is a unit of Mass. Mass and Weight are two different quantities Gravitational & Inertial Mass In Newton’s Laws, the mass is the inertial mass and measures the resistance to a change in the object’s motion In the gravitational force, the gravitational mass is determining the gravitational attraction between the object and the Earth Experiments show that gravitational mass and inertial mass have the same value 5.6 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 F12 = ̶ F21 (5.7) Note on notation: FAB is the force exerted by A on B Newton’s Third Law, 2 Forces always occur in pairs A single isolated force cannot exist The action force is equal in magnitude to the reaction force and opposite in direction One of the forces is the action force, the other is the reaction force It doesn’t matter which is considered the action and which the reaction Newton’s Third Law, 3 The action and reaction forces must: act on different objects. be of the same type Forces exerted BY a body DO NOT (directly) influence its motion!! Forces exerted ON a body (BY some other body) DO influence its motion!! When discussing forces, use the words “BY” and “ON” carefully. Example: 5.1 Action-Reaction The force F12 exerted BY object 1 ON object 2 is equal in magnitude and opposite in direction to F21 exerted BY object 2 ON object 1 F12 = – F21 Example: 5.2 Action-Reaction The force Fhn exerted BY the hammer ON the nail is equal in magnitude and opposite in direction to Fnh exerted BY the nail ON the hammer Fhn = – Fnh Example: 5.3 Action-Reaction We can walk forward because when one foot pushes backward against the ground, the ground pushes forward on the foot. – FPG ≡ FGP Example: 5.4 Action-Reaction (Normal Force) Does the force of gravity stop? OF COURSE NOT But, object does not move: 2nd Law ∑F = m a = 0 There must be some other force acting besides gravity (weight) to have ∑F = 0. The Normal Force : FN = n Normal is math term for Perpendicular () FN is to the surface & opposite to the weight (in this simple case only) Example: 5.4 Normal Force, 2 The normal force: n (table on monitor) is the reaction of the force the monitor exerts ON the table The action (Fg, Earth on monitor) force is equal in magnitude and opposite in direction to the reaction force, the force the monitor exerts ON the Earth Caution: The normal force is not always = & opposite to the weight!! As we’ll see! Free Body Diagram In a free body diagram, you want the forces acting ON a particular object The normal force and the force of gravity are the forces that act ON the monitor Normal Force Where does the Normal Force come from? From the other body!!! Does the normal force ALWAYS equal to the weight ? NO!!! Weight and Normal Force are not ActionReaction Pairs!!! Example 5.5 Normal Force m = 10 kg Weight: Fg = mg = 98.0N The normal force is equal to the weight!! Only this case FN = n = mg = 98.0N Example 5.6 Normal Force, 2 m = 10 kg Weight: Fg = mg = 98.0N Pushing Force = 40.0N The normal force is NOT ALWAYS equal to the weight!! FN = n = 40.0N mg= 138.0N Example 5.7 Normal Force, final m = 10 kg Weight: Fg = mg = 98.0N Puling Force = 40.0N The normal force is NOT ALWAYS equal to the weight!! FN = n = 98.0N – 40.0N= 58.0N Example 5.8 Accelerating the box m = 10 kg Fg = 98.0N From Newton’s 2nd Law: ∑F = ma FP – mg = m a m a = 2.0N The box accelerates upwards because FP > m g Example 5.9 Weight Loss (Example 5.2 Text Book) Apparent weight loss. The lady weights 65kg = 640N, the elevator descends with a = 0.2m/s2. What does the scale read (FN)? From Newton’s 2nd law: ∑F = ma FN – mg = – m a FN = mg – m a FN = 640N – 13N = 627N = 52kg Upwards! FN is the force the scale exerts on the person, and is equal and opposite to the force she exerts on the scale. Example 5.9 Weight Loss, 2 What does the scale read when the elevator descends at a constant speed of 2.0m/s? From Newton’s 2nd law: ∑F = 0 FN – mg = 0 FN = mg = 640N = 65kg The scale reads her true mass! NOTE: In the first case the scale reads an “apparent mass” but her mass does not change as a result of the acceleration: it stays at 65 kg 5.7 Some Applications of Newton’s Law OBJECTS IN EQUILIBRIUM If the acceleration of an object that can be modeled as a particle is zero, the object is said to be in EQUILIBRIUM!! Mathematically, the net force acting on the object is zero F 0 F 0 and F x y 0 Example 5.10 Equilibrium A lamp is suspended from a chain of negligible mass The forces acting on the lamp are Force of gravity (Fg) Tension in the chain (T) Equilibrium gives F y 0 T Fg 0 T Fg Example 5.10 Equilibrium, final The forces acting on the chain are T’ and T” T” is the force exerted by the ceiling T’ is the force exerted by the lamp T’ is the reaction force to T Only T is in the free body diagram of the lamp, since T’ and T” do not act on the lamp Example 5.11 Traffic Light at Rest Example 5.4 (Text Book) This is an equilibrium problem No movement, so a = 0 Upper cables are not strong as the lower cable. They will break if the tension exceeds 100N. Will the light remain or will one of the cables break? mg =122N Example 5.11 Traffic Light at Rest, 2 Apply Equilibrium Conditions: ΣFy = 0 T3 – Fg = 0 T3 = Fg = 122N Find Components: T1x = – T1cos37 T1y = T1sin37 T2x = T2cos53 T2y = T2sin53 T3x = 0 T3y = – 122N Apply Newton’s 2nd Law: ΣFx = 0 ΣFy = 0 T2y T1y -T1x T2x Example 5.11 Traffic Light at Rest, final ΣFx = – T1cos37 + T2cos53 = 0 (1) ΣFy = T1sin37 + T2sin53 – 122N = 0 (2) Solving for T1 or T2: From Eqn (1) solve for T2 T2 = (cos37/cos53)T1 =1.33T1 Substituting this value into Eqn (2) T1sin37 + (1.33T1)sin53 =122N T1 = 74.4 N and T2 = 97.4N Both values are less than 100N, so the cables will not break!!! T2y T1y -T1x T2x Objects Experiencing a Net Force, Examples If an object that can be modeled as a particle experiences an acceleration (a ≠ 0), there must be a nonzero net force (F ≠ 0) acting on it. Draw a free-body diagram Apply Newton’s Second Law in component form Example 5.12 Net Force FR F12 F22 F 1 tan 2 F 1 1002 1002 141N tan 1(1) 45 Forces on The Crate Forces acting on the crate: A tension, the magnitude of force T The gravitational force, Fg The normal force, n, exerted by the floor Forces on The Crate, 2 Apply Newton’s Second Law in component form: F x F y T max n Fg 0 n Fg ONLY in this case n = Fg Solve for the unknown(s) If T is constant, then a is constant and the kinematic equations can be used to more fully describe the motion of the crate Example 5.13 Normal Force The normal force, FN is NOT always equal to the weight!! m = 10.0 kg Fg = 98.0N Find: ax ≠ 0? FN? if ay = 0 m=10kg Example 5.13 Normal Force, final FPy = FPsin(30) = 20.0N FPx = FPcos(30) = 34.6N ΣFx = FPx = m ax ax = 34.6N/10.0kg ax = 3.46m/s2 ΣFy = FN + FPy – mg = m ay FN + FPy – mg = 0 FN = mg – FPy= 98.0N – 20.0N FPy FPx FN = 78.0N Example 5.14 Conceptual Example: The Hockey Puck Moving at constant velocity, with NO friction. Which free-body diagram is correct? (b) Inclined Planes Choose the coordinate system with x along the incline and y perpendicular to the incline Forces acting on the object: The normal force, n, acts perpendicular to the plane The gravitational force, Fg , acts straight down Example 5.15 The Runaway Car (Example 5.6 Text Book) Replace the force of gravity with its components: Fgx = mgsin Fgy = mgcos With: ay = 0 & ax ≠ 0 (A). Find ax Using Newton’s 2nd Law: y-Direction ΣFy = n – mgcos = may = 0 n = mgcos Example 5.15 The Runaway Car, final x-direction ΣFx = mgsin = max ax = gsin Independent of m!! (B) How long does it take the front of the car to reach the bottom? 2 1 x f xi vxi t 2 axt d 12 axt 2 t 2d t ax 2d g sin (C). What is the car’s speed at the bottom? vxf 2 vxi 2 2ax d vxf 2 2( g sin )d vxf 2 gd sin