5.5 The Gravitational Force and Weight

The Gravitational Force and Weight The gravitational force: The gravitational force: F g ≡ The Earth F g ≡ Force that The Earth exerts on an object toward the center This force is directed toward the center of the earth. magnitudeTHE WEIGHT Its magnitude is called THE WEIGHT of the object Weight ≡ |F g | ≡ mg (5.6) gthe weight Because it is dependent on g, the weight varies with location gthe weight g, and therefore the weight, is less at higher altitudes Weight is not an inherent property of the object

Gravitational Force, 2 FREE FALL ∑F = m a Object in FREE FALL. Newton’s 2 nd Law: ∑F = m a F g ∑F y = m a y If no other forces are acting, only F g acts (in vertical direction). ∑F y = m a y F g = mg(down, of course)(5.6) Or: F g = mg (down, of course) (5.6) g = 9.8 m/s 2 Where: g = 9.8 m/s 2 SI Units:Newton SI Units: Newton (just like any force!). If m = 1 kg  F g = (1kg)(9.8m/s 2 ) = 9.8N If m = 1 kg  F g = (1kg)(9.8m/s 2 ) = 9.8N

Gravitational Force, final REMARKS!!! REMARKS!!! F g g F g Depends on g, then it varies with geographic location. F g F g Decreases from Sea level to a higher altitude. losewithout dietclimb a high mountain. You want lose Weight without diet  climb a high mountain. mgravitational mass m in Equation 5.6 is called the gravitational mass The kilogramis notWeight Mass The kilogram is not a unit of Weight is a unit of Mass. Mass and Weight are two different quantities Mass and Weight are two different quantities

Gravitational & Inertial Mass inertial massresistance to a change In Newton’s Laws, the mass is the inertial mass and measures the resistance to a change in the object’s motion the gravitational massgravitational attraction In the gravitational force, the gravitational mass is determining the gravitational attraction between the object and the Earth gravitational mass inertial masssame value Experiments show that gravitational mass and inertial mass have the same value

5.6 Newton’s Third Law F 12 byon F 21 byon If two objects interact, the force F 12 exerted by object 1 on object 2 is equal in magnitude and opposite in direction to the force F 21 exerted by object 2 on object 1 F 12 = ̶ F 21 (5.7) F AB by on Note on notation: F AB is the force exerted by A on B

Newton’s Third Law, 2 Forces always occur in pairs Forces always occur in pairs A single isolated force cannot exist A single isolated force cannot exist The action force is equal in magnitude to the reaction force and opposite in direction The action force is equal in magnitude to the reaction force and opposite in direction action force reaction force One of the forces is the action force, the other is the reaction force action reaction It doesn’t matter which is considered the action and which the reaction

Newton’s Third Law, 3 actionreaction The action and reaction forces must: different objects. act on different objects. same type be of the same type BYDO NOT itsmotion!! Forces exerted BY a body DO NOT (directly) influence its motion!! ONBY DOitsmotion!! Forces exerted ON a body (BY some other body) DO influence its motion!! BY ON When discussing forces, use the words “BY” and “ON” carefully.

Example: 5.1 Action-Reaction F 12 BYON F 21 BY ON The force F 12 exerted BY object 1 ON object 2 is equal in magnitude and opposite in direction to F 21 exerted BY object 2 ON object 1 F 12 = – F 21 F 12 = – F 21

Example: 5.2 Action-Reaction F hn BYON F nh BY ON The force F hn exerted BY the hammer ON the nail is equal in magnitude and opposite in direction to F nh exerted BY the nail ON the hammer F hn = – F nh F hn = – F nh

Example: 5.3 Action-Reaction backward forward We can walk forward because when one foot pushes backward against the ground, the ground pushes forward on the foot. – F PG ≡ F GP

OF COURSE NOT Does the force of gravity stop? OF COURSE NOT does not move But, object does not move: ∑F = m a = 0 2 nd Law  ∑F = m a = 0 some other force weight∑F = 0.  There must be some other force acting besides gravity (weight) to have ∑F = 0. The Normal ForceF N = n The Normal Force : F N = n Normal Perpendicular(  ) Normal is math term for Perpendicular (  ) F N  F N is  to the surface & opposite weight in this simple case only to the weight (in this simple case only) Example: 5.4 Action-Reaction (Normal Force)

The normal force:n ON The normal force: n (table on monitor) is the reaction of the force the monitor exerts ON the table The actionF g, Earth on monitor the reaction ON The action (F g, 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!! Caution: The normal force is not always = & opposite to the weight!! As we’ll see! Example: 5.4 Normal Force, 2

Free Body Diagram ON In a free body diagram, you want the forces acting ON a particular object The normal force force of gravity ON The normal force and the force of gravity are the forces that act ON the monitor

Normal Force Normal Force Where does the Normal Force come from? From the other body!!! From the other body!!! normal forceALWAYS weight Does the normal force ALWAYS equal to the weight ?NO!!! Weight and Normal Force are not Action- Reaction Pairs!!!

Example 5.5 Normal Force m = 10 kg Weight: F g = mg = 98.0N The normal forceis equal force is equal weight!! to the weight!! Only this case F N = n = mg = 98.0N

Example 5.6 Normal Force, 2 m = 10 kg Weight: F g = mg = 98.0N Pushing Force = 40.0N The normal forceNOT The normal force is NOT ALWAYS weight!! ALWAYS equal to the weight!! F N = n = 40.0N mg= 138.0N

Example 5.7 Normal Force, final m = 10 kg Weight: F g = mg = 98.0N Puling Force = 40.0N The normal forceNOT The normal force is NOT ALWAYS weight!! ALWAYS equal to the weight!! F N = n = 98.0N – 40.0N= 58.0N

Example 5.8Accelerating the box Example 5.8 Accelerating the box m = 10 kg  F g = 98.0N From Newton’s 2 nd Law: ∑F = ma F P – mg = m a  m a = 2.0N accelerates upwards The box accelerates upwards because F P > m g

Example 5.9 Weight Loss Example 5.2 Text Book) Example 5.9 Weight Loss (Example 5.2 Text Book) Apparent weight loss Apparent weight loss. The lady weights 65kg = 640N, the elevator descends with a = a = 0.2m/s 2. F N What does the scale read (F N )? ∑F = ma From Newton’s 2 nd law: ∑F = ma F N – mg = – m a  F N = mg – m a F N 52kgUpwards! F N = 640N – 13N = 627N = 52kg Upwards! F N F N 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 constant speed What does the scale read when the elevator descends at a constant speed of 2.0m/s? ∑F = 0 From Newton’s 2 nd law: ∑F = 0 F N – mg = 0  F N = mg =65kg F N – mg = 0  F N = mg = 640N = 65kg The scale reads her true mass! NOTE: “apparent mass”does not 65 kg 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 OBJECTS IN EQUILIBRIUM acceleration zero EQUILIBRIUM!! If the acceleration of an object that can be modeled as a particle is zero, the object is said to be in EQUILIBRIUM!! net force zero Mathematically, the net force acting on the object is zero

Example 5.10 Equilibrium A lamp is suspended from a chain of negligible mass The forces acting on the lamp are Force of gravity(F g ) Force of gravity (F g ) Tension in the chain (T) Tension in the chain (T) Equilibrium gives

Example 5.10 Equilibrium, final T’T” The forces acting on the chain are T’ and T” T” T” is the force exerted by the ceiling T’ T’ is the force exerted by the lamp T’T T’ is the reaction force to T T T’ 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) Example 5.4 (Text Book) This is an equilibrium problem No movementa = 0 No movement, so a = 0 100N 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? Will the light remain or will one of the cables break? mg =122N

Example 5.11 Traffic Light at Rest, 2 Apply Equilibrium Conditions: Apply Equilibrium Conditions: ΣF y = 0 T 3 – Fg = 0 ΣF y = 0  T 3 – Fg = 0  T 3 = Fg = 122N Find Components: Find Components: T 1x = – T 1 cos37 T 1y = T 1 sin37 T 2x = T 2 cos53 T 2y = T 2 sin53 T 3x = 0 T 3y = – 122N Apply Newton’s 2 nd Law: Apply Newton’s 2 nd Law: ΣF x = 0 ΣF y = 0 ΣF x = 0 ΣF y = 0 -T 1x T 2x T 2y T 1y

Example 5.11 Traffic Light at Rest, final ΣF x = – T 1 cos37 + T 2 cos53 = 0 (1) ΣF y = T 1 sin37 + T 2 sin53 – 122N = 0 (2) Solving for T 1 or T 2 : Solving for T 1 or T 2 : From Eqn (1) solve for T 2 T 2 = (cos37/cos53)T 1 =1.33T 1 (2) Substituting this value into Eqn (2) T 1 sin37 + (1.33T 1 )sin53 =122N T 1 sin37 + (1.33T 1 )sin53 =122N  T 1 = 74.4 N and T 2 = 97.4N Both values are less than 100N, so the cables will not break!!! -T 1x T 1y T 2y T 2x

Objects Experiencing a Net Force, Examples experiences an acceleration (a ≠ 0),nonzero net force (F ≠ 0) 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. free-body Draw a free-body diagram Apply Newton’s Second Law Apply Newton’s Second Law in component form

Example 5.12 Net Force

Forces on The Crate Forces acting on the crate: Forces acting on the crate: T A tension, the magnitude of force T F g The gravitational force, F g n The normal force, n, exerted by the floor

Forces on The Crate, 2 Apply Newton’s Second Law in component form: Apply Newton’s Second Law in component form: ONLYn = F g ONLY in this case n = F g Solve for the unknown(s) Ta 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, F N is NOT always equal to the weight!! m = 10.0 kg  F g = 98.0N Find: a x ≠ 0? F N ? a y = 0 if a y = 0 m=10kg

Example 5.13 Normal Force, final F Py = F P sin(30) = 20.0N F Px = F P cos(30) = 34.6N ΣF x = F Px = m a x  a x = 34.6N/10.0kg  a x = 3.46m/s 2 ΣF y = F N + F Py – mg = m a y  F N + F Py – mg = 0  F N = mg – F Py = 98.0N – 20.0N F N = 78.0N F N = 78.0N F Px F Py

Example 5.14 Conceptual Example: The Hockey Puck constant velocity,NO friction. Moving at constant velocity, with NO friction. Which free-body diagram is correct? (b)

Inclined Planes Choose the coordinate systemx along the incliney perpendicular to the incline Choose the coordinate system with x along the incline and y perpendicular to the incline Forces acting on the object: Forces acting on the object: n The normal force, n, acts perpendicular to the plane F g The gravitational force, F g, acts straight down

Example 5.15 The Runaway Car (Example 5.6 Text Book) force of gravity Replace the force of gravity with its components: F gx = mgsin  F gy = mgcos  a y = 0 a x ≠ 0 With: a y = 0 & a x ≠ 0 (A). a x (A). Find a x Using Newton’s 2nd Law: y-Direction y-Direction ΣFy = n – mgcos  = ma y = 0  n =mgcos  ΣFy = n – mgcos  = ma y = 0  n = mgcos 

Example 5.15 The Runaway Car, final x-direction x-direction ΣF x = mgsin  = ma x  a x = gsin  Independent of m!! Independent of m!! (B) (B) How long does it take the front of the car to reach the bottom? (C). (C). What is the car’s speed at the bottom?