 # FORCE A force is any influence that can change the velocity of a body. Forces can act either through the physical contact of two objects (contact forces:

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FORCE A force is any influence that can change the velocity of a body. Forces can act either through the physical contact of two objects (contact forces: push or pull) or at a distance (field forces: magnetic force, gravitational force).

FREE BODY DIAGRAMS In all but the simplest problems that involve forces, it is helpful to draw a free body diagram (FBD) of the situation. This is a vector diagram that shows all the forces that act on the body whose motion is being studied. Forces that the body exerts on anything else should not be included, since such forces do not affect the body's motion.

FT force due to tension: a rope or cord
4.1 Complete the free body diagram showing all of the forces acting on the mass M. Be sure to show the direction of each force as an arrow and label each force clearly! FT force due to tension: a rope or cord FN force normal: acting perpendicular to a surface Ff force of friction: opposes motion Fg gravitational force or weight, always downward Fa applied force: push or pull Fs force exerted by a spring

Example: FN Fa Ff Fg

Fs Fg

FN1 FN2 Fa Ff FT FT Fg1 Fg2

FN Fa Ff Fg

FN Fa Ff Fg

FN2 FN1 FT Ff Ff Fa Fg1 Fg1+ Fg2

FIRST LAW OF MOTION According to Newton's First Law of Motion:
According to Newton's First Law of Motion: " If no net force acts on it, a body at rest remains at rest and a body in motion remains in motion at constant speed in a straight line." Isaac Newton ( )

MASS The property 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 SI unit of mass is the kilogram (kg).

Motion tends to continue unchanged.
The elephant at rest tends to remain at rest. Tablecloth trick: Too little force, too little time to overcome "inertia" of tableware.

SECOND LAW OF MOTION According to Newton's Second Law of Motion, the net force acting on a body equals the product of the mass and the acceleration of the body. The direction of the force is the same as that of the acceleration. In equation form: F = ma  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.

Net force is sometimes designated F.
The second law of motion is the key to understanding the behavior of moving bodies since it links cause (force) and effect (acceleration) in a definite way.

F = ma = 2 m/s2 vf = vo + at = 2(5) = 10 m/s
4.2 A force of 3000 N is applied to a 1500-kg car at rest. a. What is its acceleration? F = 3000 N m = 1500 kg vo = 0 m/s F = ma = 2 m/s2 b. What will its velocity be 5 s later? vf = vo + at = 2(5) = 10 m/s

4. 3 A 1000 kg car goes from 10 to 20 m/s in 5 s
4.3 A 1000 kg car goes from 10 to 20 m/s in 5 s. What force is acting on it? m = 1000 kg vo = 10 m/s vf = 20 m/s t = 5 s F = ma = 2000 N

4.4 A 60-g tennis ball approaches a racket at 15 m/s, is in contact with the racket for s, and then rebounds at 20 m/s. Find the average force exerted by the racket. m = 0.06 kg vo = 15 m/s t = s vf = - 20 m/s F = ma = N

= - 3 m/s2 = 10 s x = vot+½at2 = 30(10)+ ½ (-3)(10)2 = 150 m
4.5 The brakes of a 1000-kg car exert 3000 N. a. How long will it take the car to come to a stop from a velocity of 30 m/s? m = 1000 kg F = N vo = 30 m/s vf = 0 m/s = - 3 m/s2 = 10 s b. How far will the car travel during this time? x = vot+½at2 = 30(10)+ ½ (-3)(10)2 = 150 m

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: "The action force one object exerts on the other, and the equal but opposite reaction force the second object exerts on the first." Action and reaction forces never balance out because they act on different objects.

Action-Reaction Pair Examples

The forces FN and Fg are balanced.
4.6 A book rests on a table. a. Show the forces acting on the table and the corresponding reaction forces. b. Why do the forces acting on the table not cause it to move? ACTION: FTB REACTION: FBT The forces FN and Fg are balanced.

WEIGHT The weight of a body is the gravitational force with which the Earth attracts the body. 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. From the Second Law of Motion: W = mg Units: Newtons (N)

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.

= 1020.4 kg = 3.92 m/s2 vf = vo + at = 0 +3.92(8) = 3.14 m/s
4.7 A net horizontal force of 4000 N is applied to a car at rest whose weight is 10,000 N. What will the car's speed be after 8 s? Fa = 4000 N Fg = 10,000 N t = 8s = kg = 3.92 m/s2 vf = vo + at = (8) = 3.14 m/s

Fg = mg = 5(9.8) = 49.1 N FT ΣFy = FT - Fg = ma FT = ma + Fg
4.8 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 Fg = mg = 5(9.8) = 49.1 N ΣFy = FT - Fg = ma FT = ma + Fg = 5(0.3) = 50.5 N FT a (+) Fg

FN Fa ΣFx = Fa – Fgx = ma = 0 Fa = Fgx = Fgsin 30 = 200(sin30) Fgy
4.9 A 200-N wagon is to be pulled up a 30 incline at constant speed. How large a force parallel to the incline is needed if friction is negligible? Fg = 200 N FN Fa ΣFx = Fa – Fgx = ma = 0 Fa = Fgx = Fgsin 30 = 200(sin30) = 100 N Fgy θ Fg Fgx

4. 10 A cord passing over a frictionless pulley has a 7
4.10 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 FT FT a (+) a (+) Fg2 Fg1

FT FT a (+) a (+) ΣF = FT – Fg1 -FT + Fg2 = mTa Fg1 Fg2 = 1.22 m/s2

Use either side of the pulley: you get the same answer! FT – Fg1 = m1a
b. Find the tension of the cord a (+) a (+) Use either side of the pulley: you get the same answer! FT – Fg1 = m1a FT = m1a+ Fg1 = m1(a+g) = 7( ) = 77.1 N Fg1 Fg2

FN Fg = mg = 45(9.8) FT = 441 N ΣFy = FN – Fgy = 0 FN = Fgy=Fg(cos35º)
4.11 A crate, which has a mass of m = 45.0 kg, is being pulled up a frictionless inclined plane, at an angle of 35˚ by a rope. a. What will be the magnitude of the normal force (FN) acting on the crate? m = 45.0 kg FN Fg = mg = 45(9.8) = 441 N ΣFy = FN – Fgy = 0 FN = Fgy=Fg(cos35º) = 441cos 35º = N FT Fgy θ Fg Fgx

ΣFx = FT – Fgx = 0 FT = Fgx=Fg(sin35˚) = 441(sin 35˚) = 252.9 N FN FT
b. What will be the magnitude of the tension force (FT) in the rope? ΣFx = FT – Fgx = 0 FT = Fgx=Fg(sin35˚) = 441(sin 35˚) = N FN FT Fgy θ Fg Fgx

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.

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 (perpendicular) force N 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: Static friction Kinetic friction

FN Fa Ff ΣFy = FN – Fg = 0 FN = Fg =mg = 60(9.8) Fg = 588 N
4.12 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? FN Fa = 140 N m = 60 kg Fa Ff ΣFy = FN – Fg = 0 FN = Fg =mg = 60(9.8) = 588 N Fg ΣFx = Fa – Ff = ma = 0 Fa = Ff = μFN = 0.24

FN Ff Fg ΣFx = – Ff = ma = - 1.75 m/s2
4.13 A 400-g block with an initial speed of 80 cm/s slides along a tabletop against a friction force of 0.70 N. a. How far will it slide before stopping? FN m = 400x10-3 kg vo = 0.8 m/s Ff = 0.7 N Ff Fg ΣFx = – Ff = ma = m/s2

FN Ff Fg = 0.18 m

ΣFy = FN – Fg = 0 FN = Fg =mg = 400x10-3(9.8) = 3.92 N
b. What is the coefficient of friction between the block and the tabletop? ΣFy = FN – Fg = 0 FN = Fg =mg = 400x10-3(9.8) = 3.92 N ΣFx = – Ff = ma = 0 Ff = μFN = 0.18

4.14 A 600-kg go-cart is moving on a level road at 30 m/s.
a. How large a retarding force is required to stop it in a distance of 70 m? m = 600 kg vo = 30 m/s x = 70 m vf= 0 m/s = m/s2 F = ma = 600(-6.42) = N

ΣFy = FN – Fg = 0 FN = Fg =mg = 600(9.8) = 5880 N Ff = μFN = 0.65
b. What is the minimum coefficient of friction between the tires and the road? ΣFy = FN – Fg = 0 FN = Fg =mg = 600(9.8) = 5880 N Ff = μFN = 0.65

ΣFy = FN – Fg = 0 FN = Fg =mg = 70(9.8) = 686 N ΣFx = Fa – Ff = ma
4.15 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. m = 70 kg Fa = 400 N μ = 0.5 ΣFy = FN – Fg = 0 FN = Fg =mg = 70(9.8) = 686 N ΣFx = Fa – Ff = ma Ff = μFN = (0.5)(686) = 343 N = 0.81 m/s2

FN Fa Ff Fax = 400 cos 30 = 346.4 N Fg Fay = 400 sin 30 = 200 N
4.16 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 m = 70 kg Fa = 400 N, 30 μ = 0.5 FN Fa Ff Fax = 400 cos 30 = N Fay = 400 sin 30 = 200 N Fg Ff = μFN = (0.5)(486) = 243 N ΣFy = FN + Fay - Fg = 0 FN = Fg - Fay = 70(9.8) - 200 = 486 N

FN Fa Ff ΣFx = Fax – Ff = ma Fg = 1.47 m/s2

FN Ff Fa Fg Fax = 400 cos 50 = 257.1 N Fay = 400 sin 50 = 306.4 N
4.17 A force of 400 N pushes on a 25-kg box at an angle of 50. Starting from rest, the box achieves a velocity of 2.0 m/s in a time of 4.0 s. Find the coefficient of friction between the box and the floor. FN Fa = 400 N, 50 m = 25 kg vo = 0 m/s vf = 2 m/s t = 4 s Ff Fa Fg Fax = 400 cos 50 = N Fay = 400 sin 50 = N = 0.5 m/s2

FN ΣFy = FN - Fay - Fg = 0 FN = Fay +Fg = (9.8) = N Ff Fa Fg Ff = μFN ΣFx = Fax - Ff = ma Ff = Fax - ma = (0.5) = N = 0.44

FN Ff Fg = 20(9.8) Fgy =196 N Fg Fgx = 196 cos 30 = 98 N Fgx
4.18 A 20-kg 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. m = 20 kg θ = 30 μ = 0.3 FN Ff Fg = 20(9.8) =196 N Fgy θ Fg Fgx = 196 cos 30 = 98 N Fgy = 196 sin 30 = N Fgx

FN Ff ΣFy = FN - Fgy = 0 FN = Fg = 169.7 N Ff = μFN = (0.3)(169.7) Fg
θ Ff = μFN = (0.3)(169.7) = 50.9 N Fg Fgy Fgx ΣFx = Fgxx – Ff = ma = 2.35 m/s2

FN1 FN2 Fg1 = FN1 Fg2 = FN2 Ff1 F Ff2 Fg1 Fg2
4.19 Two blocks m1 (300 g) and m2 (500 g), are pushed by a force F. If the coefficient of friction 0.40. a. What must be the value of F if the blocks are to have an acceleration of 200 cm/s2? FN1 FN2 Fg1 = FN1 Fg2 = FN2 m1 = 0.3 kg m2 = 0.5 kg μ = 0.4 a = 2 m/s2 Ff1 F Ff2 Fg1 Fg2 ΣFx = F - Ff1 - Ff2 = mTa F = mTa + Ff1 + Ff2 = mTa + FNT μ = mTa +mTg μ = 0.8(2) (9.8)(0.4) = 4.7 N

m2 alone ΣFx = m2a F12 - Ff2 = m2a F12 = Ff2 + m2a
b. How large a force does m1 then exert on m2? m2 alone ΣFx = m2a F12 - Ff2 = m2a F12 = Ff2 + m2a = 0.5 (9.8) (0.4) (2) = 2.96 N

FN Ff FT FT FgA FN = FgA FgB = 25(9.8) = 245 N FfA = μFN = 0.2 (245)
4.20 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. If the coefficient of friction is 0.20 between the table and block A, how far will block B drop in the first 3.0 s after the system is released? FN Ff FT FT FgA FN = FgA = 25(9.8) = 245 N FfA = μFN = 0.2 (245) = 49 N FgB mA = 25 kg mB = 15 kg μ = 0.2 t = 3 s

ΣF = FT - FfA + FgB - FT = mTa
FN Ff FT FT = 2.45 m/s2 FgA y = ½at2 = ½ (2.45)(3)2 = 11 m FgB

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