PHU 205 Mechanics for Life Sciences

Chapter 4:

The Laws of Motion

Prof. Liliana Braescu & Prof. Nouredine Zettili

Required text College Physics Raymond A. Serway and Chris Vuille 9th Edition, 2012 BROOKS/COLE CENGAGE Learning ISBN 10: 1-111-42745-3 ISBN 13: 978-1-111-42745-0

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Course Outline Chapter 4

The Laws of Motion 4.1 Forces 4.2 Newton’s First Law 4.3 Newton’s Second Law 4.4 Newton’s Third Law 4.5 Applications of Newton’s Laws 4.6 Forces of Friction

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Chapter 4 The Laws of Motion 4.1 Forces 4.2 Newton’s First Law 4.3 Newton’s Second Law 4.4 Newton’s Third Law 4.5 Applications of Newton’s Laws 4.6 Forces of Friction

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Classical Mechanics u Describes the relationship between the motion of objects in our everyday world and the forces acting on them. u Conditions when Classical Mechanics does not apply:

✔ very tiny objects (< atomic sizes) ✔ objects moving near the speed of light

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4.1 Forces u Commonly imagined as a force as a push or pull on some object. u A vector quantity (it is defined by both magnitude and direction). u May be a contact force or a field force: ✔ Contact forces result from physical contact between two objects. ✔ Field forces act between disconnected objects.

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4.1 Forces u Force – the interaction of an object with its environment. ✔ Contact forces – result f rom a phys ica l contact between two objects (i.e. push/pull forces). ✔ Act ion -a t -a -d is tance forces – do not involve physical contact between two objects. Also called field forces (i.e. magnetic, electric, gravitational, nuclear, and radioactive forces).

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Fundamental Forces u Types:

✔ Strong nuclear force ✔ Electromagnetic force ✔ Weak nuclear force ✔ Gravity

u Characteristics: ✔ All field forces ✔ Listed in order of decreasing strength ✔ The strong and weak nuclear forces operate only on the nuclear scale. ✔ Classical physics deals only with gravitational and electromagnetic forces (infinite range).

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The Three Laws of Sir Isaac Newton

u1642 – 1727 u Formulated basic concepts and laws of mechanics u Universal Gravitation u Calculus u Light and optics

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Chapter 4 The Laws of Motion 4.1 Forces 4.2 Newton’s First Law 4.3 Newton’s Second Law 4.4 Newton’s Third Law 4.5 Applications of Newton’s Laws 4.6 Forces of Friction

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4.2 Newton’s First Law u If there is no external force acting on an object, the object moves with zero acceleration and constant velocity:

u From this, the object is either at rest or is moving on a straight line at a constant speed: ✔ → object at rest ✔ → moving with constant speed

Fext = 0⇒ !a = 0⇒ !v = constant∑

v = 00v C= ≠

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4.2 Newton’s First Law u Every object has both inertia and mass. u Inertia – tendency of an object to resist motion u Mass – a measurement of inertia u Mass is an inherent property (due to the object’s

composition) that is independent of the surrounding environment.

u SI unit of mass → kilogram (kg) u Consider two boxes of equal size: one made of

wood and one made of steel. The box of steel will be harder to lift; thus, it will be more successful at resisting motion and will have more inertia.

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Chapter 4 The Laws of Motion 4.1 Forces 4.2 Newton’s First Law 4.3 Newton’s Second Law 4.4 Newton’s Third Law 4.5 Applications of Newton’s Laws 4.6 Forces of Friction

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4.3 Newton’s Second Law uThe acceleration of an object is directly

proportional to the net force acting on it and inversely proportional to its mass:

u SI unit of force → Newton (N) u Weight – measure of the amount of gravitational

force exerted on an object by the planet: W = mg

!a =

!Fext∑

m⇒

!Fext = m!a∑

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Units of Force

u SI unit of force is a Newton (N): u US Customary unit of force is a pound (lb) 1 N = 0.225 lb

1N ≡ 1kg ms2

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Remarks:

! Since acceleration due to gravity (g) changes from one planet to another, so the weight of an object changes accordingly. u Mass, however, does not change. (W=mg!) u Examples: A 70 kg person weighs three different amounts when on Earth, the moon, and the sun:

(70)(9.8) 686 154.4E EW mg N lb= = = =

(70)(1.6) 112 25.2Moon MoonW mg N lb= = = =(70)(270) 18900 4252.5Sun SunW mg N lb= = = =

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Chapter 4 The Laws of Motion 4.1 Forces 4.2 Newton’s First Law 4.3 Newton’s Second Law 4.4 Newton’s Third Law 4.5 Applications of Newton’s Laws 4.6 Forces of Friction

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4.4 Newton’s Third Law uIf two objects interact, the force exerted on

object 1 by 2 is equal in magnitude but opposite in direction to the force exerted on object 2 by 1:

uAn object sitting on a table exerts a force on the

table equal to its weight: W=mg. uThe table reacts with an equal, opposite force

called the normal force:

!F12 = −

!F21

!N = −

!W

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Problems

u Problem 4.1 A 6.0-kg object undergoes an acceleration of 2.0 m/s2. Questions: (i)  What is the magnitude of the resultant force

acting on it? (ii) If this same force is applied to a 4.0-kg object, what acceleration is produced?

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Problems u Problem 4.1

Solution

(i) FR = ma = (6.0 kg)(2.0 m/s2) = 12 N.

(ii) a = FR/m = 12N/4.0kg = 3.0 m/s2.

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Chapter 4 The Laws of Motion 4.1 Forces 4.2 Newton’s First Law 4.3 Newton’s Second Law 4.4 Newton’s Third Law 4.5 Applications of Newton’s Laws 4.6 Forces of Friction

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4.6 Applications of Newton’s Law u When examining Newton’s laws, we are only interested in the external forces acting upon an object; internal forces are not taken into consideration. uThe best method for applying Newton’s laws to an

object is to draw the object and identify all of the external forces acting on it (actions and reactions).

uThis is known as a free-body diagram. uIn the following slides, we will see examples of

free-body diagrams for two types of systems: objects at rest and objects in motion.

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Free Body Diagram u Must identify all the forces acting on the object of interest u Choose an appropriate coordinate system u If the free body diagram is incorrect, the solution will likely be incorrect

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Free Body Diagram Example u The force is the tension acting on the box; The tension is the same at all points along the rope u are the forces exerted by the earth and the ground

!n and

!F

g

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Free Body Diagram Equilibrium Example

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Free Body Diagram Multiple Objects Example

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4.6 Applications of Newton’s Law (a)  Object at Rest uAn object at rest has and

(i.e. a book resting on a table). uTo solve a problem which involves an object in

equilibrium, follow next steps: ✔ Make a sketch of the situation to be studied. ✔ Draw a free-body diagram and isolate the object. ✔ Find Fx and Fy.

uApply Newton’s First Law and . Solve the system of equations.

!a = 0 !Fext = 0∑

Fx = 0∑ Fy = 0∑

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Problems

u Problem 4.2 Find the tension in each cable supporting the 600-N cat burglar in the figure.

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Problems u Problem 4.2

Solution

✔ Beginning with this free-body diagram, use ΣFy = 0 to find

T2sin37.0° - 600 N = 0, from which T2 = 997 N. ✔Now, use ΣFx = 0 to find

T2cos37.0° - T1 = 0. Substituting the value T2 = 997 N, we find T1 = 796 N.

x

y

37°

T2

T1

W = 600 N

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4.6 Applications of Newton’s Law (b) Object in Motion When presented with a problem involving a moving object, follow these steps to assess the involved forces: u Draw a figure. u Isolate the object. u Sketch a free-body diagram. u Select a coordinate system; then, use it to find Fx and Fy. u Apply Newton’s Second Law to each axis:

& Solve the system of equations.

Fx = max∑ Fy = may∑

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Inclined Planes u Choose the coordinate system with x along the incline and y perpendicular to the incline. u Replace the force of gravity with its components. u a=g sin(theta)

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Inclined Planes

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Problems

u Problem 4.3 A mass, m1 = 5.00 kg, resting on a frictionless horizontal table is connected to a cable that passes over a pulley and then is fastened to a hanging mass, m2 = 10.0 kg, as in the figure. Q u e s t i o n : F i n d t h e acceleration of each mass and the tension in the cable.

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Problems u Problem 4.3

Solution

✔ Each block will be analyzed using its own free-body diagram. ✔First consider the block moving along the horizontal. -  The only force in the direction of

movement is T. ΣFx = ma => T = (5.00 kg)a

(Equation I)

T

+X

5 kg

+Y

T

10 kg

W = 98 N

(a) (b)

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T

+X

5 kg

+Y

T

10 kg

W = 98 N

(a) (b)

Problems Solution (continuation)

✔ Next consider the block which moves vertically. -The forces on it are the tension T and its weight, 98 N. ΣFy = ma = 98 N - T = (10.0 kg)a

(Equation II) u Note that both blocks must have the same magnitude of acceleration. u Equations I and II can be solved simultaneously to give a = 6.53 m/s2 and T = 32.7 N.

T = (5.00 kg)a (Equation I)

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More Problems u Problem 4.4

A 150-N bird feeder is supported by three cables as shown in Figure P4.17. Question: Find the tension in each cable.

Figure  P4.17

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More Problems u Problem 4.5

The leg and cast in the Figure weigh 220 N (w1). Question: Determine the weight w2 and the angle α needed so that no force is exerted on the hip joint by the leg plus the cast.

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More Problems u Problem 4.6

A 1000-kg car is pulling a 300-kg trailer. Together, the car and trailer have an acceleration of 2.15 m/s2 in the forward direction. Questions: Neglecting frictional forces on the trailer, determine (a)  the net force on the car, (b)  the net force on the trailer, (c)  the force exerted by the trailer on the car, and (d) the resultant force exerted by the car on the road.

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Chapter 4 The Laws of Motion 4.1 Forces 4.2 Newton’s First Law 4.3 Newton’s Second Law 4.4 Newton’s Third Law 4.5 Applications of Newton’s Laws 4.6 Forces of Friction

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4.7 Forces of Friction u Friction force occurs at the surface of contact between two objects. u This force has a direction opposite to the direction of motion because friction impedes the motion. u Two types of friction force: ✔ static force of friction – before motion starts ✔ kinetic force of friction – after the object starts to move u N is the normal force u µ is the coefficient of friction, a dimensionless constant which depends only on the two surfaces which are coming in contact

fs = µsN

fk = µk N

0 < µ ≤1

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More About Friction

u Friction is proportional to the normal force. u The force of static friction is generally greater than the force of kinetic friction. u The coefficient of friction (µ) depends on the surfaces in contact. u The direction of the frictional force is opposite the direction of motion. u The coefficients of friction are nearly independent of the area of contact.

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Static Friction, fs

u Static friction acts to keep the object from moving. u If F increases, so does ƒs. u If F decreases, so does ƒs u fs=µsn

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Kinetic Friction, fk

u The force of kinetic friction acts when the object is in motion. u fk=µkn Variations of the coefficient with speed will be ignored. ! Note especially where ƒs = F

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Some Coefficients of Friction

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Remember: Inclined Planes Friction is considered No Friction

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Example: Block on Ramp Example u Axes are rotated as usual on an incline. u The direction of impending motion would be down the plane. u Friction acts up the plane – It opposes the motion. u Apply Newton’s Laws and solve equations.

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Remarks: u On a horizontal surface, N=mg. u On an incline (non-horizontal surface): In above formulas, f can be fs or fk , depending on the problem.

f N mgµ µ= =

cosN mg θ=

cosf N mgµ µ θ= =

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Problems

u Problem 4.7 Masses m1 = 10.0 kg and m2 = 5.00 kg are connected by a light string that passes over a fr ict ionless pul ley (refer to the f igure accompanying Problem 4.3). Question: If, when the system starts from rest, m2 falls 1.00 meter in 1.20 seconds, determine the coefficient of kinetic friction between m1 and the table.

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Problems u Problem 4.7 Solution

u Each block will be analyzed using its own free-body diagram. Similarly, it was the case of the Problem 4.3, but….here we have friction!

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Problems u Problem 4.7 Solution

✔The acceleration of the system is found from y = v0t + (1/2)at2, or 1.00 = 0 + (1/2)a(1.20)2 which gives a = 1.39 m/s2.

✔Using the free body diagram of m2, the second law gives 5.00(9.80) - T = 5(1.39), or T = 42.1 N. ✔Applying the second law to the horizontal forces on m1, 42.1 - f = 10.0(1.39), or f = 28.2 N. ✔ Since N = m1g = 98.0 N, we have

µk = f/N = 28.2/98.0 = 0.288.

fT

m2gmg

N = mg Ta

a

m1 10 kg

m2 5 kg

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Chapter 4 Ends

Quiz 4 is coming ….. before Chapter 5 For review, Study the following problems: 4.12 page 107; 4.13 page

108; 4.14 page 109; 4.15 page 110. Solve the following problems: 41, 47 pages 118-119; 59, 60,

61 page 120; 73 page 122. Textbook “College Physics” Raymond A. Serway and Chris Vuille 9th Edition, 2012