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## PowerPoint Slideshow about ' SECOND LAW OF MOTION' - nash-smith

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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 Motionstatesthat 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:

In the SI system, the unit for force is the newton(N):

A newtonis that net force which, when applied to a 1-kg mass, gives it an acceleration of 1 m/s2.

Pushing the cart with twice the force produces twice the acceleration. Three times the force triples the acceleration.

Acceleration and Force With Zero Friction ForcesF

a/2

a

Pushing two carts with same force Fproduces one-half the acceleration. The acceleration varies inverselywith the mass.

Acceleration and Mass Again With Zero FrictionAccording 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 totwo 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.

Third Law:For every action force, there must be an equal and opposite reaction force. Forces occur in pairs.

Action

Reaction

Action

Reaction

Newton’s Third LawUse the words by and on to study action/reaction forces below as they relate to the hand and the bar:

Action

Reaction

Acting and Reacting ForcesThe action force is exerted by the _____ on the _____.

hands

bar

The reaction force is exerted bythe _____ on the _____.

bar

hands

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(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 theacceleration of gravity g.

FG= mgUnits: Newtons (N)

g = acceleration due to gravity

= 9.8 m/s2

Anormal forceis 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 = kgFG = 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 axSFy = 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?

5kg

N2L

m = 5 kg

a = 0.3 m/s2

ΣFy = FT - FG = ma

FT = m(a + g)

= 5(0.3 + 9.8)

= 50.5 N

FT

a (+)

FG

9 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

FT

a (+)

a (+)

FG1

FG2

FT

b. Find the tension of the cord

a (+)

a (+)

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

FG1

FG2

Theactual weight of a body is the gravitational force that acts on it. The body\'s apparent weightis the force the body exerts on whatever it rests on. Apparent weight can be thought of as thereading on a scalea 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

FN

m = 75 kg

FG = 735 N

ΣFy = FN - FG = 0

FN = FG

= 735 N

FG

b. With constant downward speed of 5 m/s

FN= 735 N

ΣF = FN - FG = ma

FN = ma + FG

= 75 (2.45) + 735

= 919 N

b. Going up with an

acceleration of

0.25 g

a = 2.45 m/s2

FG

N2L

ΣF = FG - FN = ma

FN = FG - ma

= 735 - 75 (2.45)

= 551 N

c. Going down with an

acceleration of 0.25 g

a = 2.45 m/s2

FG

N2L

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.

The frictional force between two surfaces depends on thenormalforce FNpressing them together and on the natures of the surfaces. The latter factor is expressed quantitatively in thecoefficient of friction(mu)whose value depends on the materials in contact. The frictional force is experimentally found to be:

Static friction

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

ΣFy = FN – FG = 0

FN = FG =mg

= 60(9.8)

= 588 N

FG

ΣFx = Fa – Ff = 0

Fa = Ff = μFN

= 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

ΣFx = Fa – Ff = ma

Ff = μFN

= (0.5)(686)

= 343 N

FG

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

m = 70 kg

Fa= 400 N, 30

μ = 0.5

Fa

Ff

FG

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

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

ΣFy = FN - FGy = 0

FN = FGy

FN

Ff

ΣFx = FGx – Ff = ma

Ff = μFN

ΣFx = FGx –μ FGy= ma

θ

FGy

FG

FGx

θ = 30

μ = 0.3

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 acceleration of 200 cm/s2?

N2L

FG1 = FN1

FG2 = FN2

FN1

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 = 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

ΣFx = m2a

F12 - Ff2 = m2a

F12 = Ff2 + m2a

= μ m2 g + m2a

= 0.4(0.5) (9.8) + 0.5 (2)

= 2.96 N

FN2

Ff2

F12

FG2

15 kg

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

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

N2L

b.What is the tension on the string?

ΣF = FT - Ff = mAa

FT = mAa + Ff

= 25 (2.45) + 49

= 110.25 N

FN

Ff

FT

FT

FGA

FGB

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.

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.

c. Determine the value of the suspended mass M that allows blocks 1 and 2 to move with constant velocity down the plane.

FN2

FT

N1L

FN1

2f

FT

FT

f

FT

m2 g

m1 g

Mg

d. The string between blocks 1 and 2 is now cut. Determine the acceleration of block 1 while it is on the inclined plane.

FN1

f

m1 g

N2L

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

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 axSFy = m ay
- Solve for unknown quantities.

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