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All figures taken from Vector Mechanics for Engineers: Dynamics, Beer and Johnston, 2004

ENGR 214Chapter 12

Kinetics of Particles:Newton’s Second Law

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Newton’s Second Law of Motion

• If the resultant force acting on a particle is not zero, the particle will have an acceleration proportional to the magnitude of resultant and in the direction of the resultant.

• We must use a Newtonian frame of reference, i.e., one that is not accelerating or rotating.

• If no force acts on particle, particle will not accelerate, i.e., it will remain stationary or continue on a straight line at constant velocity.

F ma

• If particle is subjected to several forces:

F ma

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Linear Momentum of a Particle

dvF ma m

dtd d

mv Ldt dt

L mv

Linear momentum

F L Sum of forces = rate of change of linear momentum

0F

If linear momentum is constant

Principle of conservation of linear momentum

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Equations of Motion• Newton’s second law amF

• Convenient to resolve into components:

zmFymFxmF

maFmaFmaF

kajaiamkFjFiF

zyx

zzyyxx

zyxzyx

• For tangential and normal components:

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t t n n

t n

F ma F ma

dv vF m F m

dt

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Dynamic Equilibrium

• Alternate expression of Newton’s law:

0F ma

• If we include inertia vector, the system of forces acting on particle is equivalent to zero. The particle is said to be in dynamic equilibrium.

• Inertia vectors are often called inertia forces as they measure the resistance that particles offer to changes in motion.

ma

inertia vector

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Sample Problem 12.2

An 80-kg block rests on a horizontal plane. Find the magnitude of the force P required to give the block an acceleration of 2.5 m/s2 to the right. The coefficient of kinetic friction between the block and plane is k = 0.25.

SOLUTION:

• Draw a free body diagram

• Apply Newton’s law. Resolve into rectangular components

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Sample Problem 12.2

80 9.81 785

0.25k

W mg N

F N N

:maFx

cos30 0.25 80 2.5

200

P N

:0 yF

sin30 785 0N P

sin30 785

cos30 0.25 sin30 785 200

N P

P P

534.7P N

Pcos30

Psin30

Solve for P and N

1052.4N N

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Sample Problem 12.3

The two blocks shown start from rest. The horizontal plane and the pulley are frictionless, and the pulley is assumed to be of negligible mass. Determine the acceleration of each block and the tension in the cord.

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Sample Problem 12.3

x A AF m a 1 100 AT a

y B BF m a

2

2

2

300 9.81 300

2940- 300

B B B

B

B

m g T m a

T a

T a

y C CF m a 02 12 TT

x

y

O

• Kinematic relationship: If A moves xA to the right, B moves down 0.5 xA

1 12 2B A B Ax x a a

Draw free body diagrams & apply Newton’s law:

12940- 300 2 0Ba T 2940- 300 200 0B Aa a

2940- 300 2 200 0B Ba a 24.2 /Ba m s 28.4 /Aa m s 1 840T N 2 1680T N

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Sample Problem 12.4

The 12-lb block B starts from rest and slides on the 30-lb wedge A, which is supported by a horizontal surface.

Neglecting friction, determine (a) the acceleration of the wedge, and (b) the acceleration of the block relative to the wedge.

Block

Wedge

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N1

aBn

aBtWBcos

WBsin

N1

WB

N1cos

N1sin

sinB B BtW m a 212

12 0.5 16.1 /32.2 Bt Bta a ft s

aA

1 sin A AN m a 1

300.5

32.2 AN a

1 2cos AN W N 1 cosB B BnN W m a

But sinBn Aa a Same normal acceleration (to maintain contact)

1 cos sinB B AN W m a 1

12 0.510.39

32.2 AN a

25.08 /Aa ft s 22.54 /Bna ft s

Draw free body diagrams for block & wedge

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N1

aBn

aBtWBcos

WBsin

N1

WB

N1cos

N1sin

2cos sin 12.67 /Bx Bt Bna a a ft s

aA

2sin cos 10.25 /By Bt Bna a a ft s /B A B Aa a a

/ 12.67 10.25 5.08

17.75 10.25

B Aa i j i

i j

20.5

30°

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Sample Problem 12.5

The bob of a 2-m pendulum describes an arc of a circle in a vertical plane. If the tension in the cord is 2.5 times the weight of the bob for the position shown, find the velocity and acceleration of the bob in that position.

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Sample Problem 12.5

Resolve into tangential and normal components:

:tt maF

30sin

30sin

ga

mamg

t

t

2sm9.4ta

:nn maF

30cos5.2

30cos5.2

ga

mamgmg

n

n

2sm03.16na

• Solve for velocity in terms of normal acceleration.

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sm03.16m2 nn avv

a

sm66.5v

mgsin30

mgcos30

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Sample Problem 12.6

Determine the rated speed of a highway curve of radius = 400 ft banked through an angle = 18o. The rated speed of a banked highway curve is the speed at which a car should travel if no lateral friction force is to be exerted at its wheels.

SOLUTION:

• The car travels in a horizontal circular path with a normal component of acceleration directed toward the center of the path.The forces acting on the car are its weight and a normal reaction from the road surface.

• Resolve the equation of motion for the car into vertical and normal components.

• Solve for the vehicle speed.

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Sample Problem 12.6

SOLUTION:

• The car travels in a horizontal circular path with a normal component of acceleration directed toward the center of the path.The forces acting on the car are its weight and a normal reaction from the road surface.

• Resolve the equation of motion for the car into vertical and normal components.

:0 yF

cos

0cos

WR

WR

:nn maF

2sin

cos

sin

v

g

WW

ag

WR n

• Solve for the vehicle speed.

18tanft400sft2.32

tan2

2 gv

hmi1.44sft7.64 v

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Angular Momentum

OH r mv

Derivative of angular momentum with respect to time:

O

O

H r mv r mv v mv r ma

r F

M

Sum of moments about O = rate of change of angular momentum

and r mv

L mv

From before, linear momentum:Now angular momentum is defined as the moment of momentum

OH

is a vector perpendicular to the plane containing

Resolving into radial & transverse components:2

OH mv r mr

Moment of about OF

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Equations of Motion in Radial & Transverse Components

rrmmaF

rrmmaF rr

2

2

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Central Force

When force acting on particle is directed toward or away from a fixed point O, the particle is said to be moving under a central force.

Since line of action of the central force passes through O:

0O OM H

constantOr mv H

O = center of force

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Sample Problem 12.7

A block B of mass m can slide freely on a frictionless arm OA which rotates in a horizontal plane at a constant rate.0

a) the component vr of the velocity of B along OA, and

b) the magnitude of the horizontal force exerted on B by the arm OA.

Knowing that B is released at a distance r0 from O, express as a function of r

SOLUTION:

• Write the radial and transverse equations of motion for the block.

• Integrate the radial equation to find an expression for the radial velocity.

• Substitute known information into the transverse equation to find an expression for the force on the block.

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Sample Problem 12.7

Write radial and transverse equations of motion:

r rF m a

F m a

dr

dvv

dt

dr

dr

dv

dt

dvvr r

rrr

r

2 2 2 20 0rv r r

1 22 2 20 02F m r r

20

2

m r r

F m r r

2r r

But rv r2 r

r

dvr v

dr 2

r rr dr v dv

2

0

r

o

v r

r r o

r

v dv r dr

1 22 20 0rv r r