Kinetics of a Particle

Force and Acceleration

Chapter Two

Sir Isaac Newton at aged 46

F = ma

Have you seen these equations before?

Equation of MotionNewton’s Second Law of Motion

Normal & Tangential coordinates

Rectangular coordinates

F = ma

Fx = max Fy = may

Fz = maz

Fn = man

Ft = mat

Applied Forces, F Weight, W

Friction, FfSpring, Fs

Normal Force, FN

F = ma

Kinetics of a Particle

Force and Acceleration:Rectangular Coordinates

Fxi + Fyj + Fzk = m(axi + ayj + azk)Fxi + Fyj + Fzk = m(axi + ayj + azk)

F = maF = ma

Fx = max Fy = may Fz= maz

Fx = max Fy = may Fz= maz

Scalar equations

Create FBD

Apply Equations of Motion, F=ma

Apply Kinematics

Procedure for Analysis

Velocity Position

Example 2.1:The 50-kg crate rests on a horizontal plane for which the coefficient of kinetic friction is μk = 0.3. If the crate is subjected to a 400-N towing force,

determine the velocity of the crate in 3 s starting

from rest.

Example 2.1:The 50-kg crate rests on a horizontal plane for which the coefficient of kinetic friction is μk = 0.3. If the crate is subjected to a 400-N towing force,

determine the velocity of the crate in 3 s starting

from rest.

To find velocity, v

To know acceleration, a

To apply equation of motion, F=ma

Solution 2.1:

FBD Weight = mg = 50 (9.81) = 490.5 N

Friction force, Ff = k Nc

= 0.3 Nc

Solution 2.1: continued

400 cos 30

400 sin 30

Fx (N) Fy (N)400 cos 30 400 sin 30

- 0.3 Nc - 490.5

- Nc

x

y

Solution 2.1: continued

Equations of Motion.

2/19.5

5.290

030sin4005.490;

503.030cos400;

sma

NN

NmaF

aNmaF

C

Cyy

Cxx

Solving for the two equations yields

Solution 2.1: continuedKinematics. Acceleration is constant, since the applied force P is constant. Initial velocity is zero, the velocity of the crate in 3 s is

sm

tavv c

/6.15

)3(19.500

Example 2.2:The 400 kg mine car is hoisted up the incline.

The force in the cable is F = (3200t2) N. The car has an initial velocity of 2 m/s at t = 0.

Find:The velocity when t = 2 s.

Solution 2.2:

=

x

y

W = mg F

N

maFBD

Since the motion is up the incline, rotate the x-y axes.

= tan-1(8/15) = 28.07°

Motion occurs only in x-direction.

Solution 2.2: continued

Equation of motion in x-direction:

+ Fx = max => F – mg(sin ) = max

=> 3200t2 – (400)(9.81)(sin 28.07°) = 400a

=> a = (8t2 – 4.62) m/s2

Solution 2.2: continuedUse kinematics to determine the velocity:

a = dv/dt => dv = a dt

dv = (8t2 – 4.62) dt,

v – 2 = (8/3t3 – 4.62t) = 12.10

=> v = 14.1 m/s

v

2

2

0

2

0

Solution 2.2: continuedDetermine the distance it moves up the plane when t = 2 s:

a = dv/dt => dv = a dt

dv = (8t2 – 4.62) dt,

v – 2 = (8/3t3 – 4.62t)

ds = 8/3t3 – 4.62t + 2 dt,

s = 8/12t4 – 4.62/2t2 + 2t => = 5.43 m

v

2

t

0

s

0

t

0

Example 2.3

A 2 kg block is released from rest at A and slides down the inclined plane. If the coefficient of kinetic friction between the plane and the block is k = 0.3, determine the speed of the block after it slides 3 m down the plane.

= 30

Solution 2.3

FBD

Fx (N) Fy (N)19.62 sin 30 -19.62 cos 30

- 0.3 NB NB

Solution 2.3 (continued)

2/36.2

2)99.16(3.081.9

23.030sin62.19

99.16

030cos62.19

sma

a

aN

maF

NN

N

maF

x

x

xB

xx

B

B

yy

Equations of motion

Solution 2.3 (continued)

smv

v

savv xo

/76.3

)3)(36.2(2

222

Kinematics

Kinetics of a Particle

Force and Acceleration:Normal & Tangential

Coordinates

ApplicationsThe car goes

around a curve

Wheel-ride

The moon and satellites held in their orbit

                               

                                        

Ftut + Fnun + Fbub = mat + man

Ftut + Fnun + Fbub = mat + man

F = maF = ma

ΣFt, ΣFn, ΣFb represent the sums of all the force components acting on the particle in the

tangential, normal and binormal directions.

Ft = mat Fn = man

Fb= 0

Ft = mat Fn = man

Fb= 0

The particle is constrained to move along the path, so there is no motion in the binormal

direction

The net force and the acceleration are always in the same direction!

F = maF = ma

If the object moves along the circular path

has constant speed

Ft = 0, since at is 0(uniform circular motion)

Ft = 0, since at is 0(uniform circular motion)

Create FBD

Equations of Motion, F=ma

Kinematics

Procedure for Analysis

at=dv/dt

at=vdv/dsan=v2/

Example 2.4 (13-69):

Given: A 200 kg snowmobile with rider is traveling down the hill. When it is at point A, it is traveling at 4 m/s and increasing its speed at 2 m/s2.

Find:The resultant normal force and resultant frictional force exerted on the tracks at point A.

Solution 2.4:

W = mg = weight of snowmobile and passengerN = resultant normal force on tracksFf = resultant friction force on tracks

1) The n-t coordinate system can be established on the snowmobile at point A.

• Treat the snowmobile and rider as a particle and draw the free-body and kinetic diagrams:

tn

matman

t

n

N

Ff

W

=

Solution 2.4: continued2) Apply the equations of motion in n-t directions:

Ft = mat => W sin – Ff = mat

Fn = man => W cos – N = man

Using W = mg and an = v2/ = (4)2/

=> (200)(9.81) cos – N = (200)(16/)

=> N = 1962 cos – 3200/

Using W = mg and at = 2 m/s2 (given)

=> (200)(9.81) sin – Ff = (200)(2)

=> Ff = 1962 sin – 400 (2)

+

+

Solution 2.4: continued3) Determine by differentiating y = f(x) at x = 10 m:

Determine from the slope of the curve at A:

y = -5(10-3)x3 => dy/dx = (-15)(10-3)x2 => d2y/dx2 = -30(10-3)x

tan = dy/dx

= tan-1 (dy/dx) = tan-1 (-1.5) = 56.31°

x = 10 m

dy

dx

= =[1 + ( )2]3/2

dydx

d2ydx2

[1 + (-1.5)2]3/2

-0.3x = 10 m

= 19.53 m

Solution 2.4: continued

From Eq.(1): N = 1962 cos(56.31) – 3200/19.53 = 924 N

From Eq.(2): Ff = 1962 sin(56.31) – 400 = 1232 N

Example 2.5:Design of the ski requires knowing the type of forces that will be exerted on the skier and his approximate trajectory. In the case as shown, determine the normal force and acceleration on the 600 N skier the instant he arrives at the end of jump, A, where his velocity is 9 m/s.

Solution 2.5:Free-Body Diagram. Since the path is curved, there are two components of acceleration, an and at. Since an can be calculated, the unknown are at and NA.

Equations of Motions.

ttt

Ann

amaF

NmaF

81.9600

0;

981.9

600600;

2

Solution 2.5: continuedThe radius of curvature ρ for the path must be determined at point A(0, -15 m). Here

301

301

15601

2

2

2

dx

yd

xdxdy

xy

Solution 2.5: continuedSo at x = 0,

m

dxyd

dxdy

30

/

)/(122

2/32

X=0

Solving for NA,

NN

N

N

A

A

A

765

30

9

81.9

600600

9

81.9

600600

2

2

Solution 2.5: continued

Kinematics. With at = 0

2

22

/7.2

/7.2

smaa

smv

a

nA

n

TutorialProblems (12th Edition)

13-4, 13-12, 13-53, 13-62, 13-72