CHAPTER 13

Kinetics of a Particle:Force and acceleration ( Newton’s method)

v = vo + act

s = so + vot + act2

v2 = vo2 + 2ac(s-so)

21

2van

22tn aaa

vat vdvdsa t

Normal & Tangential Components

Rectilinear motion

constant acceleration

Curvilinear motion

KINEMATICS

KINETICS

Newton’s MethodWork and

Energy MethodImpulse and momentum

CHAPTER 13

A particle originally at rest, will remain in rest

Newton’s first law

F=0

Newton’s second law

A particle acted upon by an unbalanced force,experiences acceleration that has the samedirection as the force and a magnitude that is directlyproportional to the force

F=ma

STATICS

DYNAMICS

Free body diagram Kinetics diagram

The crate has a mass of 50 kg. If the crate is subjectedto a 400[N] towing force as shown, determine thevelocity of the crate in 3[s] starting from rest. s= 0.5, k= 0.3,

Example 1

Equations of Motion :

Kinematics : The acceleration is constant, P is constant

The crate has a mass of 80 kg. If the magnitude of P isincreased until the crate begins to slide, determine thecrate’s initial acceleration. s= 0.5, k= 0.3,

Example 2

20o

T

80(9.81) = 784.8[N]

N

Ff = 0.5N

s= 0.5k= 0.3

[solution]

: verge of slipping: impending motion

Equations of equilibrium :

Fx=0 ;

Fy=0 ;

Tcos20o – 0.5N = 0 ……..(i)

N + Psin20o – 784.8 = 0 ……...(ii)

T = 353.29 [N] , N = 663.79 [N]

20o

353.29 [N]

784.8 [N]

N

Ff = 0.3N

a

[solution]

Equations of Motion :

Fx=max ;

Fy=may ;N – 784.8 + 353.29sin20o = 80(0) N = 663.97 [N]

353.29cos20o – 0.3(663.97) = 80a a = 1.66 [m/s2]

Equations of Motion:Normal and Tangential Coordinates

When a particle moves along a curved path, it may be more convenient to write the equation of motion in

terms of normal and tangential coordinates.

o t

FtutFnun

o t

b

Fbub

Ft = mat

Fn = man

Fb = 0

Ft : tangential force

Fn : centripertal force

Fb : binormal force

the force components acting on the particle

Example 1

The 3 kg disk is D is attached to the end of a cord as shown.The other end of the cord is attached to a ball-and-socket joint located at the center of the platform. If the platform is rotating rapidly,and the disk is placed on it and released from rest as shown, determinethe time it takes for the disk to reach a speed great enough to break the cord. The maximum tension the cord can sustain is 100 N. k = 0.1

Frictional force : F = kND

Direction : oppose the relative motion of the disk with respect to the platform

Equations of Motion :

Fn=man ;

Ft=mat ;

Fb=0;

T = m(v2/) = 3v2

0.1ND = 3at

ND – 29.43 = 0

Setting T = 100 N to get critical speed vcr

ND = 29.43 Nat = 0.981 m/s2

vcr = 5.77 m/s

Kinematics :

Since at is constant, vcr = vo + att5.77 = 0 + (0.981)tT = 5.89 s

At the instant = 60o, the boys center of mass is momentarily at rest.

Determine the speed and the tension in each of the two supporting cords of the swing when = 90o.

The boy has a weight of 300 N ( ≈ 30 kg).Neglect his size and the mass of the seat and cords

Example 2

W

n

t

2T

Free-body diagram

mat

n

t

man

Kinetic diagram

At = 60o, v = 0At = 90o, v = ?

T = ?

Equations of Motion :

Fn=man ;

Ft=mat ;

2T – Wsin = man

2T – 300sin = 30(v2/3) ……(i)

an = v2/= v2/3

Wcos = mat

300cos = 30at

10cos = at ……(ii)

n

mat

t

man

W t

2T

n

θ

2T – 300sin = 30(v2/3) ……(i)

10cos = at ……(ii)

Kinematics : (to relate at and v)

vdv = at dsvdv = 10cos d

v = 2.68 [m/s]

s = ds = d

v

0

90

60

d)cos10(vdv

Solving, we get T = 186 [N]

The speed of the boy at = 90o

The 400 [kg] mine car is hoisted up the incline using the cable and motor M. For a short time, the force in the cable is F = 3200t 2 [N], where t is in seconds. If the car has an initial velocity v1=2 [m/s] at s=0 and t=0, determine the distance it moves up the plane when t=2 [s].

Ans = 5.43 m

Quiz 1

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