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CHAPTER 13
Kinetics of a Particle:
Force and acceleration ( Newton’s method)
KINEMATICS
Rectilinear motion
Curvilinear motion
s = so + vot +
v2
a t ds  vdv
a t  v
v = vo + act
1
2
a
t
2 c
2
= vo + 2ac(s-so)
constant acceleration
an 
v2

a  an  at
2
2
Normal & Tangential
Components
KINETICS
Newton’s Method
CHAPTER 13
Work and
Energy Method
Impulse and
momentum
STATICS
Newton’s first law
A particle originally at rest, will remain in rest
SF=0
Newton’s second law
A particle acted upon by an unbalanced force,
SF=ma
experiences acceleration that has the same
direction as the force and a magnitude that is directly
proportional to the force
DYNAMICS
Free body diagram
Kinetics diagram
Example 1
The crate has a mass of 50 kg. If the crate is subjected
to a 400[N] towing force as shown, determine the
velocity of the crate in 3[s] starting from rest.
ms= 0.5, mk= 0.3,
Equations of Motion :
Kinematics : The acceleration is constant, P is constant
Example 2
The crate has a mass of 80 kg. If the magnitude of P is
increased until the crate begins to slide, determine the
crate’s initial acceleration. ms= 0.5, mk= 0.3,
[solution]
ms= 0.5
mk= 0.3
80(9.81) = 784.8[N]
T
20o
Ff = 0.5N
: verge of slipping
: impending motion
Equations of equilibrium :
SFx=0 ;
SFy=0 ;
N
Tcos20o – 0.5N = 0
N + Psin20o – 784.8 = 0
T = 353.29 [N] ,
……..(i)
……...(ii)
N = 663.79 [N]
[solution]
784.8 [N]
353.29 [N]
a
20o
Ff = 0.3N
N
Equations of Motion :
SFy=may ;
N – 784.8 + 353.29sin20o = 80(0)
N = 663.97 [N]
SFx=max ;
353.29cos20o – 0.3(663.97) = 80a
a = 1.66 [m/s2]
Equations of Motion:
Normal and Tangential Coordinates
o
t
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.
b
SFbub
o
SFnun
t
SFtut
the force components acting on the particle
SFt = mat
SFn = man
SFb = 0
Ft : tangential force
Fn : centripertal force
Fb : binormal force
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, determine
the 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. mk = 0.1
Frictional force :
Direction :
F = mkND
oppose the relative motion of the disk with
respect to the platform
Equations of Motion :
SFn=man ;
T = m(v2/)
= 3v2
SFt=mat ;
0.1ND = 3at
SFb=0;
ND – 29.43 = 0
Setting T = 100 N to get critical speed vcr
ND = 29.43 N
at = 0.981 m/s2
vcr = 5.77 m/s
Kinematics :
Since at is constant,
vcr = vo + att
5.77 = 0 + (0.981)t
T = 5.89 s
Example 2
At the instant q = 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 q = 90o.
The boy has a weight of 300 N ( ≈ 30 kg).
Neglect his size and the mass of the seat and cords
At q = 60o,
At q = 90o,
v=0
v=?
Free-body diagram
n
2T
q
T=?
Kinetic diagram
n
man
mat
W
t
t
2T
n
n
q
man
θ
W
mat
t
t
Equations of Motion :
SFn=man ;
SFt=mat ;
2T – Wsinq = man
2T – 300sinq = 30(v2/3)
Wcosq = mat
300cosq = 30at
10cosq = at
an = v2/
= v2/3
……(i)
……(ii)
2T – 300sinq = 30(v2/3)
10cosq = at
……(i)
……(ii)
Kinematics : (to relate at and v)
vdv = at ds
vdv = 10cosq dq
v
90
0
60
s = q
ds = dq
 vdv   ( 10 cos q )dq
v = 2.68 [m/s]
Solving, we get
The speed of the boy at q = 90o
T = 186 [N]
Quiz 1
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
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