CHAPTER 13 Kinetics of a Particle: Force and acceleration ( Newton’s method)
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Transcript of CHAPTER 13 Kinetics of a Particle: Force and acceleration ( Newton’s method)
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CHAPTER 13
Kinetics of a Particle:Force and acceleration ( Newton’s method)
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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
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KINETICS
Newton’s MethodWork and
Energy MethodImpulse and momentum
CHAPTER 13
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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
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Free body diagram Kinetics diagram
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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
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Equations of Motion :
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Kinematics : The acceleration is constant, P is constant
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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
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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]
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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]
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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
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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
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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
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Frictional force : F = kND
Direction : oppose the relative motion of the disk with respect to the platform
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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
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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
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W
n
t
2T
Free-body diagram
mat
n
t
man
Kinetic diagram
At = 60o, v = 0At = 90o, v = ?
T = ?
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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
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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
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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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