emm 222 - dynamics & mechanism - chapter 2
TRANSCRIPT
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