Download 13.1 - Newton`s Law of Motion

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CHAPTER 13 : KINEMATICS OF A PARTICLE ‐ FORCE AND ACCELERATION
Dr Mohamed AICHOUNI
This presentation is an adapted version from : R. C. Hibbeler, « Engineering Mechanics –
Dynamics », 12th Ed.
13.1 - Newton’s Law of Motion
First Law:
A particle originally at rest
rest, or moving in a straight line with
a constant velocity, will remain in this state provided the
particle is not subjected to an unbalanced force.
Second Law:
A particle acted upon by an unbalanced force F
experiences
p
an acceleration a that has the same direction
as the force and a magnitude that is directly proportional to
the force.
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Newton’s Law of Motion (13.1)
Third Law: The mutual forces of action and reaction
between two particles are equal, opposite and collinear.
Equation of motion:
F = ma
Newton’s Law of Gravitational Attraction. A law
governing the mutual attractive gravitational force acting
between them.
F =G
m1m2
r2
G : universal constant of gravitation
Newton’s Law of Motion (13.1)
Mass and Weight.
Mass is a property of matter by which we can compare the
response of one body with that of another. It is an absolute
quantity since the measurement can be made at any
location. Weight of a body is not absolute since it is
measured in a gravitational field, hence its magnitude
depends
p
on the location measured.
W = mg
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Newton’s Law of Motion (13.1)
SI System of Units.
The mass of the body is specified in kilograms (kg) and the
weight must be calculated using the equation of motion,
F = ma
W = mg
g
((N))
(g = 9.81 m/s2)
13.2 - The Equation of Motion
The equation of motion may be written as
∑ F = ma
Consider particle P of mass m and subjected to the action of two forces, F1 and
F2.
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The Equation of Motion (13.2)
From the free body diagram, the resultant of these forces produces the vector
ma it magnitude and direction can be represented graphically on the kinetic
ma,
diagram.
∑ F = ma
Note that if FR = ΣF = 0, then acceleration is zero, so that the particle will either
remain at rest or moves along a straight line with a constant velocity.
Such a condition is called static equilibrium, Newton’s First Law of Motion
13.3 - Equation of Motion for a System
of Particles
Consider a system of n particles isolated within an enclosed region in space. •
Arbitrary ith particle having a mass of mi •
is subjected to a system of internal forces
which resultant force is represented by fi
and a resultant external force Fi.
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Equation of Motion for a System of
Particles (13.3)
The free body diagram for the ith particle are shown. Applying equation of •
motion yields
Fi + fi = miai
ΣF = ma;
If equation of motion is applied to each of the other •
particles, these equations can be added together
vectorially,
y,
ΣFi + Σ fi = Σmiai
Equation of Motion for a System of
Particles
• Since internal forces between particles all occur in equal but opposite collinear
pairs the summation of these internal forces will equal zero
pairs,
zero.
ΣFi = Σmiai
• If rG is a position vector which locates the center of mass G of the particles, then
mrG = Σmiri where m = Σmi is the total mass of all the particles
• Differentiating twice w.r.t time yields
• Therefore,
maG = Σm
Σ iai
ΣF = maG
• The sum of the external forces acting on the system of particles is equal to
the total mass of the particles times the acceleration of its center of mass G.
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13.4 - Equation of Motion: Rectangular
Coordinates
• When a particle is moving relative to an inertial x, y, z frame of reference, the
forces acting on the particle
particle, and its acceleration may be expressed in term of
their i, j, k components
ΣF = ma
ΣFxi + ΣFyj + ΣFzk = m(axi + ayj + azk)
• We may write the following three scalar equations:
∑ Fx = max
∑ Fy = ma y
∑ Fz = maz
Example 13.1
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.
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Example 13.1 - Solution
Free-Body Diagram.
The weight of the crate is W = mg = 50 (9.81) =
490.5 N. The frictional force has a magnitude F =
μkNC and acts to the left, since it opposes the motion
of the crate. The acceleration a is assumed to act
horizontally, in the positive x direction. There are 2
unknowns, namely NC and a.
Example 13.1
Equations of Motion.
+
→ ∑ Fx = max ;
400 cos 30o − 0.3 NC = 50a
+ ↑ ∑ Fy = ma y ;
NC − 490.5 + 400 sin 30o = 0
Solving for the two equations yields
NC = 290.5 N
a = 5.19m / s 2
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Example 13.1
Kinematics.
Acceleration is constant, since the applied force P is constant. Initial velocity is
zero, the velocity of the crate in 3 s is
v = v0 + act
= 0 + 5.19(3)
= 15.6m / s →
13.5 - Equations of Motion: Normal and
Tangential Coordinates
When a particle moves over a curved path which is known, the equation of
motion for the particle may be written in the tangential
tangential, normal and binormal
directions. We have
ΣF = ma
ΣFtut + ΣFnun + ΣFbub = mat +man
Here ΣFt, ΣFn, ΣFb represent the sums of all the force components acting on
the particle in the tangential, normal and binormal directions.
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Equations of Motion: Normal and
Tangential Coordinates
Since the particle is constrained to move along the path, there is no motion in
the binormal direction
∑ Ft = mat
∑ Fn = man
∑ Fb = 0
• at (=dv/dt) represents the time rate of change in the magnitude of velocity
• Therefore if ΣFt acts in the direction of motion, the particle’s speed will
increase. If it acts in the opposite direction, the particle will slow down.
• an (=v2/ρ) represents the time rate of change in the velocity’s direction.
Equations of Motion: Normal and
Tangential Coordinates
Since this vector always acts in the positive n direction, i.e. toward the path’s •
center of curvature
curvature, then ΣFn, which causes an, also act in this direction.
direction
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Example 13.6
Determine the banking θ for the race track so that the wheels of the racing cars
will not have to depend upon friction to prevent any car from sliding up or down
the track. Assume the cars have negligible size a mass m, and travel around the
curve of radius ρ with a speed v.
Example 13.6
Free-Body Diagram.
There is no frictional force acting on the car
car. Here NC represents the
resultant of the ground on all four wheels. Since an can be calculated, the
unknown are NC and θ.
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Example 13.6
Equations of Motion.
+
Using the n, b axes shown,
v2
→
∑ Fn = man ;
NC sin θ = m
+↑
∑ Fb = 0;
NC cosθ − mg = 0
ρ
Solving
g the 2 equations,
q
eliminating
g NC and m,
v2
tan θ =
gρ
−1⎛⎜
v 2 ⎞⎟
θ = tan ⎜ ⎟
⎝ gρ ⎠
13.6 - Equations of Motion:
Cylindrical Coordinates
When all forces acting on a particle are resolved into
cylindrical components
components, ii.e.
e along the unit
unit-vector
vector directions
ur, uθ, uz, the equation of motion may be expressed as:
ΣF = ma
ΣFrur + ΣFθuθ + ΣFzuz =
marur +maθuθ+mazuz
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Equations of Motion: Cylindrical
Coordinates
We may write the following three scalar equations of motion:
∑ Fr = mar
∑ Fθ = maθ
∑ Fz = maz
Equations of Motion: Cylindrical
Coordinates
Tangential and Normal Forces.
• Determination of the resultant force components ΣFr, ΣFθ, ΣFz causing a
particle to move with a known acceleration.
• If acceleration is not specified at given instant, directions or magnitudes of the
forces acting on the particle must be known or computed to solve.
• Consider the force P that causes the particle to move along a path r = f(θ)
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Equations of Motion: Cylindrical
Coordinates
•
The normal force N which the path exerts on the particle is always
perpendicular to the tangent of the path
path.
•
Frictional force F always acts along the tangent in the opposite direction of
motion.
Equations of Motion: Cylindrical
Coordinates
• The directions of N and F can be specified relative to the radial coordinate by
using the angle ψ,
ψ which is defined between the extended radial line and the
tangent to the curve.
tanψ =
13
r
dr / dθ
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Equations of Motion: Cylindrical
Coordinates
• If ψ is positive, it is measured from the extended radial line to the tangent in a
CCW sense or in the positive direction θ
• If it is negative, it is measured in the opposite direction to positive θ
Example 13.10
The 2-kg block moves on a smooth horizontal track such that its path is specfied
in polar coordinates by the parametric equations r = (3t2) m and θ = (0.5t)
(0 5t) rad
where t is in seconds. Determine the magnitude of the tangential force F
causing the motion at the instant t = 1 s.
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Example 13.10
Free-Body Diagram.
ψ
The normal force N, and the tangential force F are located at an angle
from
the r and θ axes. By expressing r = f(θ), we yield r = 12θ2. When t = 1 s, θ = 0.5
rad.
Example 13.10
tanψ =
r
12θ 2
=
dr / dθ 12(2θ )
= 0.25
θ = 0.5rad
ψ = 14.04o
ψ
Because
is a p
positive q
quantity,
y it is measured counterclockwise from the r
axis to the tangent (same direction as θ). There are four unknowns: F, N, ar and
aθ
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Example 13.10
Equations of Motion.
+ ↓ ∑ Fr = mar ;
F cos14.04o − N sin 14.04o = 2ar
↑ + ∑ Fθ = maθ ;
F sin 14.04o + N cos14.04o = 2aθ
Kinematics.
r = 3t 2
t =1s
= 3m
θ = 0.5t t =1s = 0.5rad
r& t =1s = 6m / s, &r& t =1s = 6m / s 2 , θ& = 0.5rad / s, θ&& = 0
Example 13.10
ar = &r& − rθ& 2 = 6 − 3(0.5) 2 = 5.25m / s 2
aθ = rθ&& + 2r&θ& = 3(0) + 2(6)(0.5) = 6m / s 2
Substituting into the two equations of motion and solving,
F = 13.10
13 10 N
N = 9.22 N
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Example 13.11
A smooth 2-kg cylinder C has a peg P through
its center which passes through the slot in arm
arm.
If the arm rotates in the vertical plane at a
constant rate
, determine the
force that the arm exerts on the peg at the
instant θ = 60°
θ& = 0.5rad / s
Example 13.11
Free-Body Diagram.
The force on the peg, Fp, acts perpendicular to the
slot in the arm.
arm ar and aθ are assumed to act in the directions of positive r and θ
respectively.
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Example 13.11
Equations of Motion.
+
+
∑ Fr = mar ;
∑ Fθ = maθ ;
Kinematics.
19.62 sin θ − NC sin θ = 2ar
(1)
19.62 cosθ + FP − NC cosθ = 2aθ
(2)
From the FBD, r can be related to θ by the equation,
r=
0.4
= 0.4 cscθ
sin θ
Example 13.11
θ& = 0.5
θ&& = 0
r = 0.4 cscθ
r& = −0.4(cscθ cot θ )θ&
= −0.2 cscθ cot θ
&r& = −0.2( − cscθ cot θ )(θ&) cot θ
− 0.2 cscθ (− csc2 θ )θ&
= 0.1cscθ (cot 2 θ + csc2 θ )
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Example 13.11
Evaluating the formulas at θ = 60°, we get,
θ& = 0.5
θ&& = 0
r = 0.462
r& = −0.133
&r& = 0.192
ar = &r& − rθ& 2 = 0.0770
aθ = rθ&& + 2r&θ& = −0.133
Substituting these results into Eqs. 1 and 2 with θ = 60°,
NC = 19.4 N
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FP = -0.356 N