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2 .1
Force and Acceleration
2.1 Kinetics of Particle
(tvid 2.2.a)
(F=ma)
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If the lines of action of more than one force acting on an object coincide at the same point, the total moment of the forces is zero and this object can be examined as a particle.
2.1 Direct application of equation F = ma
(2.1.a)
The force F acting on an object gives the object an acceleration in its own direction. There is a relationship between force, acceleration and mass as follows:
This independent equation is called Newton's 2nd law.
Resultant Force
Since the acceleration and the resultant force are in the same direction, this equation between them can also be written as a scalar:
(2.1b)
(2.2.a)
(2.2.b)
(2.3.a)
(2.3.b)
(2.3.c)
Relationships between components:
The relationship between resultant force and resultant acceleration:
Figure 2.1
Figure 2.2
Figure 2.3
2.1 Kinetics of Particle / Force and Acceleration (F=ma)
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(2.3.a)
(2.3.b)
(2.3.c)
Effective forces system: m.a (mass x acceleration) terms are called effective forces. They are drawn on the right.
Effective forces system
2.1.1 D’Alembert’s Principle:
It is a different interpretation of the equation F = ma. It is the method we will use in solving problems.
External forces system: It is the figure showing the external forces acting on the object. Or it is the Free Body Diagram of the object.
As a result, for each of the x, y and z directions, the external forces system at a time t is equalized to the effective forces system. From equations 2.3, the unknown values (force or acceleration) are obtained.
External forces system
Tip 2.1 : In the external forces system and the effective forces system, the positive directions are chosen the same.
Tip 2.2 : It is necessary to be aware that the calculations are made for a moment t.
a-) In Cartesian Coordinates
2.1 Kinetics of Particle / Force and Acceleration (F=ma)
Figure 2.4
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External forces system
Effective forces system
D’Alembert’s Principle in Other Coordinate Systems:
b-) Tangential-Normal Coordinates
c-) Cylindrical Coordinates
(2.4.b)
(2.4.a)
(2.5.a)
(2.5.b)
(2.5.c)
2.1 Kinetics of Particle / Force and Acceleration (F=ma)
Figure 2.5.a
Figure 2.5.b
Figure 2.6.a
Figure 2.6.b
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2.1.2 D’alembert’s Principle for a Particle System:
B
C
A
C
A
B
External forces system
Effective forces system
A system consisting of more than one particle is called a particle system. D’Alembert’s Principle can also be applied to this system.
Vectorial:
Scalar:
Note-1: 2nd solution alternative: Instead of D'Alembert's principle, the solution is reached by transferring the terms on the right side of the equations 2.3 to the left. In this case, the product ma is the inertia force.
Dynamic equilibrium equations
Inertial forces
(2.6)
(2.7.a)
(2.7.b)
(2.7.c)
(2.8.a)
(2.8.b)
(2.8.c)
2.1 Kinetics of Particle / Force and Acceleration (F=ma)
* In these notes, we will make solutions using the D’Alembert principle.
* While applying this method for rigid bodies, we should keep in mind that moment equations will also come into play.
Figure 2.7.a
Figure 2.7.b
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Example 2.1.1
A 2kN force F acts on a 400 N crate at rest, making an angle of 60o with the horizontal. The crate starts to move horizontally. The coefficient of kinetic friction between the crate and the surface during the movement is 0.4. Accordingly,
a-) calculate the acceleration of the crate and
b-) calculate how far the crate has moved from its initial position after 10 seconds and its velocity at this moment.
60o
F
μ= 0.4
Solution:
External forces system
Effective forces system
60o
F
G
NA
W
FsA
max
G
may=0
Friction force:
t =10s
b-) Constant acceleration rectilinear motion:
0
0
away
a-)
0
2.1 Kinetics of Particle / Force and Acceleration (F=ma)
velocity:
Figure 2.8
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As seen in the figure, a ball with mass m is tied to a rope to form a pendulum. When the pendulum passes through the 300 position, the force T in the rope is 2.5 times the weight of the ball. Accordingly, calculate the acceleration components and velocity of the ball at this position. 𝑔=9.8m/s2
=
=
=
=
Example 2.1.2
Solution
2.1 Kinetics of Particle / Force and Acceleration (F=ma)
Figure 2.9
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Example 2.1.3
A 40 kg child is swinging on a swing with a 2 meter chain. When the swing is moving down, the angle between the chain and the vertical is 60o and the velocity of the swing is 4m/s. Calculate the total acceleration of the child and the force in one of the swing chains in this position. Neglect the weight of the swing board. (g = 10m/s2 )
Solution:
n
t
mat
man
W
P
n
t
60o
W.Cos600
W.Sin600
In the Tangential-Normal (t-n) Axis Set
normal acceleration:
2.1 Kinetics of Particle / Force and Acceleration (F=ma)
force in one of the chains
Total acceleration:
Figure 2.10
tangential acceleration:
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x
y
The system in the figure is released while at rest. Neglecting the weight of pulleys C and D and friction;
a-) Find the accelerations of blocks A and B and the forces in the ropes.
b-) Calculate how much block A has been dragged after 2 seconds.
=
=
=
at any time t
Tip 2.3: There is only one tension force in a rope. ADCE is a single rope and the tension force is T1. CB is a different rope and the tension force on it is taken as T2.
Distance taken by block A:
b-)
a-)
Example 2.1.4
Solution:
2.1 Kinetics of Particle / Force and Acceleration (F=ma)
Figure 2.11
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A package is pushed upwards on a ramp with an inclination of 20° with an initial velocity of 8 m/s. After the package reaches point B on the ramp, it slides back to point A. If the distance d between points A and B is 7 m;
a-) Calculate the coefficient of friction between the ramp and the package,
b-) Calculate the velocity of the package when it reaches A on the way back.
A
B
d
200
μ=?
Solution:
=
W=mg
Fs=μN
N=mgcos200
200
200
mgsin200
mgcos200
A
B
7m
Since there is no movement in the y-direction, static equilibrium is achieved in that direction:
N=mgcos200
The formula for the timeless velocity valid for linear motion with constant acceleration (equation 1.9) :
moving upwards
Example 2.1.5
2.1 Kinetics of Particle / Force and Acceleration (F=ma)
coefficient of kinetic friction
Let's place the axis set at point A. We write external forces and effective forces by paying attention to their directions and signs:
B
A
a-)
Acceleration is accepted in the +x direction. If you pay attention, its sign is negative. This shows that it is in the opposite direction of the direction we have chosen. The operations are written assuming that the acceleration due to gravity is positive.
Figure 2.12
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=
W=mg
Fs=μN
N=mgcos200
200
200
mgsin200
mgcos200
A
B
7m
during moving downwards
2.1 Kinetics of Particle / Force and Acceleration (F=ma)
During moving downwards, all forces will be at the same value, only the Fs force will be in the opposite direction to the motion, that is, in the +x direction.
B
A
We apply the timeless velocity formula (equation 1.9), this time with the initial position being B and the final position being A:
b-)
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Cevat
O
A
Question 2.1.1 (*)
2.1 Kinetics of Particle / Force and Acceleration (F=ma)
Answers:
Figure 2.13
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a
b
c
e
d
Question 2.1.2 (*)
2.1 Kinetics of Particle / Force and Acceleration (F=ma)
Figure 2.14
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A car with a mass of 800 kg is moving with a constant acceleration of 3m/s2 to the right on a parabolic curved surface whose equation is given in the figure, increasing its speed. If its speed is 9m/s as it passes through point A, calculate the normal force and friction force on the car from the ground at point A.
Information: The values of 𝜌 and 𝜃 at any point i (x, y) of a curve are:
i (x, y)
slope angle :
radius of curvature :
Question 2.1.3 (*)
2.1 Kinetics of Particle / Force and Acceleration (F=ma)
;
Answer: Ν=6.73kN , Fs=1.11kN
Figure 2.15