dynamics & newton’s law topics: force newton’s law of motion application of newton’s...
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1. Kinematics:is the branch of mechanics which studies a
motion that is observed based on physical quantities such as position, distance, displacement, speed, velocity and acceleration with no consider on what causes the motion (force).
Examples: Uniform rectilinear motion of a particle Accelerated uniform rectilinear motion Vertical motion Parabolic motion etc
2. Dynamics:is the branch of mechanics concernings
motion of object with regard to what causes the motion (force).
Examples: Motion of object on a smooth plane Motion of object on a rough plane Motion of object on an inclined plane Motion of object that connected by a pulley Lift etc
Forceis a quantity that can change the state of motions
(e.g., golf game), forms (e.g., bread batter making) and sizes (e.g.,spring) of an object.
force can be a push or a pull upon an object.
The first Newton’s Law
In the absence of external forces, an object at rest remains at rest and an object
in motion continues in motion with a constant velocity (that is, with a constant
speed in a straight line).
0F
V= 0
a = 0
At rest
GLB
Sample problem
What is the acceleration resulted when a force resultant of 12 N is applied to an object which is 6 kg in mass?
Solution:
m.aF
m
Fa
22m/s6kg
12N
3. Newton’s Third Law
If two objects interact, the force F12 exerted by object 1 on object 2 is equal in magnitude and opposite in direction to the force F21 exerted by object 2 on object 1:
Faction = - F reaction
application
Normal force
Weight
On horisontal plane :
N = W
θ
ww cosθ
θw sinθ
N
On inclined plane :
N = W cosθ
There two kinds of frictional forceStatic frictional force: is a frictional force exerts on the
object which is at rest.Kinetic frictional force: is a frictional force works on the
moving object.
Nμf ss Nμf kk
fs = static frictional force (N)
fk = kinetic frictional force (N)
μs = coefficient of static friction
μk = coefficient of kinetic friction
μs > μk,
So … fs > fk
0 ≤ μ ≤ 1
μ = 0 (smooth)
μ > 0 (rough)
Friction force
F
fs maks
If F < fsmaksat rest
If F = fsmaks Exactly will be move
If F > fsmaks Object is moving
m
fF
m
Fa k
fsmaks = μs N
Sample Problem Calculate the
acceleration of system , if Force applied is : (g=10m/s2 )a. 50 Nb. 52 Nc. 60 N d. 70 Ne. 61 N
12 kg
F
μS= 0,5 μk = 0,3
Sample problem
A car is 800 kg in mass, from rest it is accelerated with a constant acceleration, after 2 seconds the car travels at a distance of 20 m. Determine the force resulted by the car if during the motion it experiences the frictional force of 200 N!
Solution:
m.aF
m.afF Because a = constant, then the car moves in accelerated uniform rectilinear motion, therefore the acceleration (a) can be determined by;
Sample problem:
Determine the normal force of cube A, B, and C, if m = 10 kg and g = 9.8 m/s2!
N=?
w
m
A
N=?
w
m
B
N=?
w
m
C
20 N 15 N
Cube A
Solution:
Because on cube A exerts no outside force,
N = w = m.g
= (10kg)(9.8m/s2) = 98 N
Cube B
Because on cube B works an outside force of 20 N,
N = ƩF = w + F = m.g + F
= (10kg)(9.8m/s2) + 20 N = 118 N
Cube C
Because on cube C works an outside force of 15 N,
N = ƩF = w - F = m.g - F
= (10kg)(9.8m/s2) - 15 N = 83 N
Problem 1
3 kg
7 kg
Calculate the acceleration of system and stress of string (percepatan system dan tegangan tali ) if flour is smooth ( jika lantai licin ) g = 10 m/s2
Answer
3 kg
7 kg
w=mg
= 30N
W2=70N
NBecause the floor is smooth , μ=0, fs =0
2
21
1
/3
73
30
sma
a
mm
wa
m
Fa
Problem
3 kg
7 kg
Calculate the acceleration of system and stress of string (percepatan system dan tegangan tali ) if flour is rough ( jika lantai kasar ) g = 10 m/s2
μs=0,4 μk=0,3
Answer
3 kg
7 kg
w=mg
= 30N
W2=70N
N
μs=0,4, fsmaks =0,4.70
= 28N
2
21
1
/9,0
10
9
73
2130
sma
a
mm
fwa
m
fFa
k
k
μs=0,4 μk=0,3
fk=μk.N
= 0,3.70=21N
Problem 2
Determine the acceleration (a) and tension of each string !
m1= 4 Kg
m2 = 6 Kg
There isn’t
Friction between string and pulley
Determine the acceleration (a) and tension of each string !
If the inclined plane is smooth. ( g = 10 m/s2)
3Kg
2Kg
30o
Answer
3Kg
2Kg
30oW1
T
W2
w2sin30o
w2cos30o
T2
21
2121
/4
520
51030
)23(.2030
)(30sin
sma
a
a
a
ammww
maFo
Tension of string (T)
3Kg
2Kg
30oW1
T
W2
w2sin30o
w2cos30o
T
NT
T
T
amwT
atau
NT
T
T
amTw
maF
18
810
4.2.20
30sin
18
1230
4.330
21
22
11