COLLECTIONS OF THE PROBLEM AND FULL DISCUSSION OF THE CATROL SYSTEM
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link : COLLECTIONS OF THE PROBLEM AND FULL DISCUSSION OF THE CATROL SYSTEM
link : COLLECTIONS OF THE PROBLEM AND FULL DISCUSSION OF THE CATROL SYSTEM
COLLECTIONS OF THE PROBLEM AND FULL DISCUSSION OF THE CATROL SYSTEM
Here are examples of problems and discussion of the dynamics dynamics pulley system.
Example 1
In a pulley whose mass is ignored, a string is strung on the ends of the P and Q loads as shown, if mp = 6 kg, mQ = 14 kg and g = 10 m / s². Specify:
a. Great acceleration of each object
b. T1 and T2 rope voltages
c. If the first object is still, calculate the velocity of the object after moving 2 seconds!
Discussion:
Is known:
mp = 6 kg
mQ = 14 kg
g = 10 m / s²
To make it easier to work, draw lines of style as a tool.
Since the Q load is larger then Q will move down and the P load moves up
a. Looking for acceleration of each object
Review the object P:
Unidirectional forces with the direction of the movement of positive value (+) and the negative negative (-)
ΣF = mp.a
T1 - wp = mp.a
T1 - mp. g = mp.a
T1 - 6 (10) = 6a
T1 = 6a + 60
Review the object Q:
ΣF = mQ.a
wQ - T2 = mQ.a
mQ. g - T2 = mQ.a
14 (10) - T2 = 14a
T2 = 140 - 14a
Because the mass of the pulley is negligible, the rope tension is the same (T1 = T2)
T1 = T2
6a + 60 = 140 - 14a
6a + 14a = 140 - 60
20a = 80
a = 4 m / s²
so the acceleration of each object is 4 m / s²
b. T1 and T2 rope voltages
T1 = 6a + 60
T1 = 6 (4) + 60 = 84 N
T1 = T2, then the strap voltage T2 = 84 N
c. The first object is silent then the initial velocity of the object = 0. After moving for 2 seconds the final velocity of the object is:
vt = v0 + a.t
vt = 0 + 4 (2)
vt = 8 m / s
Example 2
Two objects A and B of 2 kg and 3 kg respectively are connected by a string through the pulley as shown below.
If the friction between the rope and the pulley is negligible as well as the object B moves down, calculate the acceleration and the tension voltage of T if:
a. Friction object A with negligible floor (slippery floor)
b. The coefficient of kinetic friction between block A and floor 0.2
Discussion:
Is known:
mA = 2 kg
mB = 3 kg
g = 10 m / s²
a. When friction object A and floor are ignored
Notice the decomposition of forces in the system below, the mass of the B object is larger so that B objects move down. (a directional force with the direction of the movement of positive value (+) and the opposite negative (-))
Review object A:
ΣF = mA.a
T1 = mA.a
T1 = mA.a
T1 = 2a
T1 = 2a
Review object B:
ΣF = mB.a
wB - T2 = mB.a
mB. g - T2 = mB.a
3 (10) - T2 = 3a
30 - T2 = 3a
T2 = 30 - 3a
Because the mass of the pulley is negligible, the rope tension is the same (T1 = T2)
T1 = T2
2a = 30 - 3a
2a + 3a = 30
5a = 30
a = 6 m / s²
so the big acceleration of the object when friction between object A and floor is negligible is 6 m / s²
T1 and T2 rope voltages
T1 = 2a
T1 = 2 (6) = 12 N
T1 = T2, then the rope voltage = 12 N
b. Acceleration and strain voltage if coarse floor (coefficient of kinetic friction of object A with floor 0,2)
Review object A:
ΣF = mA.a
T1 - fk = mA.a
T1 - μk.NA = mA.a
T1 - 0.2 (20) = 2a
T1 - 4 = 2a
T1 = 2a + 4
Review object B: <
b. Acceleration and strain voltage if coarse floor (coefficient of kinetic friction of object A with floor 0,2)
Review object A:
ΣF = mA.a
T1 - fk = mA.a
T1 - μk.NA = mA.a
T1 - 0.2 (20) = 2a
T1 - 4 = 2a
T1 = 2a + 4
Review object B:
ΣF = mB.a
wB - T2 = mB.a
mB. g - T2 = mB.a
3 (10) - T2 = 3a
30 - T2 = 3a
T2 = 30 - 3a
Because the mass of the pulley is negligible, the rope tension is the same (T1 = T2)
T1 = T2
2a + 4 = 30 - 3a
2a + 3a = 30 - 4
5a = 26
a = 5.2 m / s²
so the big acceleration of the object if the friction between object A and floor 0.2 is 5.2 m / s²
T1 and T2 rope voltages
T1 = 2a + 4
T1 = 2 (5,2) + 4 = 14.4 N
T1 = T2, then the rope voltage = 14.4 N
Example 3
A beam of mass 1 kg and beam B mass of 2 kg. Beam B first silently then move downwards so that it touches the floor like the picture beside. How long does it take to block B to touch the floor?
Discussion:
Is known:
mA = 1 kg
mB = 2 kg
The beam is initially silent, then v0 = 0 m / s
Distance of block B to floor s = 25 m
Asked: time required beam B to the floor
To know the time it takes the beam B to get to the floor then first we must know the acceleration of the beam B.
Looking for system acceleration:
Review object A:
ΣF = mA.a
T1 = mA.a
T1 = mA.a
T1 = 1a
Review object B:
ΣF = mB.a
wB - T2 = mB.a
mB. g - T2 = mB.a
2 (10) - T2 = 2a
20 - T2 = 2a
T2 = 20 - 2a
Because the mass of the pulley is negligible, the rope tension is the same (T1 = T2)
T1 = T2
a = 20 - 2a
a + 2a = 20
3a = 20
a = 6.7 m / s²
then the time it takes the beam B to get to the floor:
s = v0.t + ½ a.t²
25 = 0 + ½ (6.7) t²
25 = 3.35 t²
7.46 = t²
t = √7,46 = 2.7 seconds
Example 4
a. System acceleration
b. Rope tilt style
Discussion:
Is known:
mA = 10 kg
mB = 20 kg
g = 10 m / s²
The mass of block B is larger then the beam B will move down and the A block moves in the direction of the incline
Directional forces of acceleration of positive values (+) and opposite directions of negative value (-)
a. Search for system acceleration
Review block A:
ΣF = mA.a
T1 - mA.g sin 30 ° = mA.a
T1 - 100 (½) = 10a
T1 - 50 = 10a
T1 = 10a + 50
Review object B:
ΣF = mB.a
wB - T2 = mB.a
mB. g - T2 = mB.a
20 (10) - T2 = 20a
200 - T2 = 20a
T2 = 200 - 20a
The pulley mass is ignored, then the rope tension is the same (T1 = T2)
T1 = T2
10a + 50 = 200 - 20a
10a + 20a = 200 - 50
30a = 150
a = 3 m / s²
b. Rope voltage style
T1 = 10a + 50
T1 = 10 (3) + 50 = 80 N
So the tension is T1 = T2 = T = 80 N
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