3 tanong tungkol sa ekwasyon ng puwersa ng alitan
1. Ang bloke A na may bigat na 3 kg ay inilalagay sa mesa at pagkatapos ay itinali sa isang lubid na konektado sa bato. Ang B = 2 kg sa pamamagitan ng isang pulley gaya ng ipinapakita. Ang masa at friction ng mga pulley ay hindi pinapansin. Ang acceleration dahil sa gravity g = 10 m/s2Tukuyin ang acceleration ng sistema at ang tension sa lubid kung:
a) makinis na mesa
b) magaspang na mesa na may koepisyent ng kinetic friction na 0.4
Kilala:
Ang masa ng bloke A (mA) = 3 kilo
Ang masa ng batong B (mB) = 2 kilo
Akselerasyon dahil sa grabidad (g) = 10 m/s2
Bigat ng bloke A (wA) = mg = (3)(10) = 30 Newton
Bigat ng bato B (wB) = mg = (2)(10) = 20 Newton
Wanted: Ang acceleration ng sistema (a) at ang tensyon sa lubid (T)
solusyon:
a) makinis na mesa
Calculate the acceleration of the system using the formula for Newton’s second law:
ΣF = ma
wB = (mA +mB) Ang
20 = (3 + 2) isang
20 = 5 a
a = 20 / 5 = 4 m/s2
Calculate the tension in the rope using the formula for the tension in the rope:
The tension in the rope on block A:
ΣF = mA a
T = mA a = (3)(4) = 12 Newton
The tension in the rope on block B:
ΣF = mB a
wB – T = (2)(4)
20 – T = 8
T = 20 – 8 = 12 Newton
b) magaspang na mesa na may koepisyent ng kinetic friction na 0.4
The force of the kinetic Friction:
Fk = µk N = (0,4)(30) = 12 Newton
Calculate the acceleration of the system using the formula for Newton’s second law:
ΣF = ma
wB - fk = (mA +mB) Ang
20 – 12 = (3 + 2) isang
8 = 5 a
a = 8 / 5 = 1,6 m/s2
Calculate the tension in the rope using the formula for the tension in the rope:
The tension in the rope on block A:
ΣF = mA a
T – fk = mA a
T – 12 = (3)(1,6)
T – 12 = 4,8
T = 4,8 + 12 = 16,8 Newton
The tension in the rope on block B:
ΣF = mB a
wB – T = (2)(1,6)
20 – T = 3,2
T = 20 – 3,2 = 16,8 Newton
2. An object with a mass of 10 kg is in a horizontal plane. The coefficient of static friction is 0.4 and the coefficient of kinetic friction is 0.35. g = 10 m/s2. If an object is given a constant horizontal force of 25 N, the magnitude of the frictional force acting on the object is…
Kilala:
The mass of the object (m) = 10 kg
The coefficient of Static friction (µs) = 0.4
The coefficient of kinetic friction (µk) = 0.35
Akselerasyon dahil sa grabidad (g) = 10 m/s2
Horizontal force (F) = 25 N
The object’s gravity (w) = m g = (10)(10) = 100 Newton
Normal na puwersa (N) = w = 100 Newton
Wanted: The amount of static friction (fs) and kinetic (fk)
solusyon:
The force of the static Friction::
fs = µs N = (0,4)(100) = 40 Newton
The force of the Kinetic Friction:
fk = µk N = (0,35)(100) = 35 Newton
The horizontal force is only 25 Newton so it can’t move objects yet.
3. The masses of blocks A and B in the figure are 10 kg and 5 kg respectively. The coefficient of friction between block A and the plane is 0.2. To prevent block A from moving, the minimum mass of block C required is…
Kilala:
Ang masa ng bloke A (mA) = 10 kilo
The mass of block B (mB) = 5 kilo
Coefficient of static friction of block A (µs) = 0,2
Gravity acceleration (g) = 10 m/s2
Block weight A (wA) = mA g = (10)(10) = 100 Newton
Block weight B (wB) = mB g = (5)(10) = 50 Newton
Static friction (fs) = µs N = (0,2)(wA +wC) = (0,2)(100 + wC) = 20 + 0,2 wC
Nagtanong: The mass of block C to keep the system at rest
Sagot:
The system is at rest so the formula for Newton’s first law is used:
ΣF = 0
wB - fs = 0
50 – (20 + 0,2 wC) = 0
50 – 20 – 0,2 wC = 0
30 – 0,2 wC = 0
30 = 0,2 wC
wC = 30 / 0,2 = 300 / 2 = 150 Newton
The mass of block C = 150 / 10 = 15 Kg
