1. Two boxes are connected by a cord running over a pulley. Ignore the mass of the cord and pulley and any friction in the pulley. Masa of the box 1 = 2 kg, mass of the box 2 = 3 kg, ubrzanje zbog gravitacije = 10 m / s2, Pronaći (a) The acceleration of the system (b) The tension in the cord!
Riješenje
Poznato:
Mass of the box 1 (m1) = 2 kg
Mass of the box 2 (m2) = 3 kg
Ubrzanje zbog gravitacije (g) = 10 m/s2
Težina of the box 1 (w1) = m1 g = (2)(10) = 20 Newtona
Weight of the box 2 (w2) = m2 g = (3)(10) = 30 Newtona
rješenje:
(a) magnitude and direction of the acceleration
w2 > w1 tako da je box 2 accelerates downward and the box 1 accelerates upward.
Forces that has the same direction with acceleration (w2 i T1), its sign positive. Forces that has opposite direction with acceleration (T2 i W1), its sign negative.
ΣF = ma
w2 - T2 + T.1 - w1 = (m1 +m2) a ——-> T1 =T2 =T
w2 – T + T – w1 = (m1 +m2) i
w2 - w1 = (m1 +m2) i
30 – 20 = (2 + 3) a
10 = 5 a
a = 10 / 5
a = 2 m/s2
Magnitude of the ubrzanje je 2 m/s2.
(b) The tension force
The box 2 :
There are two forces acts on the box 2 : first, weight of the box 2 (w2), points downward so it’s positive. Second, tension force exerted on the box 2 (T2), points upward so it’s negative. Apply Newtonov drugi zakon of motion.
ΣF = ma
w2 - T2 = m2 a
30–T2 = (3)(2)
30–T2 = 6
T2 = 30 - 6
T2 = 24 Newtona
Box 1 :
There are two forces acts on the box 1. Ime, weight of the box 1 (w1), points downward so it’s negative. Drugo, the tension force exerted on the box 1 (T1) points upward so it’s positive. Apply Newton’s second law of motion :
ΣF = ma
T1 - w1 = m1 a
T1 – 20 = (2)(2)
T1 - 20 = 4
T1 = 20 + 4
T1 = 24 Newtona
Veličina sile napetosti = T1 =T2 = T = 24 Newtona
2. An object on a rough horizontal surface. Mass of the object 1 = 2 kg, mass of the object 2 = 4 kg, acceleration due to gravity = 10 m/s2, coefficient of the static friction = 0.4, coefficient of the kinetic friction = 0.3. The system is at rest or accelerated ? If the system is accelerated, find the magnitude and direction of the system’s acceleration!
Riješenje
Poznato:
Mass of the object 1 (m1) = 2 kg
Mass of the object 2 (m2) = 4 kg
Ubrzanje zbog gravitacije (g) = 10 m/s2
Koeficijent statičko trenje (μs) = 0.4
The coefficient of the kinetic friction (μk) = 0.3
Weight of the object 1 (w1) = m1 g = (2)(10) = 20 Newtona
Weight of the object 2 (w2) = m2 g = (4)(10) = 40 Newtona
normalna snaga exerted on the object 1 (N) = w1 = 20 Newtona
Force of the static friction exerted on the object 1 (fs) = μs N = (0.4)(20) = 8 Newtona
Force of the kinetic friction exerted on the object 1 (fk) = μk N = (0.3)(20) = 6 Newtona
Htjela: ubrzanje (a)
rješenje:
w2 > fs (40 Newton > 8 Newton) so the object 2 is accelerated vertically downward and the object 1 is accelerated horizontally rightward. The friction force that acts on the objects 1 is the force of the kinetic friction (fk). Apply Newton’s second law of motion :
ΣF = ma
w2 - = (m1 +m2) i
40 – 6 = (2 + 4) a
34 = 6 a
a = 34 / 6 = 17 / 3
a = 5.7 m/s2
Veličina ubrzanja = 5.7 m/s2
[wpdm_package id='484']
- Masa i težina
- normalna snaga
- Newtonov drugi zakon gibanja
- Sila trenja
- Motion on horizontal surface without friction force
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- Gibanje po nagnutoj ravnini bez sile trenja
- Gibanje po hrapavoj nagnutoj ravnini sa silom trenja
- Kretanje u liftu
- Gibanje tijela povezano je užetima i remenicama
- Dva tijela s istom brzinom ubrzanja
- Zaokruživanje ravne krivulje – dinamika kružnog gibanja
- Zaokruživanje nagnute krivulje – dinamika kružnog gibanja
- Jednoliko gibanje u horizontalnom krugu
- Centripetalna sila u jednolikoj kružnoj kretnji
