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Bodies connected by cord and pulley - application of Newton's law of motion problems and
solutions
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. Mass of box 1 = 2 kg, mass of box 2 = 3 kg, acceleration due to gravity = 10
m/s2. Find (a) Acceleration of the system (b) Tension in cord!
Solution
Known :
Mass of box 1 (m1) = 2 kg
Mass of box 2 (m2) = 3 kg
Acceleration due to gravity (g) = 10 m/s2
weight of box 1 (w1) = m1 g = (2)(10) = 20 Newton
weight of box 2 (w2) = m2 g = (3)(10) = 30 Newton
Solution :
(a) magnitude and direction of acceleration
w2 > w1 so box 2 accelerates downward and box 1 accelerates upward.
Forces that has the same direction with acceleration (w 2 and T1), its sign positive. Forces that has
opposite direction with acceleration (T2 dan w1), its sign negative.
∑F = m a
w2 – T2 + T1 - w1 = (m1 + m2) a -------> T1 = T2 = T
w2 – T + T - w1 = (m1 + m2) a
w2 – w1 = (m1 + m2) a
30 – 20 = (2 + 3) a
10 = 5 a
a = 10 / 5
a = 2 m/s2
Magnitude of acceleration is 2 m/s2.
(b) Tension force
Box 2 :
There are two forces acts on box 2 : first, weight of box 2 (w 2), points downward so it's positive.
Second, tension force exerted on box 2 (T2), points upward so it's negative. Apply Newton's second law
of motion.
∑F = m a
w2 – T2 = m2 a
30 – T2 = (3)(2)
30 – T2 = 6
T2 = 30 – 6
T2 = 24 Newton
Box 1 :
© 2018 | San Alexander
https://physics.gurumuda.net
There are two forces acts on box 1. First, weight of box 1 (w1), points downward so it's negative.
Secod, tension force exerted on box 1 (T1) points upward so it's positive. Apply Newton's second law of
motion :
∑F = m a
T1 – w1 = m1 a
T1 – 20 = (2)(2)
T1 – 20 = 4
T1 = 20 + 4
T1 = 24 Newton
Magnitude of the tension force = T1 = T2 = T = 24 Newton
2. An object on a rough horizontal surface. Mass of object 1 = 2 kg, mass of object 2 = 4 kg,
acceleration due to gravity = 10 m/s 2, coefficient of static friction = 0,4, coefficient of kinetic friction =
0,3. The system is at rest or accelerated ? If the system is accelerated, find magnitude and direction of
system's acceleration!
Solution
Known :
Mass of object 1 (m1) = 2 kg
Mass of object 2 (m2) = 4 kg
Acceleration due to gravity (g) = 10 m/s2
Coefficient of static friction (μs) = 0.4
Coefficient of kinetic friction (μk) = 0.3
weight of object 1 (w1) = m1 g = (2)(10) = 20 Newton
weight of object 2 (w2) = m2 g = (4)(10) = 40 Newton
Normal force exerted on object 1 (N) = w1 = 20 Newton
Force of static friction exerted on object 1 (fs) = μs N = (0.4)(20) = 8 Newton
Force of kinetic friction exerted on object 1 (fk) = μk N = (0.3)(20) = 6 Newton
Wanted : acceleration (a)
Solution :
w2 > fs (40 Newton > 8 Newton) so the object 2 is accelerated vertically downward and object 1 is
accelerated horizontal rightward. Friction force that acts on object 1 is force of kinetic friction (f k).
Apply Newton's second law of motion :
∑F = m a
w2 – fk = (m1 + m2) a
40 – 6 = (2 + 4) a
34 = 6 a
a = 34 / 6 = 17 / 3
a = 5.7 m/s2
Magnitude of acceleration = 5.7 m/s2
© 2018 | San Alexander
https://physics.gurumuda.net
3. Massa balok 1 adalah 2 kg, massa balok 2 adalah 4 kg, percepatan gravitasi adalah 10 m/s 2. Agar
sistem diam maka koefisien gesek statis maksimum antara balok 2 dan permukaan bidang miring
adalah...
Pembahasan
4. Massa balok 1 adalah 2 kg, massa balok 2 adalah 4 kg, koefisien gesek kinetis adalah 0.2. Tentukan
(a) besar dan arah percepatan sistem (b) besar gaya tegangan tali
Pembahasan
© 2018 | San Alexander
