College

Three sleds (a 30 kg sled connected by tension rope B to a 20 kg sled, which is connected by tension rope A to a 10 kg sled) are being pulled horizontally on frictionless ice using horizontal ropes. The pull is horizontal and has a magnitude of 143 N.

Required:

a. Find the acceleration of the system.

b. Find the tension in rope A.

c. Find the tension in rope B.

Answer :

Answer:

a) a = 2.383 m / s², b) T₂ = 120,617 N
, c) T₃ = 72,957 N

Explanation:

This is an exercise of Newton's second law let's fix a horizontal frame of reference

in this case the mass of the sleds is 30, 20 10 kg from the last to the first, in the first the horizontal force is applied.

a) request the acceleration of the system

we can take the sledges together and write Newton's second law

T = (m₁ + m₂ + m₃) a

a = T / (m₁ + m₂ + m₃)

a = 143 / (10 +20 +30)

a = 2.383 m / s²


b) the tension of the cables we think through cable A between the sledges of 1 and 20 kg

on the sled of m₁ = 10 kg

T - T₂ = m₁ a


in this case T₂ is the cable tension

T₂ = T - m₁ a

T₂ = 143 - 10 2,383

T₂ = 120,617 N

c) The cable tension between the masses of 20 and 30 kg

T₂ - T₃ = m₂ a

T₃ = T₂ -m₂ a

T₃ = 120,617 - 20 2,383

T₃ = 72,957 N

Final answer:

The acceleration of the system is 2.38 m/s^2. The tension in rope A is 142.8 N. The tension in rope B is 142.8 N.

Explanation:

(a) To find the acceleration of the system, we can use Newton's second law of motion: F = ma, where F is the net force applied to the system, m is the total mass of the system, and a is the acceleration. In this case, the net force is 143N and the total mass is 30kg + 20kg + 10kg = 60kg. Therefore, the acceleration of the system is a = F/m = 143N/60kg = 2.38 m/s^2.

(b) To find the tension in rope A, we can use the formula T = m1a + m2a + m3a, where T is the tension in the rope and m1, m2, and m3 are the masses connected by the rope. In this case, m1 = 20kg, m2 = 30kg, m3 = 10kg, and a = 2.38 m/s^2. Therefore, the tension in rope A is T = (20kg)(2.38 m/s^2) + (30kg)(2.38 m/s^2) + (10kg)(2.38 m/s^2) = 47.6 N + 71.4 N + 23.8 N = 142.8 N.

(c) To find the tension in rope B, we can use the formula T = m1a + m2a + m3a, where T is the tension in the rope and m1, m2, and m3 are the masses connected by the rope. In this case, m1 = 10kg, m2 = 20kg, m3 = 30kg, and a = 2.38 m/s^2. Therefore, the tension in rope B is T = (10kg)(2.38 m/s^2) + (20kg)(2.38 m/s^2) + (30kg)(2.38 m/s^2) = 23.8 N + 47.6 N + 71.4 N = 142.8 N.