Question

Three sleds (30kg sled connected by tension rope B to 20kg sled connected by tension rope A to 10kg sled) are being pulled horizontally on frictionless horizontal ice using horizontal ropes. The pull is horizontal and of magnitude 143N . Required:a. Find the acceleration of the system. b. Find the tension in rope A. c. Find the tension in rope B.

Answer 1

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.

Step-by-step 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.

Answer 2

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

Step-by-step 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