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.

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

hey , the answer is c

d eras and periods