A system to 10 balls each of mass 2 kg are connected via massless and unstretchable string. The system is allowed to slip over the edge of a smooth table as shown in figure. Tension on the string between the 7ᵗʰ and 8ᵗʰ ball is __________ N when 6ᵗʰ ball just leaves the table.

Solution

Physics sir · 4 steps

Find the acceleration of the whole chain first, then look only at the balls behind the 7–8 string.

1. Count the balls. Once the 6th ball has just left the table, balls 1 to 6 hang over the edge and balls 7 to 10 are still on the table. All 10 balls move together with the same acceleration a, because the string does not stretch.
2. Whole system. The table is smooth, so the only unbalanced force on the chain is the weight of the 6 hanging balls, 6mg. It accelerates the total mass 10m.

   a = 6mg/10m = 0.6g = 6 m s⁻²

3. Balls behind. The string between balls 7 and 8 is the only horizontal force on balls 8, 9 and 10, which have a total mass of 3 × 2 = 6 kg. So T = 6 × 6 = 36 N, taking g = 10 m s⁻².
4. So. The tension between the 7th and 8th balls is 36 N.

The common slip: 48. Counting ball 7 among the balls pulled by this string gives 8 × 6 = 48 N. The 7–8 string pulls only balls 8, 9 and 10.

The idea to keep. For a chain or connected bodies, get the common acceleration from the whole system, then cut the string and apply F = ma to the smaller part on one side of the cut.

Answer from the NTA final answer key. Solution by PrepWiser.