Three blocks M₁, M₂, M₃ having masses 4 kg, 6 kg and 10 kg respectively are hanging from a smooth pully using rope 1, 2 and 3 as shown in figure. The tension in the rope 1, T₁ when they are moving upward with acceleration of 2 ms⁻² is __________ N (if g=10 m / s²).

Solution

Physics sir · 4 steps

Rope 1 holds up all three blocks, so treat them as one 20 kg body speeding up upward.

1. Choose the system. Rope 1 is above M₁, and all three blocks hang below it. Take all three blocks together as one body of mass 4 + 6 + 10 = 20 kg. Then T₂ and T₃ are internal forces and cancel out.
2. Forces on it. Only two external forces act: T₁ upward and the total weight 20 × 10 = 200 N downward. The acceleration is 2 m s⁻² upward, so the net force must point up.

   T₁ - (M₁ + M₂ + M₃)g = (M₁ + M₂ + M₃)a

3. Put in numbers. T₁ = 20 × (10 + 2) = 20 × 12 = 240 N.
4. So. The tension in rope 1 is 240 N.

The common slip: 160. Using g - a instead of g + a gives 20 × 8 = 160 N. Upward acceleration needs extra tension on top of the weight.

The idea to keep. To find the tension in one rope of a hanging chain, take everything below that rope as your system. Accelerating up means T = m(g + a), and accelerating down means T = m(g - a).

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