Question

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 2ms–2 is ............... N (if g = 10 m/s2 )

Answer 1

Correct Answer: 240

Let’s analyze the forces acting on each block.

For the system as a whole (masses \(M_1\), \(M_2\), and \(M_3\) together) moving upwards with an acceleration \(a = 2 \, \mathrm{m/s^2}\):

Total mass, \(M = M_1 + M_2 + M_3 = 4 + 6 + 10 = 20 \, \mathrm{kg}.\)

Total weight, \(W = Mg = 20 \times 10 = 200 \, \mathrm{N}\)

Since the entire system is accelerating upwards, the net force \(F\) required to produce this acceleration is given by:

\(F = Ma = 20 \times 2 = 40 \, \mathrm{N}\)

Thus, the tension \(T_1\) in rope 1 must support both the weight and the additional force required for acceleration:

\(T_1 = W + F = 200 + 40 = 240 \, \mathrm{N}\)

Answer 2

Step 1: Interpret the setup
Three blocks are connected one below another by light ropes. The entire chain of masses moves upward with acceleration \(a = 2\,\text{m s}^{-2}\). We are asked for the tension in the top rope (rope 1), denoted \(T_1\). Take \(g = 10\,\text{m s}^{-2}\). The masses are \[ M_1=4\,\text{kg},\quad M_2=6\,\text{kg},\quad M_3=10\,\text{kg}. \] Hence the total mass of the hanging system is \[ M_{\text{tot}} = 4+6+10 = 20\,\text{kg}. \]

Step 2: Use a combined free-body for all three blocks
If we consider the three blocks together as one system, the only external forces on this combined system are:

• Upward tension \(T_1\) from the top rope,
• Downward total weight \(M_{\text{tot}}\,g = 20g\).

Internal tensions between blocks cancel in the combined FBD.

Applying Newton’s second law in the upward direction: \[ T_1 - 20g = M_{\text{tot}}\,a = 20 \times 2 = 40. \] Therefore \[ T_1 = 20g + 40 = 20 \times 10 + 40 = 240\,\text{N}. \]

Step 3: Quick check
Since the system accelerates upward, the top tension must exceed the total weight \(20g = 200\,\text{N}\) by the extra amount needed to accelerate the 20 kg mass at \(2\,\text{m s}^{-2}\), i.e., \(ma = 40\,\text{N}\). Thus \(T_1 = 200 + 40 = 240\,\text{N}\) is consistent.

Final answer

240