Three masses M=100 kg, m1=10 kg and m2=20 kg are arranged in a system as shown in figure. All the surfaces are frictionless and strings are inextensible and weightless. The pulleys are also weightless and frictionless. A force F is applied on the system so that the mass m2 moves upward with an acceleration of 2 ms−2 . The value of F is (Take g=10 ms−2 )

1. Identify the constraints

The rope passes once over a smooth pulley, therefore the horizontal displacement of m₁ (relative to the box) equals the upward displacement of m₂.

If we denote

• $x_{M}$ : displacement of the big block $M$
• $x_{1}$ : additional horizontal displacement of $m_{1}$ relative to $M$
• $y_{2}$ : vertical displacement of $m_{2}$ (downwards positive) then the rope‐length condition gives $x_{1}+y_{2}=\text{constant} \;\;\Longrightarrow\;\; \ddot{x}_{1}+\ddot{y}_{2}=0$

Because the problem demands m₂ to move upward with 2 m/s², we set $\ddot{y}_{2}=-2\;\text{m/s}²$ (upward negative). Therefore $\ddot{x}_{1}=+2\;\text{m/s}²$ This is the acceleration of m₁ relative to the box.

2. Forces on each mass

1. Block $m_{2}$ (20 kg)

   • Upward tension $T$
   • Downward weight $m_{2}g$
   • Upward acceleration $2\,\text{m/s}²$ $T-m_{2}g = m_{2}(+2) \;\;\Longrightarrow\;\; T = m_{2}(g+2)=20(10+2)=\boxed{240\,\text{N}}$

2. Block $m_{1}$ (10 kg) (in ground frame)

   • Only horizontal force is the same tension $T$ (to the right)
   • Its horizontal acceleration is the sum of the box’s acceleration $A$ and the relative $+2$ just found: $a_{1,x}=A+2$ $T = m_{1}(A+2) \;\;\Rightarrow\;\; 240 = 10(A+2) \;\;\Rightarrow\;\; A = 22\,\text{m/s}²$ So the big block M (and the pulley fixed on it) must accelerate rightward at 22 m/s².

3. Big block $M$ (100 kg)

   • Pulled to the right by $F$
   • Pulled to the left by the horizontal component of the rope, which is the full tension $T$ because that segment is horizontal.
   • Horizontal equation: $F-T = M A$.

3. Evaluate $F$

Instead of writing three separate equations with interaction forces, we may use the horizontal motion of all three masses, because $F$ is the only external horizontal force:

• $M$ : acceleration $A$
• $m_{1}$ : acceleration $A+2$
• $m_{2}$ : acceleration $A$ (it hangs directly under the pulley, so it shares the box’s sideways motion)

Hence $F = M A + m_{1}(A+2) + m_{2}A$

Plug numbers: $F = 100\times22 + 10\times24 + 20\times22 = 2200 + 240 + 440 = \boxed{2880\,\text{N}}$

What is happening?

Imagine a big, smooth box M (100 kg) resting on an ice-rink so it can slide without rubbing. A light pulley is fixed to the right edge of this box.

On the roof of the box we put a small block m₁ (10 kg) that can also slide smoothly. This block is tied to a hanging block m₂ (20 kg) through the pulley: the rope runs horizontally from m₁ to the pulley and then straight down to m₂.

Now we pull the big box towards the right with a force F. Because the rope cannot stretch, whenever m₁ slides 1 m to the right, m₂ rises 1 m. The question says "make m₂ rise with 2 m/s²". We must find what pull F will do that.

So we have to balance:

1. Us pulling the big box → the whole set starts moving right.
2. The rope tugging the little blocks → m₁ is dragged right, m₂ is lifted up.
3. Gravity pulling m₂ down.

Using Newton’s laws and the rope rule we finally get F = 2880 N.

Step–by–step calculation

1. Rope constraint
   $x_{1}+y_{2}=\text{constant}\;\Rightarrow\; \ddot{x}_{1}+\ddot{y}_{2}=0$
   Given $\ddot{y}_{2} = -2\;\text{m/s}²$ (upward), we have
   $\ddot{x}_{1}=+2\;\text{m/s}²$

2. Block $m_{2}$
   $T - m_{2}g = m_{2}(+2)$
   $T = 20(10+2)=240\,\text{N}$

3. Block $m_{1}$
   $T = m_{1}(A+2)$
   $240 = 10(A+2) \;\Longrightarrow\; A = 22\,\text{m/s}²$

4. Total horizontal force
   Entire system’s horizontal accelerations:
   $M\!:\!A$, $m_{1}\!:\!A+2$, $m_{2}\!:\!A$.
   $F = MA + m_{1}(A+2) + m_{2}A$
   $F = 100\times22 + 10\times24 + 20\times22$
   $F = 2200 + 240 + 440 = 2880\,\text{N}$

Final answer: $\boxed{F = 2.88 \times 10^{3}\,\text{N}}$