Three masses $$M = 100 \text{ kg}$$, $$m_1 = 10 \text{ kg}$$ and $$m_2 = 20 \text{ kg}$$ 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 $$m_2$$ moves upward with an acceleration of $$2 \text{ m s}^{-2}$$. The value of $$F$$ is (Take $$g = 10 \text{ m s}^{-2}$$)

Consider the frame of reference of the $$100\ kg$$ block. Since this block is accelerating towards the right with acceleration a, this frame is non-inertial, and hence a pseudo force ma acts on each mass towards the left.

For the $$10\ kg$$ block (which moves left relative to the big block with acceleration $$2 \text{ m s}^{-2}$$), the net acceleration in this non-inertial frame is $$2 \text{ m s}^{-2}$$ towards the left. Applying Newton’s second law along the horizontal direction,

$$10a-T=10\times2\quad$$

where 10a is the pseudo force and T is the tension acting opposite to it.

For the 20 kg hanging mass, which moves upward with acceleration  $$2 \text{ m s}^{-2}$$ applying Newton’s second law in vertical direction,

$$T−20g=20\times2⇒T−200=40$$

Adding equations (1) and (2),

$$10a−200=60⇒10a=260⇒a=26m/s^2$$

Substituting back into (2),

$$T=240N$$

Now, considering the horizontal motion of the entire system, the 100 kg block and the 20kg mass together move with acceleration a. The only external horizontal force is F, and tension T opposes motion. Hence,

$$F-T=(100+20)a$$

$$F−240=120\times26⇒F=3360N$$