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 m s⁻². The value of F is (Take g=10 m s⁻²)

Idea: the cart accelerates right with some A; the string makes m₁ slide backwards relative to the cart by exactly as much as m₂ rises, so m₁ and m₂ do not share one acceleration.

Start with the block that has the simplest equation, the hanging m₂, vertically:

T-m₂g=m₂(2)

T=20(10+2)=240 N.

The string is inextensible, so if m₂ rises at 2 m s⁻² then m₁ moves at 2 m s⁻² backwards relative to the cart. In the ground frame m₁'s acceleration is therefore A-2, and the only horizontal force on it is the tension:

T=m₁(A-2)

240=10(A-2), so A=26 m/s².

Now take the whole system horizontally. F is the only external horizontal force, and the three masses have horizontal accelerations A, A-2 and A:

F=MA+m₁(A-2)+m₂A

F=100(26)+10(24)+20(26)=2600+240+520

F=3360 N.

Treating all 130 kg as moving at A gives 3380 N — option (b), and the reason the relative motion has to be tracked.