Question

A frictionless cart A of mass 100 kg carries other two frictionless carts B and C having masses 8kg and 4kg respectively connected by a string passing over a pulley as shown in figure. What horizontal force F must be applied on the cart so that smaller carts do not move relative to it ?

• A. 560 N
• B. 150 N
• C. 630 N
• D. 340 N

Answer 1

The correct option is A 560 N

Let $a$ be the acceleration of the system in the right direction.
Since the smaller carts do not move relative to the larger one, the equation of motion of the whole system,
$(m_A+m_B+m_C)a=F$
$(100+8+4)a=F$

$⟹F=112a$

As block B experience a pseudo force which is equal to $ma$ and as the blocks are at rest so this psudo force balance the $T$.

For block B, $T=m_{B}a=8a$ ....(1)

Block C is not moving so we balance the forces in vertical direction.
For block C, $T=4g$ ....(2)

From (1) and (2), we get $8a=4g$
Thus $a=0.5g$

$a=5m/s^2$ (as $g=10m/s^2$)

$\therefore F=112\times 5=560N$