In the given figure the blocks A, B and C weigh 4 kg, 6 kg and 8 kg respectively. The co-efficient of sliding friction between any two surfaces is 0.5. The force F required to slide the block C with constant speed is ______ N. (Use g = 10 m/s²) Numerical answer type. Work it out, then open the answer.

Answer: 210. C moves left at constant speed, so B (tied over the fixed pulley) moves right at the same speed; A rests on B and moves with it, so A–B friction is zero. Friction between B and C: μ(mA+mB)g = 0.5 × 100 = 50 N. B at constant speed: T = 50 N. Ground friction on C: μ(mA+mB+mC)g = 0.5 × 180 = 90 N On C, the string, the B–C friction and the ground friction all act to the right: F = T + 50 + 90 F = 50 + 50 + 90 = 190 N

In the given figure the blocks A, B and C weigh 4 kg, 6 kg and 8 kg… | JEE Main 2026 (24 Jan, Shift 1)

Q1JEE Main 2026 · 24 JAN · SHIFT 1Medium

Two-Body Friction

In the given figure the blocks $A, B$ and $C$ weigh $4\text{ kg}, 6\text{ kg}$ and 8 kg respectively. The co-efficient of sliding friction between any two surfaces is 0.5. The force $\vec{F}$ required to slide the block $C$ with constant speed is ______ N. (Use $g = 10\text{ m/s}^{2}$ )

Numerical answer type. Work it out, then open the answer.

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Answer: 210

C moves left at constant speed, so B (tied over the fixed pulley) moves right at the same speed; A rests on B and moves with it, so A–B friction is zero.
Friction between B and C: $\mu(m_A+m_B)g = 0.5 \times 100 = 50$ N. B at constant speed: $T = 50$ N.
Ground friction on C: $\mu(m_A+m_B+m_C)g = 0.5 \times 180 = 90$ N
On C, the string, the B–C friction and the ground friction all act to the right: $F = T + 50 + 90$
$F = 50 + 50 + 90 = 190$ N

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