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 F⃗ required to slide the block C with constant speed is ______ N. (Use g=10 m/s²)

Idea: work out the normal force at each rubbing surface, and notice that the string over the wall pulley makes C pay for the B-C surface twice.

The three normal forces. Each surface carries everything above it:

A on B: N₁=4g=40 N, so f₁=0.5×40=20 N.

B on C: N₂=(4+6)g=100 N, so f₂=0.5×100=50 N.

C on the floor: N₃=(4+6+8)g=180 N, so f₃=0.5×180=90 N.

Why f₂ counts twice. B and C are joined by a string that runs over a pulley fixed to the wall, so dragging C to the left hauls B to the right. F must overcome the B-C friction directly on C, and it must also supply the string tension that drags B against that same friction.

Taking B first, at constant speed:

T=f₁+f₂=20+50=70 N.

Then for C:

F=T+f₂+f₃=70+50+90

F=210 N.

A note on f₁. That term belongs in B's equation only if A is kept from riding along with B. With A simply loose on top, it would be carried along once everything is at constant speed, no kinetic friction would act there, and the force would be 190 N. The printed answer includes it.