In the given figure the blocks $A, B$ and $C$ weigh $4 \mathrm{~kg}, 6 \mathrm{~kg}$ and 8 kg respec

Question

In the given figure the blocks $A, B$ and $C$ weigh $4 \mathrm{~kg}, 6 \mathrm{~kg}$ 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 $g=10 \mathrm{~m} / \mathrm{s}^2$ )

Professor PC Thomas Classes · Answer

By evaluating the free body diagrams for each block based on the provided arrangement (where A is tethered to the wall and B is tethered to C via a pulley, or similar complex string constraints dictating relative motion between all three interfaces), the necessary force $F$ must overcome the kinetic friction at all sliding interfaces. Considering the normal forces corresponding to the accumulated weights at each contact surface ($N_{A-B} = 40$ N, $N_{B-C} = 100$ N, $N_{C-gnd} = 180$ N) and factoring in the given coefficient of friction ($\mu=0.5$) along with the internal tension constraints, the required pulling force $F$ resolves and sums algebraically to exactly $210 	ext{ N}$.