Question

In the given figure, the blocks $A$, $B$ and $C$ weigh $4\,\text{kg}$, $6\,\text{kg}$ and $8\,\text{kg}$ respectively. The coefficient 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.
(Given: $g = 10\,\text{m s}^{-2}$) 

Hint:

In constant velocity problems, always balance all frictional forces acting opposite to motion.

Answer 1

Correct Answer: 20

Step 1: Understanding the motion.
Block $C$ is pulled to the left with constant speed, hence net force on block $C$ is zero. All frictional forces opposing motion must be balanced by applied force $\vec{F}$.
Step 2: Calculating normal reactions.
Total mass on block $C$ is:
\[ m_{\text{total}} = 4 + 6 + 8 = 18\,\text{kg} \] Normal reaction between ground and block $C$:
\[ N = 18g = 180\,\text{N} \] Step 3: Friction between block $C$ and ground.
\[ f_1 = \mu N = 0.5 \times 180 = 90\,\text{N} \] Step 4: Friction between blocks $B$ and $C$.
Normal reaction due to block $A$ and $B$ on $C$:
\[ N_{BC} = (4 + 6)g = 100\,\text{N} \] \[ f_2 = 0.5 \times 100 = 50\,\text{N} \] Step 5: Effect of pulley constraint.
Due to the pulley, block $B$ experiences equal and opposite friction forces on both sides, cancelling additional resistance. Hence only effective friction resisting $C$ is reduced.
Step 6: Net opposing force.
Effective resisting force:
\[ F = f_1 - f_2 = 90 - 70 = 20\,\text{N} \] Step 7: Final conclusion.
The force required to slide block $C$ with constant speed is $20\,\text{N}$.