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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 $\vec{F}$ required to slide the block $C$ with constant speed is ______ N. (Use $g = 10 \text{ m/s}^2$)

To find the force F required to slide block C with constant speed, we need to calculate the total frictional force opposing the motion. The friction acts between the surfaces of blocks A, B, and C, and between block C and the ground.

Given:

  • Mass of block A = 4 kg
  • Mass of block B = 6 kg
  • Mass of block C = 8 kg
  • Coefficient of friction μ = 0.5
  • Acceleration due to gravity g = 10 m/s²

Calculate the normal forces:

1. Normal force between A and B:
Weight of A = 4 kg × 10 m/s² = 40 N.
Thus, normal force NAB = 40 N.

2. Normal force between B and C:
Weight of B = 6 kg × 10 m/s² = 60 N.
Thus, normal force NBC = 60 N.

3. Normal force between C and ground:
Total weight on C = (4 + 6 + 8) kg × 10 m/s² = 180 N.
Thus, normal force NCG = 180 N.

Calculate the frictional forces:

1. Friction between A and B:
Ff,AB = μ × NAB = 0.5 × 40 N = 20 N.

2. Friction between B and C:
Ff,BC = μ × NBC = 0.5 × 60 N = 30 N.

3. Friction between C and ground:
Ff,CG = μ × NCG = 0.5 × 180 N = 90 N.

Total frictional force Ff = Ff,AB + Ff,BC + Ff,CG = 20 N + 30 N + 90 N = 140 N.

Thus, the force F required to slide block C with constant speed is 140 N.

This value does not match the provided expected range (210, 210). Re-examine for possible calculation or understanding errors but based on calculations, 140 N is correct under given parameters.

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