Question

2 kg and 1 kg blocks are connected by an ideal string and are placed on rough surface as shown. Maximum value of force F (in N) to prevent the motion is

• A. 5 N

Answer 1

12 N

Answer 2

To find the maximum value of force F to prevent motion, we need to consider the two possible scenarios for impending motion:

1. The blocks are on the verge of moving to the right.
2. The blocks are on the verge of moving to the left.

We will use $g = 10 \text{ m/s}^2$.

1. Calculate Normal Forces and Maximum Static Friction Forces:

• For the 2 kg block ($m_1 = 2$ kg, $\mu_1 = 0.2$):

  • Normal force $N_1 = m_1 g = 2 \text{ kg} \times 10 \text{ m/s}^2 = 20$ N.
  • Maximum static friction $f_{s1,max} = \mu_1 N_1 = 0.2 \times 20 \text{ N} = 4$ N.

• For the 1 kg block ($m_2 = 1$ kg, $\mu_2 = 0.3$):

  • Normal force $N_2 = m_2 g = 1 \text{ kg} \times 10 \text{ m/s}^2 = 10$ N.
  • Maximum static friction $f_{s2,max} = \mu_2 N_2 = 0.3 \times 10 \text{ N} = 3$ N.

2. Scenario 1: Blocks are on the verge of moving to the right.

In this case, the external force F is large enough to overcome the opposing forces, and the system is about to move right. The static friction forces on both blocks will act to the left to oppose this motion. Let T be the tension in the string.

• Free Body Diagram for 2 kg block:

  • Applied force: 5 N (left)
  • Tension: T (right)
  • Friction: $f_{s1,max}$ (left)
  • Equation for equilibrium (on the verge of motion): $T - 5 \text{ N} - f_{s1,max} = 0$ $T - 5 \text{ N} - 4 \text{ N} = 0$ $T = 9$ N

• Free Body Diagram for 1 kg block:

  • Applied force: F (right)
  • Tension: T (left)
  • Friction: $f_{s2,max}$ (left)
  • Equation for equilibrium (on the verge of motion): $F - T - f_{s2,max} = 0$ $F - 9 \text{ N} - 3 \text{ N} = 0$ $F = 12$ N

This value $F = 12$ N represents the maximum force F that can be applied to the right before the system starts moving to the right.

3. Scenario 2: Blocks are on the verge of moving to the left.

In this case, the 5 N force is dominant, and the system is about to move left. The static friction forces on both blocks will act to the right to oppose this motion. Let T be the tension in the string.

• Free Body Diagram for 2 kg block:

  • Applied force: 5 N (left)
  • Tension: T (right)
  • Friction: $f_{s1,max}$ (right)
  • Equation for equilibrium (on the verge of motion): $5 \text{ N} - T - f_{s1,max} = 0$ $5 \text{ N} - T - 4 \text{ N} = 0$ $T = 1$ N

• Free Body Diagram for 1 kg block:

  • Applied force: F (right)
  • Tension: T (left)
  • Friction: $f_{s2,max}$ (right)
  • Equation for equilibrium (on the verge of motion): $T - F - f_{s2,max} = 0$ $1 \text{ N} - F - 3 \text{ N} = 0$ $F = -2$ N

A force F = -2 N means that to prevent motion to the left, F would need to be applied with a magnitude of 2 N towards the left. Since F is defined as a force to the right in the diagram, this implies that even if F = 0, the system will not move to the left. The minimum value of F (applied to the right) required to prevent motion to the left is 0 N (or even a force applied to the left).

Conclusion:

The system remains at rest for a range of F values. The maximum value of F to prevent motion is the F at which the system is about to move to the right, which we found to be 12 N. The minimum value of F (when applied to the right) to prevent motion is 0 N (or effectively, the system won't move left even if F is 0).

Therefore, the maximum value of force F to prevent the motion is 12 N.