Question

The value of frictional force and acceleration of block of mass 10 kg in the figure are

• A. 10 N, 1 m/s²
• B. 20 N, 2 m/s²
• C. 20 N, 0 m/s²
• D. 10 N, 0 m/s²

Answer 1

Answer: (C)

20 N, 0 m/s²

Answer 2

To determine the frictional force and acceleration of the 10 kg block, we need to analyze the forces acting on both blocks and check for the condition of motion.

1. Identify Given Data:

• Mass of block on table, $m_1 = 10$ kg
• Mass of hanging block, $m_2 = 4$ kg
• Coefficient of friction, $\mu = 0.4$
• Angle of the string with the horizontal, $\theta = 60^\circ$
• Assume acceleration due to gravity, $g = 10 \text{ m/s}^2$.

2. Free Body Diagram (FBD) and Forces:

• For the hanging block ($m_2 = 4$ kg):

  • Weight acting downwards: $W_2 = m_2 g = 4 \times 10 = 40$ N
  • Tension acting upwards: $T$

• For the block on the table ($m_1 = 10$ kg):

  • Weight acting downwards: $W_1 = m_1 g = 10 \times 10 = 100$ N
  • Normal force acting upwards: $N_1$
  • Tension $T$ acting at an angle of $60^\circ$ above the horizontal. This tension has two components:
    • Horizontal component: $T_x = T \cos(60^\circ)$ (to the right)
    • Vertical component: $T_y = T \sin(60^\circ)$ (upwards)
  • Frictional force $f$ acting to the left, opposing the tendency of motion.

3. Check for Motion (Assume System is at Rest, $a=0$):

If the system is at rest, the acceleration $a=0$.

• Equation for the hanging block ($m_2$): Since $a=0$, the net force is zero. $T - W_2 = 0$ $T = W_2 = 40$ N

• Equations for the block on the table ($m_1$):

  • Vertical Equilibrium: The sum of upward forces equals the sum of downward forces. $N_1 + T \sin(60^\circ) - W_1 = 0$ $N_1 = W_1 - T \sin(60^\circ)$ Substitute $W_1 = 100$ N and $T = 40$ N: $N_1 = 100 - 40 \times \sin(60^\circ)$ $N_1 = 100 - 40 \times \frac{\sqrt{3}}{2}$ $N_1 = 100 - 20\sqrt{3}$ N Using $\sqrt{3} \approx 1.732$: $N_1 \approx 100 - 20 \times 1.732 = 100 - 34.64 = 65.36$ N

  • Calculate Maximum Static Friction ($f_{s,max}$): The maximum static friction that can oppose motion is $f_{s,max} = \mu N_1$. $f_{s,max} = 0.4 \times (100 - 20\sqrt{3})$ $f_{s,max} \approx 0.4 \times 65.36 = 26.144$ N

  • Horizontal Equilibrium (Required Static Friction, $f_{req}$): The force tending to move the block horizontally is the horizontal component of tension. $T \cos(60^\circ) - f_{req} = 0$ $f_{req} = T \cos(60^\circ)$ Substitute $T = 40$ N: $f_{req} = 40 \times \cos(60^\circ)$ $f_{req} = 40 \times 0.5 = 20$ N

4. Compare Required Friction with Maximum Available Friction:

• Required static friction $f_{req} = 20$ N
• Maximum available static friction $f_{s,max} = 26.144$ N

Since $f_{req} < f_{s,max}$ ($20 \text{ N} < 26.144 \text{ N}$), the force tending to move the block is less than the maximum static friction that can be exerted. Therefore, the system will remain at rest.

5. Conclusion:

• The acceleration of the block is $a = 0 \text{ m/s}^2$.
• The actual frictional force acting on the block is equal to the force tending to move it, which is $f = f_{req} = 20$ N.

The value of frictional force is 20 N and the acceleration of the block is 0 m/s².