Question

All the pulleys and strings shown in figure are massless and frictionless. Mark the correct statement(s). (Take g = 10 m/s²)

• A. The frictional force acting on 1 kg block is 2 N
• B. Acceleration of 2 kg block is 2 m/s²
• C. The tension in the string connecting 4 kg block and 1 kg block is 2 N
• D. The tension in the string connecting 1 kg block and 4 kg block is zero

Answer 1

None of the options are correct. The system is at rest. The frictional force on the 1 kg block is 6 N, the acceleration of the 2 kg block is 0 m/s², and the tension in the string connecting the 4 kg and 1 kg blocks is 6 N.

Answer 2

The problem involves a system of three blocks connected by strings and pulleys, with friction on the horizontal surface.

1. Define Variables and Forces:

   • $m_1 = 1 \text{ kg}$, $m_2 = 4 \text{ kg}$, $m_3 = 2 \text{ kg}$.
   • $g = 10 \text{ m/s}^2$.
   • Coefficient of friction for $m_1$: $\mu_1 = 0.6$.
   • Coefficient of friction for $m_2$: $\mu_2 = 0.2$.
   • Let $T_1$ be the tension in the string connecting $m_1$ and $m_2$.
   • Let $T$ be the tension in the string passing through the pulley system (connected to $m_3$ and $m_2$).
   • Let $a$ be the acceleration of $m_1$ and $m_2$ (since they are connected by a string and move together).
   • Let $a_3$ be the acceleration of $m_3$.

2. Analyze the Pulley System and Kinematic Constraint:

   If $m_2$ moves to the right by a distance $x$, the length of the string segment from the fixed pulley to $m_2$ increases by $x$. This length must be compensated by a decrease in the string length on the vertical side. The vertical string length is $2y$, where $y$ is the distance from the fixed point to the movable pulley. So, if $m_2$ moves right by $x$, the movable pulley (and $m_3$) must move up by $x/2$.

   Therefore, if $a$ is the acceleration of $m_2$ to the right, then the acceleration of $m_3$ upwards is $a/2$.

   Let $a_3$ be the magnitude of acceleration of $m_3$. If $m_3$ moves downwards, its acceleration is $-a_3$. So $a = 2a_3$.

3. Equations of Motion:

   • For $m_3$ (2 kg block):

     The string has tension $T$. The movable pulley is pulled upwards by two segments of the string, so the upward force is $2T$. The weight is $m_3 g$ downwards.

     Assuming $m_3$ moves downwards:

     $m_3 g - 2T = m_3 a_3$

     $2(10) - 2T = 2a_3 \implies 20 - 2T = 2a_3 \implies 10 - T = a_3$ (Equation 1)

   • For $m_2$ (4 kg block):

     Pulled to the right by tension $T$.

     Friction force $f_2 = \mu_2 N_2 = \mu_2 m_2 g = 0.2 \times 4 \times 10 = 8 \text{ N}$.

     Pulled to the left by tension $T_1$.

     Assuming $m_2$ moves to the right:

     $T - T_1 - f_2 = m_2 a$

     $T - T_1 - 8 = 4a$ (Equation 2)

   • For $m_1$ (1 kg block):

     Pulled to the right by tension $T_1$.

     Friction force $f_1 = \mu_1 N_1 = \mu_1 m_1 g = 0.6 \times 1 \times 10 = 6 \text{ N}$.

     Assuming $m_1$ moves to the right:

     $T_1 - f_1 = m_1 a$

     $T_1 - 6 = 1a \implies T_1 = a + 6$ (Equation 3)

4. Solve the System of Equations:

   Substitute $T_1$ from Equation 3 into Equation 2:

   $T - (a + 6) - 8 = 4a$

   $T - a - 14 = 4a$

   $T = 5a + 14$ (Equation 4)

   Now, use the kinematic constraint $a = 2a_3$, which means $a_3 = a/2$. Substitute this into Equation 1:

   $10 - T = a/2$ (Equation 5)

   Substitute $T$ from Equation 4 into Equation 5:

   $10 - (5a + 14) = a/2$

   $10 - 5a - 14 = a/2$

   $-4 - 5a = a/2$

   $-4 = 5a + a/2$

   $-4 = (10a + a)/2$

   $-4 = 11a/2$

   $11a = -8$

   $a = -8/11 \text{ m/s}^2$

5. Interpret the Result for Acceleration:

   The negative sign for $a$ indicates that our initial assumption for the direction of motion (right for $m_1, m_2$ and down for $m_3$) is incorrect. The system will not move in that direction. This means the applied forces are not enough to overcome static friction, or the system will move in the opposite direction.

   The maximum static friction forces:

   $f_{1,max} = \mu_1 m_1 g = 0.6 \times 1 \times 10 = 6 \text{ N}$

   $f_{2,max} = \mu_2 m_2 g = 0.2 \times 4 \times 10 = 8 \text{ N}$

   The driving force for the system is the weight of $m_3$, which is $m_3 g = 2 \times 10 = 20 \text{ N}$.

   The total maximum static friction force that needs to be overcome for $m_1$ and $m_2$ to move is $f_{1,max} + f_{2,max} = 6 + 8 = 14 \text{ N}$.

   If the system is in equilibrium, then $a=0$ and $a_3=0$.

   From Equation 1: $10 - T = 0 \implies T = 10 \text{ N}$.

   From Equation 3: $T_1 - 6 = 0 \implies T_1 = 6 \text{ N}$.

   From Equation 2: $T - T_1 - f_2 = 0 \implies 10 - 6 - f_2 = 0 \implies 4 - f_2 = 0 \implies f_2 = 4 \text{ N}$.

   Since all required static friction forces are less than or equal to their maximum possible values, the system will remain at rest.

   Therefore, the acceleration of all blocks is zero.

   $a = 0 \text{ m/s}^2$ and $a_3 = 0 \text{ m/s}^2$.

6. Evaluate the Options:

   • A. The frictional force acting on 1 kg block is 2 N

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

     From Equation 3: $T_1 - f_1 = 0 \implies T_1 = f_1$.

     From the equilibrium conditions derived above, $T_1 = 6 \text{ N}$. So, $f_1 = 6 \text{ N}$.

     The statement says $f_1 = 2 \text{ N}$, which is incorrect.

   • B. Acceleration of 2 kg block is 2 m/s²

     Since the system is at rest, the acceleration of the 2 kg block ($m_3$) is $a_3 = 0 \text{ m/s}^2$.

     The statement says $a_3 = 2 \text{ m/s}^2$, which is incorrect.

   • C. The tension in the string connecting 4 kg block and 1 kg block is 2 N

     This is $T_1$. From the equilibrium conditions, $T_1 = 6 \text{ N}$.

     The statement says $T_1 = 2 \text{ N}$, which is incorrect.

   • D. The tension in the string connecting 1 kg block and 4 kg block is zero

     This is $T_1$. From the equilibrium conditions, $T_1 = 6 \text{ N}$.

     The statement says $T_1 = 0 \text{ N}$, which is incorrect.