Question

Find M (in kg) for which m₂ will start sliding over m₁. Table on which m₁ and m₂ are placed is smooth, $\mu$ is the friction coefficient between m₁ and m2. (Given : m₁ = 2 kg, m2 = 1 kg, $\mu$ = 0.5, g = 10 m/s²)

Answer 1

3.00

Answer 2

To find the mass M for which m₂ will start sliding over m₁, we need to consider the forces acting on the system and the condition for relative motion.

1. Condition for m₂ to start sliding: For m₂ to move without sliding over m₁, the static friction force between m₁ and m₂ must provide the necessary acceleration to m₂. The maximum static friction force ($f_{s,max}$) between m₁ and m₂ is given by: $f_{s,max} = \mu N_2$ Where $N_2$ is the normal force exerted by m₁ on m₂. Since m₂ is on a horizontal surface (m₁), $N_2 = m_2 g$. So, $f_{s,max} = \mu m_2 g$

The maximum acceleration ($a_{max}$) that m₂ can have without sliding over m₁ is when the static friction force reaches its maximum value. From Newton's second law for m₂: $f_{s,max} = m_2 a_{max}$ $\mu m_2 g = m_2 a_{max}$ $a_{max} = \mu g$

So, m₂ will start sliding over m₁ when the acceleration of the system (m₁ and m₂ moving together) exceeds $a_{max} = \mu g$. Therefore, the critical condition for m₂ to just start sliding is when the acceleration of the system is exactly $a = \mu g$.

2. Acceleration of the system (m₁ + m₂ + M): Consider the entire system consisting of masses m₁, m₂, and M. The table on which m₁ and m₂ are placed is smooth, so there is no friction from the table. The only external horizontal force acting on the (m₁ + m₂) block is the tension from the string, which is ultimately due to the weight of mass M. The total mass being accelerated is $(m_1 + m_2 + M)$. The net force causing the acceleration is the weight of mass M, which is $Mg$. Using Newton's second law for the entire system: $F_{net} = (m_1 + m_2 + M) a$ $Mg = (m_1 + m_2 + M) a$ So, the acceleration of the system is: $a = \frac{Mg}{m_1 + m_2 + M}$

3. Equating the accelerations to find M: For m₂ to just start sliding over m₁, the acceleration of the system must be equal to $a_{max}$: $a = a_{max}$ $\frac{Mg}{m_1 + m_2 + M} = \mu g$

Cancel 'g' from both sides (assuming $g \ne 0$): $\frac{M}{m_1 + m_2 + M} = \mu$

Now, substitute the given values: m₁ = 2 kg m₂ = 1 kg μ = 0.5

$\frac{M}{2 + 1 + M} = 0.5$ $\frac{M}{3 + M} = 0.5$

Cross-multiply: $M = 0.5 (3 + M)$ $M = 1.5 + 0.5M$ $M - 0.5M = 1.5$ $0.5M = 1.5$ $M = \frac{1.5}{0.5}$ $M = 3$ kg

Thus, m₂ will start sliding over m₁ when M is 3 kg.