Solution: Stacked Blocks with Friction
1. Equilibrium Analysis
In equilibrium, the right spring (\( k \)) is stretched by \( x_0 \). To balance the system (since floor is frictionless), the left spring (\( 3k \)) must pull with equal force.
$$ F_{3k} = F_k \implies 3k x_L = k x_0 \implies x_L = \frac{x_0}{3} $$So the left spring is stretched by \( x_0/3 \).
2. Oscillation Parameters
Both blocks move together. The effective spring constant is \( K_{eff} = 3k + k = 4k \). Total mass is \( 2m \).
$$ \omega^2 = \frac{4k}{2m} = \frac{2k}{m} $$Max acceleration at amplitude \( A \): \( a_{max} = \omega^2 A = \frac{2k A}{m} \).
3. Friction Requirement
Consider the forces at the extreme right position (\( x = +A \)).
- Displacement from equilibrium is \( A \) to the right.
- Acceleration is max to the left: \( a = -2kA/m \).
- Bottom Block: The right spring force changes by \( -kA \). The net force on the system is restoring, so friction from the top block must help/oppose? Let’s verify with the Top Block.
- Top Block:
- Spring force (Left, \( 3k \)): Stretched further by \( A \). Force \( F_s = 3k(x_0/3 + A) = kx_0 + 3kA \) (Left).
- Friction \( f \): Acts to the right to oppose the excessive spring force? Let’s check Newton’s law:
- \( f – F_s = m a \implies f – (kx_0 + 3kA) = m(-2kA/m) \)
- \( f – kx_0 – 3kA = -2kA \)
- \( f = kx_0 + kA \) (Positive, so acting to the right).
The required friction is magnitude \( f = k(x_0 + A) \).
4. Limiting Condition
For no slipping, \( f \le \mu N \), where \( N = mg \).
$$ k(x_0 + A) \le \mu mg $$ $$ x_0 + A \le \frac{\mu mg}{k} $$
$$ A \le \frac{\mu mg}{k} – x_0 $$