Oscillations and Waves > The superposition of oscillations > Problem 3.4.11

Statement

$3.4.11.$ [Two balls of mass $m_1$ and $m_2$, attached to identical springs, can oscillate by sliding on a bar of mass $M$ without friction. The bar lies on a horizontal plane. The balls are connected by a thread, the tension of which is $F$. The thread is burned out. At what smallest coefficient of friction between the plane and the bar the bar will not move?]

Solution

Solution

To prevent the bar of mass $M$ from moving, the maximum horizontal force exerted on it by the springs must not exceed the maximum static friction force between the bar and the horizontal plane.

1. Before the thread is burned, the entire system is in a state of static equilibrium. The thread connects the two balls of mass $m_1$ and $m_2$, holding them under a constant tension $F$. Since the balls are initially at rest, each spring exerts an equal and opposite restoring force to perfectly balance this tension. Therefore, the initial tension force registered in each spring is:
$$F_1 = F_2 = F$$

When the thread is burned out, the constraint holding the balls is removed, and they begin to undergo simple harmonic oscillations along the bar.

2. As the balls oscillate, they exert time-dependent forces back onto the bar through the attachments of the springs. By Newton's third law, the force exerted by a spring on the bar at any instant is equal in magnitude to the force it exerts on its respective ball. The maximum force each spring can exert on the bar during the oscillation cycle occurs at the points of maximum elongation or compression. Since the system was released from rest at its maximum initial elongation, the maximum force each spring can possibly exert during the subsequent motion is exactly equal to its initial equilibrium force:
$$F_{1,\max} = F$$
$$F_{2,\max} = F$$

To find the critical boundary condition where the bar is most likely to slip, we must analyze the worst-case scenario. This occurs when both oscillating balls happen to move in phase such that both spring forces pull the bar in the exact same direction simultaneously, maximizing the net horizontal pulling force:
$$F_{\text{net},\max} = F_{1,\max} + F_{2,\max} = 2F$$

3. The vertical forces acting on the bar determine the total normal force $N$ exerted between the bar and the horizontal plane. The total downward gravitational force includes the weight of the bar itself and the weights of both balls sliding on it:
$$N = (M + m_1 + m_2)g$$

The maximum static friction force $f_{\max}$ that the horizontal plane can provide to counteract the horizontal pulling force and prevent the bar from sliding is governed by the coefficient of static friction $\mu$:
$$f_{\max} = \mu N = \mu (M + m_1 + m_2)g$$

4. For the bar to remain perfectly stationary, the maximum static friction force must be greater than or equal to the maximum net horizontal force exerted by the springs ($f_{\max} \ge F_{\text{net},\max}$):
$$\mu (M + m_1 + m_2)g \ge 2F$$

By isolating the coefficient of friction, we determine the smallest threshold value required to ensure stability:
$$\mu = \frac{2F}{(M + m_1 + m_2)g}$$

Answer

$$\mu = \frac{2F}{(M + m_1 + m_2)g}$$

Answer

$$\boxed{\mu = \frac{2F}{(M + m_1 + m_2)g}}$$