Maintaining Level

Let the masses of the two objects be $m_a$ and $m_b$ .
Let the point midway between two pivots be $x = 0$ . The pivots are at $x = \pm \frac d2$ .

We don't know where the center of mass of object A is. Suppose it is at $x = a$ . Suppose object B is at $x = b$ .

In the extreme cases just before equilibrium starts to break, the normal force at one of the pivots is $m_ag + m_bg$ , and the normal force at the other pivot is zero.

• When the normal force at the left pivot is just zero, the torque about the other pivot is $\tau = m_ag (a -\frac d2) + m_bg (b - \frac d2)$ . For equilibrium, this torque should be zero.

$\frac{m_a}{m_b} \left(a -\frac d2 \right) + \left(b - \frac d2 \right) = 0 \implies b_{max} = \frac d2 - \frac{m_a}{m_b} \left(a -\frac d2\right)$

• When the normal force at the right pivot is just zero, the torque about the other pivot is $\tau = m_ag (a + \frac d2) + m_bg (b + \frac d2)$ . For equilibrium, this torque should be zero.

$\frac{m_a}{m_b} \left(a + \frac d2 \right) + \left(b + \frac d2 \right) = 0 \implies b_{min} = - \frac d2 - \frac{m_a}{m_b} \left(a +\frac d2 \right)$

The distance between two extremes is $b_{max} - b_{min} = d + \frac{m_a}{m_b} d = (1 + \frac{m_a}{m_b}) d$

In this case, $\frac{m_a}{m_b} = \frac 53$ , so $b_{max} - b_{min} =(1 + \frac 53) \times \frac 32 L = \frac 83 \times \frac 32 L = 4 L$ .