Question

Note that xcritical, as computed in the previous part, is not necessarily positive. If xcritical<0, the bar will be stable no matter where the block of mass m2 is placed on it. Assuming that m1, d, and L are held fixed, what is the maximum block mass mmax for which the bar will always be stable? In other words, what is the maximum block mass such that xcritical≤0? Answer in terms of m1, d, and L

Answer 1

To find the maximum block mass mmax for which the bar will always be stable, substitute m₂ = mmax into the equation for x_critical and solve for mmax.

Step-by-step explanation:

To determine the maximum block mass for which the bar will always be stable, we need to find the maximum value of m₂ such that x_critical ≤ 0. From the information given, we can see that x_critical is the point where the equilibrium becomes unstable. In other words, when x_critical < 0, the bar is stable regardless of the position of m₂.

To find the maximum m₂, we need to find the value of m₂ such that x_critical is less than or equal to zero. Using the equations and information provided in the previous parts, you can substitute m₂ = mmax into the equation for x_critical and solve for mmax. This will give you the maximum block mass mmax for which the bar will always be stable.