Question: Previous Part: Xcritical=d-(m1/m2)((L/2)-d) Note that Xcritical, as co

Question

Question:

Previous Part: Xcritical=d-(m1/m2)((L/2)-d)

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?

Other information about the system:

A rigid, uniform, horizontal bar of mass m1 and length Lis supported by two identical massless strings. (Figure 1) Both strings are vertical. String A is attached at a distance d<L/2 from the left end of the bar and is connected to the ceiling; string B is attached to the left end of the bar and is connected to the floor. A small block of mass m2 is supported against gravity by the bar at a distance x from the left end of the bar, as shown in the figure.

Throughout this problem positive torque is that which spins an object counterclockwise. Use g for the magnitude of the acceleration due to gravity.

Explanation / Answer

sum the moments about x=d as follows:

M = 0 = m2*g*d - m1*g*(L/2 - d)

where m2 is the maximum block mass.

m2 = m1*(L/2 - d)/d = m1*(L/2d - 1)

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