MHB Equivalent Damping Constant for Massless Bar

[Dustinsfl]

A massless bar of length \(1\) m is pivoted at one end and subjected to a force \(F\) at the other end. Two translational dampers, with damping constants \(c_1 = 10 \ N\cdot s/m\) and \(c_2 = 15 \ N\cdot s/m\) are connected to the bar as shown in figure 1.109. Determine the equivalent damping constant, \(c_{eq}\), of the system so that the force \(F\) at point \(A\) can be expressed as \(F = c_{eq}v\), where \(v\) is the linear velocity of point \(A\).

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What I have is \(F_{d1} = c_1\dot{x} = -r_1c_1\sin(\theta)\dot{\theta}\) and \(F_{d2} = c_2\dot{x} = -r_2c_1\sin(\theta)\dot{\theta}\) where \(r_2 = r_1\frac{x_2}{x_1}\).

\(F_d = c_{eq}\dot{x} = -rc_{eq}\sin(\theta)\dot{\theta}\) so \(c_{eq} = c_1 + \frac{x_2}{x_1}c_2\).

Is this correct?

 

[DaveE]

Nope, I don't think so. You aren't really clear on your definitions of locations and such, but I think I can follow it. Anyway, your solution doesn't take into account how long the bar is. There are three important values of x.
$x_1+0.25m$, $x_2+0.75m$, and $x_3+1m$, they all affect the solution.

Try again and think about the torque applied at each of those three position versus the velocity. The linear velocity in the damper is related to the angular velocity of the rotating bar. Each of those three points will have the same angular velocity. Use the small angle approximation for the sin term. This is equivalent to assuming the dampers are well aligned with the bar (perpendicular to it).