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A uniform rod AB of length 5m and mass 6kg is pivoted at C where AC =1.5m. Calculate the mass of the particle which must be attached at A to maintain equilibrium with the rod horizontal.
help i cant get right answers for these questions. The answer is 4kg
You need the total torque to be zero.
$\displaystyle \sum \tau = 0 $
$\displaystyle 0 = mg(1.5)-M_{rod}g(1) \, \Rightarrow m = 4 kg $
Take clockwise moments and anti-clockwise moments (either side of the pivot)Quote:
Originally Posted by Sundae
so you have clockwise moment = 6kg x 1m(because mass acts through middle of the beam, because it is uniform and this is 1m from the pivot)
and mass at A = M, therefore counter-clockwise moment = 1.5m x Mkg
moment must be equal to maintain equilibrium, and so
6x1 = 1.5M ==> 6/1.5 = M = 4kg.
