1. ## Moments

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

2. You need the total torque to be zero.

$\displaystyle \sum \tau = 0$

$\displaystyle 0 = mg(1.5)-M_{rod}g(1) \, \Rightarrow m = 4 kg$

3. Originally Posted by Sundae
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
Take clockwise moments and anti-clockwise moments (either side of the pivot)
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.