ok here is the problem <br>A uniform meter stick of mass M has a half-filled can of fruit juice of mass m attached to one end. The meter stick and the can balance at a point 33.0 cm from the end of the stick where the can is attached. When the balanced stick-can system is suspended from a scale, the reading on the scale is 2.70 N. <br>(a) Find the mass of the meter stick. <br>(b) Find the mass of the can of juice. Well I am pretty sure that the side with the can on it... will be of greater mass than the other side... And I also know that the sum of the torques should equal zero, right? Well i can find the "mass" at the center of mass to be.... m = F/a m=2.70/9.81 = .259 kg what I did at first was divide the mass in half and tried to set the meter stick in portions and add the totals assuming the portions were equal in mass around, but somebody told me that was wrong... I am somewhat lost... please help!