Determine the Distance for Balancing Uneven Weights

[dyan]

does anyone no of a math formula to balance uneven weights?

say if i had a 100g (3.5oz) weight on a bar 100mm (3.9in) from a center point and another weight of 150g (5.5oz) how far will it need to be from the center to level the bar.

could it be as simple as double the weight double the distance?

thanks Dylan

 

[HallsofIvy]

Not exactly. In order to balance, the lighter weight (less mass) must be farther from the fulcrum. It is, instead "double the weight halve the distance". This is more a physics question that mathematics: two objects will balance when their 'torques' about the fulcrum are equal. And torque is equal to weight times distance from the fulcrum. The 100 gram mass has weight 100 g dynes (the "g" here is the acceleration due to gravity" 981 cm/s² in cgs units. I assume your "g" was "grams".) A distance 100 mm= 10 cm from the fulcrum, it will cause torque of 1000g ergs about the fulcrum. Similarly a mass of 150 grams will have weight 150g dynes and at distance x cm from the fulcrum has torque 150 gx ergs. Setting those equal, 150gx= 1000g so x= 1000g/150g= 100/15= 20/3= (2/3)10 cm.

Notice that 150 is NOT "double" 100- it is 3/2 of it so we get 2/3 the distance.