Balaning metre-stick - Find the mass

[NMW]

3) a metre-stick is found to balance at the 49.7cm mark when placed on a fulcrum, when a 50g mass is attached at the 10cm mark the fulcrum must be moved to the 39.2 cm mark for balance. what is the mass of the meter stick?

- just plain stuck on this, tried about 3 different ways of attempting the question, but couldn't find one that suited! anyone got any ideas on how to start it?? again, any help appreciated!

:)

 

[mukundpa]

The center of gravity of the stick is at 29.7cm mark, its mass to be considerd at this point, but the rest part is not appearing correct. It will not get balanced in second situation. Check the numerical values given.

 

[NMW]

The center of gravity of the stick is at 29.7cm mark, its mass to be considerd at this point, but the rest part is not appearing correct. It will not get balanced in second situation. Check the numerical values given.

yes sorry, you're right... it was meant to read 49.7cm, which i have now changed! sorry! i have still been trying to do it, found a number of various formulas and ways to go about it on the net, but none are helping me to find the mass of the stick, rather the masses of the objects joined to the stick...

 

[ramollari]

Consider the meter stick as a point mass at the 49.7cm mark. Then on the left you have the 50g point mass at a distance d1 = (39.2 - 10)cm from the pivot and on the right you have the c.m. of the stick at a distance d2 = (49.7 - 39.2)cm from the pivot.
So just use the equilibrium condition $$m_1gd_1 = m_2gd_2$$