SUppose 200 square nuts are placed 9cm from the fulcrum on one side of a balanced meter stick.?

Question

A. Where must an object with a mass of 75 square nuts be placed to balance the stick?

B. If an object of mass m is added to the 200 square nuts at 9cm, the object with a mass of 75square nuts must be moved away from the fulcrum to maintain the balance, call the distance the object must be moved D. solve for D in terms of M.

Answer 1

The torques must balance
A) 200*9=75*d
d=24 cm

B) (200+M)*9=75*(24+D)
1800+9*M=1800+75*D
D=9*M/75 cm

j

Answer 2

If the mass of the meter stick have been uniformly dispensed, the stick could stability on the 50 cm mark. because it is, besides the fact that, it balances on the forty six.7 mark. So we've mass m of the stick at that element. After the burden is further on the 11.5 mark, there is stability on the 35.6 mark. forty seven.7 g (35.6 - 11.5) cm = m (forty six.7 - 35.6) cm ==> 11.a million m = 1149.fifty seven ==> m = 103.6 g (answer)