two torque questions I'm trying to figure out..?

Question

1. The mass m1 is 0.63 kg and it is located at x1 = 30 cm. The pivot point is located at x = 45 cm. The mass of the meter stick (mms = 0.40 kg) is located at its geometric center, xms = 50 cm. The mass m2 is 0.35 kg and it is located at x2 = 80 cm. Calculate the net torque (in N⋅m with the proper sign) due to these three weights (Use g = 9.8 m/s2).



2. Suppose that a meter stick is balanced at its center. A 0.27-kg is positioned at 20 cm from the left end of the meter stick. Where should a 0.30 kg mass be placed to balance the 0.27 kg mass? Express your answer in terms of the position (in cm) of the 0.30-kg mass as measured from the left end of the meter stick.

Answer 1

1. So is everything mounted on a poor meter stick and pivoted at x=45?

M=r x F (vector product)
Sum of the moments then is (assume clockwise as positive)

Mt=-(.45-.30)m1g + (.80-.45)m2g + (.50-.45)mms g =
Mt=(-0.15m1 +0.35m2 + 0.5ms)g=
Mt=(-0.15 x 0.63 + 0.35x 0.35 +0.5 x 0.40)x 9.81=
Mt=4.09 mN clockwise

2. Same theoory applies

However noe Mt=0
x1m1g = x2m2g
or
x2= x1(m1/m2)= 0.20( 0.27)/0.30 = 0.18
However it is positioned 18 cm aposite or to the right from the pivot point.