Anonymous

Weight B is on the other side of the pivot from A. It is 0.8 times as heavy so it needs to be 1/0.8 times as far from the pivot as A to produce the same kind of torque. 20 cm / 0.8 = 25 cm. So B is at distance 25 cm from the pivot. Done. That obvious. Physically, for a balanced object the sum of torques about any axis must be 0. So you choose the horizontal axis going through the middle of the ruler at 90° to it and you make the torques equal to 0, with forces acting one way (say clockwise) representing positive torque and the other way being negative. 50 N * 20 cm - 40 N * d = 0 d = 25 cm

Anonymous

50 x 20 = 40 x ? ? = 25

Anonymous

44.4*50=55.5*40=2220

Anonymous

The data doesn't say which weight is what distance just that one is 20cm, It just says one is 50 and the other is 40 so: If the 40N is 20cm from the pivot and it's torque is 40*20=800 N-cm. The 50 N weight is 800/50= 16cm from the pivot and it's torque is 50*16= 800Ncm. The system is balanced. If the 50N weight is 20cm then it's torque is 50*20 =1000N-cm and the 40N weight will be 1000/40 = 25 cm from the pivot. If it said A and B are 50 and 40 Respectively then it would not be ambiguous. (maybe it does and you didn't state it.)

Anonymous

The torques must add to zero. The ruler is supported at the midpoint so it adds zero torque itself. Therefore we have a torque of 50 N * 0.2 m = 10 Nm from A to add to get zero then B must give a torque of - 10 Nm 40 * x = -10 x = -0.25 ie 25 cm on the other side of the pivot.

Anonymous

As the rule is not rotating, the net torque on it is zero. So we have the sum of torques as Aa - Bb = 0 where A = 50N and B = 40N, and a = 20 cm from the pivot. We solve for b = Aa/B = 50*20/40 = 25 cm from the pivot. ANS.