Me! a uniform rod xy of weight 2.0 n has a length of 80 cm. the rod is suspended by a thread 20 cm from end x. a weight of 5.0 n is suspended from end x. a student hangs a 6.0 n weight on the rod so that it is at equilibrium. what is the distance of the 6.0 n weight from end x.

Given: Length of rod (L) = 80 cm

Weight of rod (W) =2.0 N

From the attached pictorial diagram,

XY = 80 cm,,XC = 40 cm, XA = 20 cm, AC = 20 cm

Let AB = z cm

Apply the principle of moment about the thread,

            5.0 N × AX = (6.0 N × AB) + (2.0 N ×AC)

or,   5.0 N × 20 cm = (6.0 N × z cm) + (2.0 N ×20 cm)

or.                AB = z = 10 cm

Now, distance XB = XA + AB = 20 cm + 10 cm = 30 cm

Hence, 6.0 N weight should be suspended from X end at distance of 30 cm

a person with a mass of 106kg has a weight of 1060n