Question

Consider an 75 kg mother and her 15 kg daughter, balancing on a seesaw, which is a uniform board 3 m long and pivoted exactly at its center. If the daughter sits at the end of the seesaw (1.5 m from the center), how far from the center on the other side should the mother sit? Select one: O a. b) 0.4 m O b. d) 1 m O c. c)0.5 m O d. a) 0.3 m

Answer 1

To balance seesaw, net torque should be 0.

let,

w1=weight of the mother=75 x 9.8 N

w2=weight of the daughter=15 x 9.8 N

r1,r2 be the distance of mother and daughter from the center respectively.

therefore, t1=torque of the mother's weight at center=w1 x r1

t2=torque of the daughter's weight at center= w2 r2

to balance the seesaw,

t1 = t2

w1 x r1 =w2 x r2

75 x 9.8 x r1 = 15 x 9.8 x 1.5

r1=(15 x 1.5)/(75)=0.3

therefore mothere should sit at 0.3m distance from the center to balance the seesaw