Question

Q 1. Suppose two children are using a uniform seesaw that is 3.00 m long and pivoted at 2 m from the left edge, and a 30.0 kg child sits 1.40 m from the center of mass of the seesaw.

a. Calculate where the second 18.0 kg child must sit to balance the seesaw ?.

b. How far will the child be from the center of mass of the seesaw ?

c. How far will the child be from the left edge of the seesaw ?

Answer 1

M = 30 kg (on the left side)

m = 18 kg (on the right side)

R = distance of M from pivot point = 1.4 m

let distance of m from pivot point be d.

a) Balacing the torques produced by the weight of the childrens about pivot point.

m*d = M*1.4

=> d = (30*1.4) / 18 = 2.33 m.

18 kg body must sit at a distance 2.33 m from pivot point.

b) From center of mass of seesaw , 18 kg body is at a distance = 2.33+0.5 = 2.83 m [answer]

c) From left edge of seesaw, 18 kg body is at a distance = 2.33 + 2 = 4.33 m [answer]

Note: the values calculated is unreasonable, because the distance 2.33 m is beyond the seesaw, i.e distance of right edge from pivot is 1 m, and to balance the seesaw, one have to sit within 1m, but here it is greater than 1m.