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Question:


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 to the right  of the the center of mass of the seesaw. Seesaw weighs 5 kg.

a. Calculate where the second 18.0 kg child must sit to balance the seesaw. [ 5 points]

b. How far will the child be from the center of mass of the seesaw? [ 2 points]

c. How far will the child be from the left edge of the seesaw? [3 points] 


Answer:


(a)

Length of seesaw l = 3 m

Pivot point is at 2m from left

Centre of mass of seesaw is at 1.5 m, therefore, distance of centre of mass from pivot point is

lc = 2 - 1.5 m =0.5m

The child sits 1.40 m to the right of the the center of mass of the seesaw. So, distance of child from pivot point is

lR = 1.40 - 0.5 m = 0.9 m, towards right

mass of child at right m= 30kg

mass of seesaw,mc= 5 kg

mass of child at left, mL = 18 kg

distance of left child from pivot point = lL

Child on the left and centre of mass of seesaw provide a anticlockwise torque whereas student on the right provide a clockwise torque.

For rotational symmetry

ml + mclc = mRlR

18 * 1 + 5*0.5 = 30 * 0.9

18 *lL + 2.5 = 27

1.36т                                                           (Ans)

The child should sit at 1.36 m from the pivot point, towards left

 

(b)

The center of mass of the seesaw is at 0.5m from the pivot

Distance of child to the centre of mass is

dlc = 1.36m - 0.5m = 0.86m                    (Ans)

 

(c)

The distance of left edge to the pivot point is 2m

Distance of child from the left edge is

di = 2m – 1.36m = 0.64m                    (Ans)

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